ICSE Class 10 Circles Solutions

This chapter provides the complete ICSE Class 10 Selina Maths Circles Solutions, guiding you through one of the most important topics in geometry. You will explore the many properties and theorems related to circles, such as the relationship between the angle subtended by an arc at the centre and at any point on the remaining part of the circle. We will also cover crucial concepts like cyclic quadrilaterals, the properties of tangents from an external point, and how to apply these theorems to solve complex geometrical problems. Mastering the concepts in this Circles chapter from your Selina Concise Mathematics textbook is essential for building a strong foundation for your board exams.

If you are stuck on a specific proof or calculation in the Circles chapter, you’ve come to the right place. We understand that geometry requires precise steps, and our solutions are designed to provide exactly that. This page contains clear, step-by-step solutions for all 110 questions from Exercise 17(A), Exercise 17(B), Exercise 17(C), and the Test Yourself section. Each solution follows the exact method and format expected by the ICSE board. Here, you will find reliable guidance to verify your answers and understand the correct problem-solving approach.

Exercise 17(A)

Question 1(a)

In the given figure, O is center of the circle and ∠B = 55°. The angle A is equal to :

55°
35°
45°
50°

Answer:

Consider the triangle $\triangle OCB$ . Here, $OB = OC$ because they are radii of the same circle.

Notice that in any triangle, angles opposite to equal sides are equal. Thus, $\angle C = \angle B = 55^\circ$ .

Applying the angle sum property of triangles:

\angle C + \angle B + \angle O = 180^\circ

Substituting the known angles:

55^\circ + 55^\circ + \angle O = 180^\circ

This simplifies to:

\angle O + 110^\circ = 180^\circ

Solving for $\angle O$ :

\angle O = 180^\circ - 110^\circ = 70^\circ

We know that the angle subtended by an arc at the center of the circle is twice the angle it subtends at any point on the remaining part of the circle’s circumference. Therefore:

\angle O = 2\angle A

Thus, solving for $\angle A$ :

\angle A = \frac{\angle O}{2} = \frac{70^\circ}{2} = 35^\circ

Hence, Option 2 is the correct option.

Question 1(b)

In the given figure, O is center of the circle and angle OBA = 50°. The angle P is :

50°
80°
40°
60°

Answer:

Consider the triangle $\triangle OAB$ with $OA = OB$ since both are radii of the circle. This makes $\triangle OAB$ an isosceles triangle.

Notice that in an isosceles triangle, the angles opposite the equal sides are equal. Therefore, $\angle A = \angle B = 50^\circ$ .

Applying the angle sum property of a triangle, we have:

\angle A + \angle B + \angle O = 180^\circ

Substituting the known values:

50^\circ + 50^\circ + \angle O = 180^\circ

Simplifying, we find:

\angle O + 100^\circ = 180^\circ

\angle O = 180^\circ - 100^\circ = 80^\circ

Remember that the angle subtended by an arc at the center of the circle is twice the angle it subtends at any point on the remaining part of the circle’s circumference. Thus:

\angle O = 2 \angle P

So, $\angle P = \frac{\angle O}{2} = \frac{80^\circ}{2} = 40^\circ$ .

Hence, Option 3 is the correct option.

Question 1(c)

In the given figure, chord AB = chord PB and angle C = 50°. The angle PAB is equal to :

65°
50°
75°
60°

Answer:

Notice that angles subtended by the same chord in the same segment of a circle are equal.

∴ ∠APB = ∠ACB = 50°

Considering △ PAB, we have:

PB = AB (as given)

This implies that the angles opposite these equal sides must also be equal.

∴ ∠PAB = ∠APB = 50°.

Hence, Option 2 is the correct option.

Question 1(d)

O’ and O” are centers of two circles which intersect each other at points A and B. Then :

BC = BD
BC is larger than BD.
BC is smaller than BD.
C, B and D are collinear.

Answer:

Recall that the angle subtended by a diameter on the circumference of a circle is a right angle.

In the given figure:

Both AC and AD serve as diameters.

∴ ∠ABC = ∠ABD = 90°.

According to the figure:

⇒ ∠ABC + ∠ABD = 90° + 90° = 180°.

This implies that CBD forms a straight line, meaning points C, B, and D are collinear.

Hence, Option 4 is the correct option.

Question 1(e)

In the given figure, O is center of the circle, AB || DC and ∠ACD = 32°, ∠DAB is equal to :

122°
148°
90°
none of the above

Answer:

Connect point D to point A.

Observe the diagram:

In △DAC,

∠DAC = 90° because the angle formed in a semicircle is always a right angle.

Since AB is parallel to DC, the alternate angles are equal.

∴ ∠CAB = ∠ACD = 32°

From the diagram,

∠DAB = ∠DAC + ∠CAB = 90° + 32° = 122°.

Hence, Option 1 is the correct option.

Question 2

In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°

(i) Prove that AC is a diameter of the circle.

(ii) Find ∠ACB

Answer:

(i) Consider the triangle $\triangle ABD$ . By the angle sum property of a triangle, we have:

\angle DAB + \angle ABD + \angle ADB = 180^\circ

Substituting the given angles, we get:

65^\circ + 70^\circ + \angle ADB = 180^\circ

This simplifies to:

135^\circ + \angle ADB = 180^\circ

Subtracting $135^\circ$ from both sides, we find:

\angle ADB = 180^\circ - 135^\circ = 45^\circ

Referring to the figure, notice that:

\angle ADC = \angle ADB + \angle BDC = 45^\circ + 45^\circ = 90^\circ

Since the angle in a semicircle is a right angle, it follows that arc $ADC$ forms a semicircle, and thus $AC$ is the diameter of the circle.

Hence, proved that $AC$ is the diameter.

(ii) According to the property that angles in the same segment of a circle are equal, we have:

\angle ACB = \angle ADB = 45^\circ

Hence, $\angle ACB = 45^\circ$ .

Question 3(i)

In the following figure, O is the center of the circle. Find the value of a.

Answer:

We understand that an angle inscribed in a semicircle is a right angle.

∴ ∠BAD = 90°

Observing the figure, we have:

⇒ ∠BAD + ∠ADB + ∠DBA = 180°

Substituting the known values:

⇒ 90° + ∠ADB + 35° = 180°

This simplifies to:

⇒ ∠ADB + 125° = 180°

Rearranging gives:

⇒ ∠ADB = 180° – 125° = 55°.

Furthermore, angles subtended by the same chord at the circle are equal.

∴ a = ∠ADB = 55°.

Hence, a = 55°.

Question 3(ii)

In the following figure, O is the center of the circle. Find the value of b.

Answer:

Identify the intersection point of lines AC and BD as E:

Since AC forms a straight line,

∴ ∠AEB + ∠BEC = 180°

⇒ 120° + ∠BEC = 180°

⇒ ∠BEC = 180° – 120° = 60°.

Now, consider △BEC:

⇒ ∠BEC + ∠ECB + ∠CBE = 180°

⇒ 60° + ∠ECB + 25° = 180°

⇒ ∠ECB + 85° = 180°

⇒ ∠ECB = 180° – 85° = 95°.

Recall that angles subtended by the same chord on the circle are equal.

∴ b = ∠ACB

From the figure,

∠ACB = ∠ECB = 95°.

Hence, b = 95°.

Question 3(iii)

In the following figure, O is the center of the circle. Find the value of c.

Answer:

Given that the angle subtended by an arc at the center of a circle is twice the angle it subtends at any point on the remaining part of the circle’s circumference, we have:

∠AOB = 2∠ACB = 2 \times 50° = 100°.

Since OA and OB are radii of the circle, they are equal. Therefore, the angles opposite these equal sides are also equal, so:

∠OBA = ∠OAB = c.

In △OAB, the sum of angles is 180°:

⇒ ∠OAB + ∠AOB + ∠OBA = 180°.

Substituting the known values:

⇒ 2∠OAB + 100° = 180° $\text{since} ∠OBA = ∠OAB \text{and} ∠AOB = 100°$

⇒ 2c + 100° = 180°.

Simplifying gives:

⇒ 2c = 180° – 100°.

⇒ 2c = 80°.

Dividing both sides by 2:

⇒ c = \dfrac{80}{2} = 40°.

Thus, the value of c is 40°.

Question 3(iv)

In the following figure, O is the center of the circle. Find the value of d.

Answer:

Recall that an angle inscribed in a semicircle is a right angle.

∴ ∠APB = 90°

In △APB:

⇒ ∠APB + ∠PBA + ∠PAB = 180°

⇒ 90° + 45° + ∠PAB = 180°

⇒ ∠PAB = 180° – 135° = 45°.

Notice that angles subtended by the same chord at the circumference are equal.

∴ d = ∠PAB = 45°.

Hence, d = 45°.

Question 4

Calculate:

(i) ∠CDB,

(ii) ∠ABC,

(iii) ∠ACB.

Answer:

(i) Notice that the angles subtended by the same chord on the circle are equal. Therefore, ∠CDB is equal to ∠BAC, which is 49°.

Hence, ∠CDB = 49°.

(ii) Similarly, the angles subtended by the same chord on the circle are equal. Thus, ∠ABC is equal to ∠ADC, which is 43°.

Hence, ∠ABC = 43°.

(iii) Consider △ABC:

⇒ ∠ABC + ∠ACB + ∠BAC = 180°

⇒ 43° + ∠ACB + 49° = 180°

⇒ ∠ACB + 92° = 180°

⇒ ∠ACB = 180° – 92° = 88°.

Hence, ∠ACB = 88°.

Question 5

Given: ∠CAB = 75° and ∠CBA = 50°. Find the value of ∠DAB + ∠ABD.

Answer:

Consider triangle ∆ABC. According to the angle sum property, the sum of the angles in a triangle is 180°.

∴ ∠ACB + ∠CBA + ∠CAB = 180°

Substituting the given values:

⇒ ∠ACB + 50° + 75° = 180°

⇒ ∠ACB + 125° = 180°

Solving for ∠ACB, we get:

⇒ ∠ACB = 180° – 125° = 55°.

Recall that angles subtended by the same chord in a circle are equal.

∴ ∠ADB = ∠ACB = 55°.

Now, in triangle ∆ABD, apply the angle sum property again:

⇒ ∠DAB + ∠ABD + ∠ADB = 180°

⇒ ∠DAB + ∠ABD + 55° = 180°

Rearranging gives:

⇒ ∠DAB + ∠ABD = 180° – 55°

⇒ ∠DAB + ∠ABD = 125°

Hence, ∠DAB + ∠ABD = 125°.

Question 6

In the figure given alongside, AOB is a diameter of the circle and ∠AOC = 110°, find ∠BDC.

Answer:

To solve this, let’s connect point A to point D.

We are aware that the angle formed at the center of a circle is twice the angle formed at the circumference by the same chord. Thus, we have:

∠ADC = $\dfrac{1}{2}$ ∠AOC = $\dfrac{1}{2} \times 110° = 55°$ .

Additionally, it’s a known fact that the angle inscribed in a semicircle is a right angle.

∴ ∠ADB = 90°

From the diagram, we can deduce:

∠BDC = ∠BDA – ∠ADC = 90° – 55° = 35°.

Therefore, ∠BDC = 35°.

Question 7

In the following figure, O is the centre of the circle, ∠AOB = 60° and ∠BDC = 100°. Find ∠OBC.

Answer:

Recall that the angle formed at the centre of a circle is twice the angle formed at the circumference by the same arc.

Thus, $\angle ACB = \dfrac{1}{2} \angle AOB = \dfrac{1}{2} \times 60^\circ = 30^\circ$ .

Now, consider the triangle $\triangle BDC$ :

\angle BDC + \angle DCB + \angle CBD = 180^\circ

Using the given information, substitute the values:

100^\circ + 30^\circ + \angle CBD = 180^\circ

This simplifies to:

\angle CBD + 130^\circ = 180^\circ

⇒ $\angle CBD = 180^\circ - 130^\circ = 50^\circ$ .

According to the figure, we have:

⇒ $\angle OBC = \angle CBD = 50^\circ$ .

Therefore, $\angle OBC = 50^\circ$ .

Question 8

In cyclic quadrilateral ABCD, ∠DAC = 27°; ∠DBA = 50° and ∠ADB = 33°. Calculate :

(i) ∠DBC,

(ii) ∠DCB,

(iii) ∠CAB.

Answer:

(i) Consider the property that angles subtended by the same arc in a circle are equal. Therefore, we have:

∠DBC = ∠DAC = 27°.

Hence, ∠DBC = 27°.

(ii) Again, using the property of angles in the same segment, we find:

∠ACB = ∠ADB = 33°.

Additionally,

∠ACD = ∠ABD = 50°.

From the diagram, we can see that:

∠DCB = ∠ACD + ∠ACB = 50° + 33° = 83°.

Hence, ∠DCB = 83°.

(iii) In the cyclic quadrilateral ABCD, the sum of the measures of opposite angles is 180°.

∴ ∠DAB + ∠DCB = 180°

Substituting the known values:

∠DAC + ∠CAB + ∠DCB = 180°

27° + 83° + ∠CAB = 180°

⇒ ∠CAB + 110° = 180°

⇒ ∠CAB = 180° – 110° = 70°.

Hence, ∠CAB = 70°.

Question 9

In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80° and ∠CDE = 40°, find the number of degrees in :

(i) ∠DCE,

(ii) ∠ABC.

Answer:

(i) Consider the angle ∠CED. Since it is an angle in a semicircle, it measures 90°.

In triangle CED, apply the angle sum property:

∴ ∠CED + ∠CDE + ∠DCE = 180°

Substitute the known values:

⇒ 90° + 40° + ∠DCE = 180°

This simplifies to:

⇒ ∠DCE + 130° = 180°

⇒ ∠DCE = 180° – 130° = 50°.

Thus, ∠DCE = 50°.

(ii) For triangle BOC, remember that the exterior angle is equal to the sum of the two opposite interior angles. Here, ∠AOC serves as the exterior angle.

⇒ ∠AOC = ∠OCB + ∠OBC

Rearrange to find ∠OBC:

⇒ ∠OBC = ∠AOC – ∠OCB

Since ∠OCB = ∠DCE, substitute:

⇒ ∠OBC = 80° – 50° = 30°.

In the figure, ∠ABC corresponds to ∠OBC, thus:

Therefore, ∠ABC = 30°.

Question 10

In the figure given alongside, AB || CD and O is the center of the circle. If ∠ADC = 25°; find the angle AEB. Give reasons in support of your answer.

Answer:

To solve this, we start by connecting points AC, BD, and CB.

Since the angle subtended by a semicircle is a right angle, we have:

∴ ∠CAD = 90° and ∠CBD = 90°

Given that AB is parallel to CD, we know:

∠BAD = ∠ADC = 25° because alternate angles are equal.

Looking at the figure, we calculate:

∠BAC = ∠BAD + ∠CAD = 25° + 90° = 115°.

In a cyclic quadrilateral, the sum of opposite angles is 180°.

Thus, in quadrilateral ACDB, we have:

⇒ ∠CDB + ∠BAC = 180°

⇒ ∠CDA + ∠ADB + ∠BAC = 180°

⇒ 25° + ∠ADB + 115° = 180°

Solving for ∠ADB gives:

⇒ ∠ADB = 180° – 115° – 25° = 40°.

Angles in the same segment of a circle are equal, therefore:

∴ ∠AEB = ∠ADB = 40°.

Hence, ∠AEB = 40°.

Question 11

AB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines. Find :

(i) ∠PRB,

(ii) ∠PBR,

(iii) ∠BPR.

Answer:

(i) Notice that angles subtended by the same arc in a circle are equal. Therefore, $\angle PRB = \angle PAB = 35^\circ$ .

Hence, $\angle PRB = 35^\circ$ .

(ii) Observe from the diagram that $\angle BPA = 90^\circ$ since an angle in a semicircle is a right angle.

⇒ $\angle BPA + \angle BPQ = 180^\circ$ because they form a linear pair.

⇒ $90^\circ + \angle BPQ = 180^\circ$

⇒ $\angle BPQ = 180^\circ - 90^\circ = 90^\circ$ .

In a triangle, the exterior angle equals the sum of the two opposite interior angles.

⇒ $\angle PBR = \angle BPQ + \angle BQP = 90^\circ + 25^\circ = 115^\circ$ .

Hence, $\angle PBR = 115^\circ$ .

(iii) In $\triangle ABP$ , the angles add up to 180°.

⇒ $\angle BPA + \angle PAB + \angle ABP = 180^\circ$

⇒ $90^\circ + 35^\circ + \angle ABP = 180^\circ$

⇒ $125^\circ + \angle ABP = 180^\circ$

⇒ $\angle ABP = 180^\circ - 125^\circ = 55^\circ$ .

From the diagram, calculate $\angle ABR$ as follows:

⇒ $\angle ABR = \angle PBR - \angle ABP = 115^\circ - 55^\circ = 60^\circ$ .

Since angles in the same segment are equal, $\angle APR = \angle ABR = 60^\circ$ .

In $\triangle BPR$ , the sum of angles is 180°.

⇒ $\angle BPR + \angle PRB + \angle PBR = 180^\circ$

⇒ $\angle BPR + 35^\circ + 115^\circ = 180^\circ$

⇒ $150^\circ + \angle BPR = 180^\circ$

⇒ $\angle BPR = 180^\circ - 150^\circ = 30^\circ$ .

Hence, $\angle BPR = 30^\circ$ .

Question 12

In the given figure, A is the center of the circle, ABCD is a parallelogram and CDE is a straight line. Prove that : ∠BCD = 2∠ABE.

Answer:

We know that $\angle BAD = 2\angle BED$ because the angle at the center is twice the angle at the circumference for the same chord.

Since CDE forms a straight line and CD is parallel to AB, it follows that AB is also parallel to ED.

This implies $\angle BED = \angle ABE$ due to the property of alternate angles being equal when two lines are parallel.

By multiplying the above equation by 2, we obtain:

2\angle BED = 2\angle ABE

Thus, $\angle BAD = 2\angle ABE$ ……………(1)

Given that ABCD is a parallelogram, we know:

$\angle BAD = \angle BCD$ because opposite angles in a parallelogram are equal ……….(2)

Combining equations (1) and (2), we conclude:

$\angle BCD = 2\angle ABE$ .

This proves that $\angle BCD = 2\angle ABE$ .

Question 13

In the given figure, I is the incenter of △ABC, BI when produced meets the circumcircle of △ABC at D. Given, ∠BAC = 55° and ∠ACB = 65°; calculate :

(i) ∠DCA,

(ii) ∠DAC,

(iii) ∠DCI,

(iv) ∠AIC.

Answer:

(i) Join IA, IC and CD.

In triangle △ABC, we have:

⇒ ∠ABC + ∠BAC + ∠ACB = 180°

⇒ ∠ABC + 55° + 65° = 180°

⇒ ∠ABC + 120° = 180°

⇒ ∠ABC = 180° – 120° = 60°.

Since I is the incenter, IB bisects ∠ABC.

∠ABD = $\dfrac{1}{2}$ ∠ABC = $\dfrac{1}{2}$ × 60° = 30°.

Angles in the same segment are equal.

∴ ∠DCA = ∠ABD = 30°.

Hence, ∠DCA = 30°.

(ii) Since IB is the bisector of ∠ABC, ∠CBD = ∠ABD = 30°.

Angles in the same segment are equal.

∴ ∠DAC = ∠CBD = 30°.

Hence, ∠DAC = 30°.

(iii) CI bisects ∠ACB, so:

∠ACI = $\dfrac{1}{2}$ ∠ACB = $\dfrac{1}{2}$ × 65° = 32.5°.

From the figure:

∠DCI = ∠DCA + ∠ACI = 30° + 32.5° = 62.5°.

Hence, ∠DCI = 62.5°.

(iv) Since AI is the angle bisector of ∠BAC:

∠IAC = $\dfrac{1}{2}$ ∠BAC = $\dfrac{1}{2}$ × 55° = 27.5°.

In triangle △AIC:

⇒ ∠IAC + ∠ACI + ∠AIC = 180°

⇒ 27.5° + 32.5° + ∠AIC = 180°

⇒ 60° + ∠AIC = 180°

⇒ ∠AIC = 180° – 60° = 120°.

Hence, ∠AIC = 120°.

Question 14

In the given figure, AB = AC = CD and ∠ADC = 38°. Calculate :

(i) Angle ABC

(ii) Angle BEC

Answer:

To solve this, draw the line BE.

(i) Since AC = CD, we have:

∠DAC = ∠ADC = 38° because angles opposite equal sides are equal.

Consider △ACD:

⇒ ∠DAC + ∠ADC + ∠ACD = 180°

⇒ 38° + 38° + ∠ACD = 180°

⇒ 76° + ∠ACD = 180°

⇒ ∠ACD = 180° – 76° = 104°

Now, notice from the figure:

⇒ ∠ACB + ∠ACD = 180° since BCD is a straight line.

⇒ ∠ACB + 104° = 180°

⇒ ∠ACB = 180° – 104° = 76°.

Given that AB = AC, it follows that:

∴ ∠ABC = ∠ACB = 76° because angles opposite equal sides are equal.

Hence, ∠ABC = 76°.

(ii) Now, in △ABC:

⇒ ∠BAC + ∠ACB + ∠ABC = 180° by the angle sum property of triangles.

⇒ ∠BAC + 76° + 76° = 180°

⇒ ∠BAC + 152° = 180°

⇒ ∠BAC = 180° – 152° = 28°.

Since angles in the same segment are equal:

⇒ ∠BEC = ∠BAC = 28°.

Hence, ∠BEC = 28°.

Question 15

In the given figure, AC is the diameter of circle, centre O. Chord BD is perpendicular to AC. Write down the angles p, q and r in terms of x.

Answer:

Recall that the angle formed at the center, ∠AOB, is twice the angle made at the circumference by the same chord, which gives us:

⇒ ∠AOB = 2∠ACB

This implies:

⇒ x = 2q

Thus, q = $\dfrac{x}{2}$ .

Notice that angles in the same segment are equal, so:

∴ ∠ADB = ∠ACB = q = $\dfrac{x}{2}$

From the diagram, we see that:

∠ADC = 90° because the angle in a semicircle is a right angle.

Therefore, ∠BDC can be calculated as:

∠BDC = ∠ADC – ∠ADB = 90° – $\dfrac{x}{2}$ .

∴ r = 90° – $\dfrac{x}{2}$ .

Considering △EBC, the angle sum property states:

⇒ ∠EBC + ∠CEB + ∠ECB = 180°

Substituting the known values gives:

⇒ ∠EBC + 90° + q = 180°

⇒ ∠EBC = 90° – q

Referring back to the diagram:

∠DBC = ∠EBC = 90° – q = 90° – $\dfrac{x}{2}$ .

Again, using the fact that angles in the same segment are equal, we have:

∴ ∠DAC = ∠DBC

Thus, p = 90° – $\dfrac{x}{2}$ .

Hence, p = 90° – $\dfrac{x}{2}$ , q = $\dfrac{x}{2}$ and r = 90° – $\dfrac{x}{2}$ .

Question 16

In the given figure, AOB is a diameter and DC is parallel to AB. If ∠CAB = x°; find (in terms of x) the values of :

(i) ∠COB,

(ii) ∠DOC,

(iii) ∠DAC,

(iv) ∠ADC.

Answer:

(i) Notice that the angle at the centre of a circle is twice the angle at the circumference when subtended by the same chord. Therefore, we have:

⇒ ∠COB = 2∠CAB = 2x

Hence, ∠COB = 2x.

(ii) Given that DC is parallel to OB, it follows that:

⇒ ∠OCD = ∠COB = 2x [Alternate angles]

In triangle △OCD, since OC and OD are radii of the same circle, they are equal.

⇒ ∠ODC = ∠OCD = 2x [Angles opposite to equal sides are equal]

Using the angle sum property in △OCD, we have:

⇒ ∠ODC + ∠OCD + ∠DOC = 180°

⇒ 2x + 2x + ∠DOC = 180°

⇒ ∠DOC = 180° – 4x.

Hence, ∠DOC = 180° – 4x.

(iii) Again, using the property that the angle at the centre is twice the angle at the circumference subtended by the same chord:

⇒ ∠DOC = 2∠DAC

⇒ ∠DAC = $\dfrac{1}{2}$ ∠DOC = ( \dfrac{1}{2} \times (180° – 4x) ) = 90° – 2x.

Hence, ∠DAC = 90° – 2x.

(iv) Since DC is parallel to AO, we have:

∴ ∠ACD = ∠OAC = x (Alternate angles are equal)

In △ADC, applying the angle sum property:

⇒ ∠ADC + ∠DAC + ∠ACD = 180°

⇒ ∠ADC + 90° – 2x + x = 180°

⇒ ∠ADC + 90° – x = 180°

⇒ ∠ADC = 180° – 90° + x

⇒ ∠ADC = 90° + x

Hence, ∠ADC = 90° + x.

Question 17

In the given figure, PQ is the diameter of the circle whose center is O. Given, ∠ROS = 42°, calculate ∠RTS.

Answer:

To tackle this problem, let’s draw a line from P to S.

Notice that ∠PSQ forms a right angle, 90°, because it is an angle in a semi-circle.

Recall the property: the angle at the center of a circle is twice the angle at the circumference subtended by the same chord. Thus, we have:

∴ ∠ROS = 2∠SPR

⇒ ∠SPR = $\dfrac{1}{2}$ ∠ROS = $\dfrac{42}{2}$ = 21°

Looking at the figure, it follows that:

⇒ ∠SPT = ∠SPR = 21°.

Also from the figure, we know:

⇒ ∠PSQ = 90° because it is an angle in a semi-circle.

Since QST is a straight line, we can say:

⇒ ∠PSQ + ∠PST = 180°

⇒ 90° + ∠PST = 180°

⇒ ∠PST = 90°.

Now, in △PST, applying the angle sum property:

⇒ ∠PTS + ∠PST + ∠SPT = 180°

⇒ ∠PTS + 90° + 21° = 180°

⇒ ∠PTS + 111° = 180°

⇒ ∠PTS = 180° – 111° = 69°.

From the diagram, it is clear that:

∠RTS = ∠PTS = 69°.

Hence, ∠RTS = 69°.

Question 18

The given figure shows a circle with center O and ∠ABP = 42°. Calculate the measure of :

(i) ∠PQB

(ii) ∠QPB + ∠PBQ

Answer:

Let’s connect point A to point P.

(i) Recall that the angle formed in a semicircle is always a right angle. Thus, we have:

∠APB = 90°.

Considering triangle △APB, we apply the angle sum property:

∴ ∠APB + ∠ABP + ∠BAP = 180°

⇒ 90° + 42° + ∠BAP = 180°

⇒ ∠BAP + 132° = 180°

⇒ ∠BAP = 180° – 132° = 48°.

Observing the figure, we see that:

∠PQB = ∠BAP = 48° because angles in the same segment are equal.

Therefore, ∠PQB = 48°.

(ii) Now, consider triangle △BQP. Using the angle sum property again, we have:

∴ ∠QPB + ∠PBQ + ∠PQB = 180°

⇒ ∠QPB + ∠PBQ + 48° = 180°

⇒ ∠QPB + ∠PBQ = 180° – 48°

⇒ ∠QPB + ∠PBQ = 132°.

Thus, ∠QPB + ∠PBQ = 132°.

Question 19

In the given figure, M is the centre of the circle. Chords AB and CD are perpendicular to each other. If ∠MAD = x and ∠BAC = y :

(i) express ∠AMD in terms of x.

(ii) express ∠ABD in terms of y.

(iii) prove that : x = y.

Answer:

Identify the point where chords AB and CD intersect as L.

(i) Consider △AMD:

Since MA = MD (both are radii of the circle),

∴ ∠MAD = ∠MDA = x.

Applying the angle sum property in △AMD:

⇒ ∠MAD + ∠MDA + ∠AMD = 180°

⇒ x + x + ∠AMD = 180°

⇒ ∠AMD = 180° – 2x.

Thus, ∠AMD = 180° – 2x.

(ii) Assume that AB and CD intersect at L.

From the diagram, we know:

∠ALC = 90°.

In △ALC:

⇒ ∠LAC + ∠LCA + ∠ALC = 180°

⇒ y + ∠DCA + 90° = 180°

⇒ ∠DCA = 180° – 90° – y

⇒ ∠DCA = 90° – y

According to the diagram:

∠ABD = ∠DCA (since angles in the same segment are equal)

∴ ∠ABD = 90° – y.

Thus, ∠ABD = 90° – y.

(iii) Recall the property:

The angle subtended by an arc at the center is twice that subtended at any point on the circle’s circumference.

∴ ∠AMD = 2∠ABD

Thus, ∠ABD = $\dfrac{1}{2}$ ∠AMD = (\dfrac{1}{2} \times (180° – 2x)) = 90° – x.

We have:

∠ABD = 90° – y and ∠ABD = 90° – x

⇒ 90° – y = 90° – x

⇒ x = y.

Hence, proved that x = y.

Exercise 17(B)

Question 1(a)

ABCD is a trapezium with AD parallel to BC. Side BC is produced to point E and angle DCE = 95°. Angle B is equal to :

85°
105°
95°
175°

Answer:

In the trapezium ABCD, we have AD parallel to BC. Since BC is extended to point E, it follows that AD is also parallel to BE.

From the diagram, notice that ∠ADC and ∠DCE are alternate angles. Given that ∠DCE = 95°, it follows that ∠ADC = 95°.

In a cyclic quadrilateral, the sum of opposite angles is 180°. Therefore,

∴ ∠ADC + ∠CBA = 180°

Substituting the known value, we have:

⇒ ∠CBA = 180° – ∠ADC = 180° – 95° = 85°.

Hence, Option 1 is the correct option.

Question 1(b)

In the given figure, ABC is an equilateral triangle. Angle ADC is :

60°
100°
80°
120°

Answer:

Given that triangle ABC is equilateral, each interior angle measures 60°.

Notice that in a cyclic quadrilateral, the sum of the measures of opposite angles is 180°.

∴ ∠ABC + ∠ADC = 180°

⇒ 60° + ∠ADC = 180°

⇒ ∠ADC = 180° – 60° = 120°.

Hence, Option 4 is the correct option.

Question 1(c)

In the given figure, O is the center of the circle. ∠OAB and ∠OCB are 30° and 40° respectively. ∠AOC is equal to :

70°
80°
150°
140°

Answer:

Consider connecting points A and C with a line segment.

In the triangle △AOC, notice that since OA and OC are radii of the circle, they are equal. Thus, ∠OAC and ∠OCA are equal, so let them be x.

Using the angle sum property of triangles, we have:

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ x + x + ∠AOC = 180°

⇒ ∠AOC = 180° – 2x

Next, in triangle △AOC, apply the angle sum property:

⇒ ∠BAC + ∠ACB + ∠CBA = 180°

⇒ (30° + x) + (40° + x) + ∠CBA = 180°

⇒ ∠CBA + 70° + 2x = 180°

⇒ ∠CBA = 180° – 70° – 2x

⇒ ∠CBA = 110° – 2x

Recall that the angle subtended by an arc at the center of a circle is twice the angle it subtends at any point on the circumference.

∴ ∠AOC = 2∠CBA

⇒ 180° – 2x = 2(110° – 2x)

⇒ 180° – 2x = 220° – 4x

⇒ 4x – 2x = 220° – 180°

⇒ 2x = 40°

⇒ x = $\dfrac{40°}{2}$ = 20°.

Finally, substitute back to find ∠AOC:

⇒ ∠AOC = 180° – 2x

⇒ ∠AOC = 180° – 2(20°) = 180° – 40° = 140°.

Hence, Option 4 is the correct option.

Question 1(d)

In the given figure APB and CQD are two straight lines, then :

AB || CD
AC || PQ
PQ || BD
AC || BD

Answer:

Consider the angles ∠BPQ = x and ∠DQP = y.

Recall that in a cyclic quadrilateral, an exterior angle is equal to the opposite interior angle.

In the cyclic quadrilateral APQC, observe that:

∠A = ∠DQP = y and ∠C = ∠BPQ = x.

Since APB is a straight line, we have:

∴ ∠APQ + ∠BPQ = 180°

⇒ ∠APQ + x = 180°

⇒ ∠APQ = 180° – x

Similarly, CQD being a straight line gives us:

∴ ∠CQP + ∠DQP = 180°

⇒ ∠CQP + y = 180°

⇒ ∠CQP = 180° – y

Now, in the cyclic quadrilateral PQDB, we note:

∠B = ∠CQP = 180° – y and ∠D = ∠APQ = 180° – x

This leads to:

⇒ ∠A + ∠B = y + (180° – y) = 180°

⇒ ∠C + ∠D = x + (180° – x) = 180°

Remember that the sum of adjacent angles in a trapezium is 180°.

∴ ABDC forms a trapezium.

∴ AC || BD.

Hence, Option 4 is the correct option.

Question 1(e)

In the figure, given below, ∠ABC is equal to :

105°
75°
90°
45°

Answer:

Consider that the sum of co-interior angles in a trapezium is always 180°.

∴ ∠A + ∠D = 180°

Given that ∠A is 105°, we have:

⇒ 105° + ∠D = 180°

Solving for ∠D, we find:

⇒ ∠D = 180° – 105° = 75°

Next, recall that in a cyclic quadrilateral, opposite angles sum up to 180°.

For cyclic quadrilateral ABCD:

⇒ ∠D + ∠B = 180°

Substituting the value of ∠D:

⇒ 75° + ∠B = 180°

Solving for ∠B gives:

⇒ ∠B = 180° – 75° = 105°.

Hence, Option 1 is the correct option.

Question 2

In the given figure, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°; find :

(i) ∠ACB,

(ii) ∠OBC,

(iii) ∠OAB,

(iv) ∠CBA.

Answer:

Given that ∠AOB = 140° and ∠OAC = 50°.

(i) Notice that the angle at the center of the circle is twice the angle at the circumference subtended by the same chord.

∴ ∠ACB = $\dfrac{1}{2}$ Reflex ∠AOB

= $\dfrac{1}{2}$ (360° – 140°)

= $\dfrac{1}{2} \times 220°$ = 110°.

Hence, ∠ACB = 110°.

(ii) Recall that the sum of the angles in a quadrilateral is 360°.

For quadrilateral OBCA:

∠OBC + ∠ACB + ∠OAC + ∠AOB = 360°

⇒ ∠OBC + 110° + 50° + 140° = 360°

⇒ ∠OBC + 300° = 360°

⇒ ∠OBC = 360° – 300° = 60°.

Hence, ∠OBC = 60°.

(iii) Let’s join AB.

In triangle ∆AOB, since OA = OB (both are radii of the circle), we have:

∠OBA = ∠OAB (angles opposite to equal sides are equal)

Using the angle sum property of a triangle:

⇒ ∠OBA + ∠OAB + ∠AOB = 180°

⇒ 2∠OAB + 140° = 180°

⇒ 2∠OAB = 40°

⇒ ∠OAB = $\dfrac{40}{2}$ = 20°

Hence, ∠OAB = 20°.

(iv) We have already determined that ∠OBC = 60°.

Thus, ∠OBC = ∠CBA + ∠OBA

⇒ 60° = ∠CBA + 20°

⇒ ∠CBA = 60° – 20° = 40°

Hence, ∠CBA = 40°.

Question 3

In the figure, given below, ABCD is a cyclic quadrilateral in which ∠BAD = 75°; ∠ABD = 58° and ∠ADC = 77°. Find :

(i) ∠BDC,

(ii) ∠BCD,

(iii) ∠BCA.

Answer:

(i) Consider the triangle $\triangle ABD$ . We know that the sum of angles in any triangle is $180^\circ$ . Therefore, we have:

\angle ADB + \angle ABD + \angle DAB = 180^\circ

Substituting the given angles:

\angle ADB + 58^\circ + 75^\circ = 180^\circ

This simplifies to:

\angle ADB + 133^\circ = 180^\circ

Thus, solving for $\angle ADB$ gives:

\angle ADB = 180^\circ - 133^\circ = 47^\circ

Now, observe that:

\angle BDC = \angle ADC - \angle ADB = 77^\circ - 47^\circ = 30^\circ

Hence, $\angle BDC = 30^\circ$ .

(ii) In a cyclic quadrilateral, the sum of opposite angles is $180^\circ$ . Therefore:

\angle BAD + \angle BCD = 180^\circ

Substituting the given angle:

\angle BCD = 180^\circ - 75^\circ = 105^\circ

Hence, $\angle BCD = 105^\circ$ .

(iii) By joining AC, we can apply the property that angles subtended by the same chord are equal. Thus:

\angle BCA = \angle ADB = 47^\circ

Hence, $\angle BCA = 47^\circ$ .

Question 4

In the following figure, O is the centre of the circle and ΔABC is equilateral.

Find:

(i) ∠ADB, (ii) ∠AEB.

Answer:

(i) In an equilateral triangle, each angle measures 60°.

∴ ∠ACB = 60°

Since angles in the same segment of a circle are equal, we have:

∴ ∠ADB = ∠ACB = 60°.

Hence, ∠ADB = 60°.

(ii) Let’s connect OA and OB.

We know that the angle at the center of a circle is twice the angle at the circumference subtended by the same chord.

∴ ∠AOB = 2∠ACB = 2 \times 60° = 120°.

Thus, ∠AEB is half of the reflex angle ∠AOB:

∠AEB = \dfrac{1}{2} \text{ Reflex } ∠AOB

= \dfrac{1}{2}(360° – 120°) = \dfrac{1}{2} \times 240°

= 120°.

Hence, ∠AEB = 120°.

Question 5

ABCD is a cyclic quadrilateral in a circle with centre O. If ∠ADC = 130°, find ∠BAC.

Answer:

Recall that the angle subtended by a diameter in a circle is a right angle. Thus, $\angle ACB = 90^\circ$ .

Also, in a cyclic quadrilateral, the sum of each pair of opposite angles is $180^\circ$ .

∴ $\angle ABC = 180^\circ - \angle ADC = 180^\circ - 130^\circ = 50^\circ$ .

Now, consider $\triangle ACB$ :

\angle ACB + \angle CBA + \angle BAC = 180^\circ \quad \text{(Angle sum property of a triangle)}

Substituting the known values:

90^\circ + 50^\circ + \angle BAC = 180^\circ

This simplifies to:

\angle BAC + 140^\circ = 180^\circ

Solving for $\angle BAC$ , we find:

\angle BAC = 180^\circ - 140^\circ = 40^\circ

Hence, $\angle BAC = 40^\circ$ .

Question 6(a)

In the following figure,

(i) if ∠BAD = 96°, find ∠BCD and ∠BFE.

(ii) Prove that AD is parallel to FE.

Answer:

(i) Since ABCD forms a cyclic quadrilateral, the sum of opposite angles is 180°. Thus, $\angle BAD + \angle BCD = 180^\circ$ .

⇒ $\angle BCD = 180^\circ - \angle BAD$

= $180^\circ - 96^\circ$

= $84^\circ$ .

Considering DCE as a straight line, we have:

∴ $\angle BCE = 180^\circ - \angle BCD$

= $180^\circ - 84^\circ = 96^\circ$ .

As BCEF is also a cyclic quadrilateral, $\angle BCE + \angle BFE = 180^\circ$ .

⇒ $\angle BFE = 180^\circ - \angle BCE$

= $180^\circ - 96^\circ$

= $84^\circ$ .

Thus, $\angle BCD = 84^\circ$ and $\angle BFE = 84^\circ$ .

(ii) Notice that $\angle BAD + \angle BFE = 96^\circ + 84^\circ = 180^\circ$ .

Since these are interior angles on the same side of the lines AD and FE, it follows that:

AD is parallel to FE.

Question 6(b)

ABCD is a parallelogram. A circle through vertices A and B meets side BC at point P and side AD at point Q. Show that quadrilateral PCDQ is cyclic.

Answer:

In the given diagram, parallelogram ABCD is shown with a circle passing through vertices A and B, intersecting side BC at P and side AD at Q. We are given that ∠BAD = 96°.

Recall that for a cyclic quadrilateral, an exterior angle is equal to the opposite interior angle.

∴ ∠1 = ∠A ……. (i)

In addition, we know that:

∠A = ∠C ………..(ii) [Since opposite angles of a parallelogram are equal.]

Combining equations (i) and (ii), we derive:

∠1 = ∠C ………(iii)

Furthermore, we have:

∠C + ∠D = 180° [The sum of co-interior angles in a parallelogram equals 180°]

This implies:

⇒ ∠1 + ∠D = 180°

Hence, proved that PCDQ is cyclic.

Question 7

Prove that:

(i) the parallelogram, inscribed in a circle, is a rectangle.

(ii) the rhombus, inscribed in a circle, is a square.

Answer:

(i) Consider a parallelogram ABCD that is inscribed in a circle.

Notice that in a parallelogram, opposite angles are equal, so ∠A = ∠C and ∠B = ∠D. Since ABCD is a cyclic quadrilateral, the sum of opposite angles is 180°, giving us ∠A + ∠C = 180° and ∠B + ∠D = 180°.

This leads to 2∠A = 180° and 2∠B = 180°, which simplifies to ∠A = 90° and ∠B = 90°. Consequently, ∠A = ∠C = 90° and ∠B = ∠D = 90°.

Additionally, opposite sides of a parallelogram are equal, so AB = CD and AD = BC.

Hence, proved that ABCD is a rectangle.

(ii) Now consider a rhombus ABCD that is inscribed in a circle.

In a rhombus, opposite angles are equal, so ∠BAD = ∠BCD. Since ABCD is a cyclic quadrilateral, the opposite angles are supplementary, giving us ∠BAD + ∠BCD = 180°.

This results in 2∠BAD = 180°, so ∠BAD = ∠BCD = 90°.

Similarly, for the other pair of opposite angles, ∠ABC = ∠ADC, and we have ∠ABC + ∠ADC = 180°, leading to 2∠ABC = 180°.

Thus, ∠ABC = ∠ADC = 90°.

Since all sides of a rhombus are equal, and each angle is 90°, ABCD is a square.

Hence, proved that ABCD is a square.

Question 8

In the given figure, AB = AC. Prove that DECB is an isosceles trapezium.

Answer:

We know that AB = AC. Therefore, the angles opposite these equal sides must also be equal, which implies that ∠B = ∠C. Let’s label this as equation (1).

Observing the figure, we see that DECB forms a cyclic quadrilateral. In such a quadrilateral, the sum of a pair of opposite angles is always 180°. Thus, ∠B + ∠DEC = 180°.

By substituting from equation (1), we get ∠C + ∠DEC = 180°. This is the sum of interior angles on the same side of a transversal, which implies DE is parallel to BC.

Since DE is parallel to BC, corresponding angles are equal, so ∠ADE = ∠B and ∠AED = ∠C.

This leads to ∠ADE = ∠AED, indicating that AD = AE.

Subtracting these equal segments from equal sides AB and AC, we find AB – AD = AC – AE, which simplifies to BD = CE.

Therefore, with DE || BC and BD = CE, we conclude that DECB is an isosceles trapezium.

Question 9

The figure given below, shows a circle with centre O.

Given : ∠AOC = a and ∠ABC = b.

(i) Find the relationship between a and b.

(ii) Find the measure of angle OAB, if OABC is a parallelogram.

Answer:

(i) Notice that the angle formed at the centre of a circle is twice the angle formed at the circumference by the same chord.

∴ ∠ABC = $\dfrac{1}{2}$ Reflex ∠COA

⇒ b = $\dfrac{1}{2}$ (360° – a)

⇒ 2b = 360° – a

⇒ a + 2b = 360° $\ldots$ (1)

Thus, the relationship between a and b is expressed by the equation: a + 2b = 360°.

(ii) Referring back to equation (1), we know:

⇒ a + 2b = 360°

⇒ a = 360° – 2b

Since OABC is a parallelogram, opposite angles must be equal.

Thus, a = b

⇒ 360° – 2b = b

⇒ 3b = 360°

⇒ b = $\dfrac{360°}{3}$ = 120°

Assume ∠OAB = x and ∠OCB = x.

⇒ ∠OAB + ∠OCB + ∠AOC + ∠ABC = 360°

⇒ x + x + a + b = 360°

⇒ 2x + 120° + 120° = 360°

⇒ 2x + 240° = 360°

⇒ 2x = 120°

⇒ x = $\dfrac{120°}{2}$ = 60°.

Therefore, ∠OAB = 60°.

Question 10

In the figure given, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°.

Calculate :

(i) ∠RNM,

(ii) ∠NRM.

Answer:

(i) Let’s connect RN and MS in the diagram.

Notice that ∠RMS is 90° because the angle in a semicircle is always a right angle.

Considering triangle RMS, we have:

∠RMS + ∠RSM + ∠MRS = 180°
∴ 90° + ∠RSM + 29° = 180°
⇒ ∠RSM + 119° = 180°
⇒ ∠RSM = 180° – 119° = 61°.

In a cyclic quadrilateral, the sum of opposite angles is 180°.

Thus, ∠RNM + ∠RSM = 180°
⇒ ∠RNM + 61° = 180°
⇒ ∠RNM = 180° – 61° = 119°.

Therefore, ∠RNM = 119°.

(ii) Given RS is parallel to NM, we know:

∠NMR = ∠MRS = 29° because alternate angles are equal.

From the diagram, we find:

∠NMS = ∠NMR + ∠RMS = 29° + 90° = 119°.

Again, applying the property of opposite angles in a cyclic quadrilateral:

In quadrilateral NMSR,

∠NRS + ∠NMS = 180°
⇒ ∠NRM + ∠MRS + ∠NMS = 180°
⇒ ∠NRM + 29° + 119° = 180°
⇒ ∠NRM = 180° – 119° – 29°
⇒ ∠NRM = 32°.

Hence, ∠NRM = 32°.

Question 11

In the given figure, SP is the bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that :

SQ = SR.

Answer:

Given that PQRS is a cyclic quadrilateral, the sum of its opposite angles equals 180°. Thus, we have:

∠QRS + ∠QPS = 180° \ ………..(1)

Since QPT forms a straight line, we also know that:

∠QPS + ∠SPT = 180° \ [QPT is a straight line] ……..(2)

By subtracting equation (2) from equation (1), we derive:

⇒ ∠QRS + ∠QPS – (∠QPS + ∠SPT) = 180° – 180°

⇒ ∠QRS – ∠SPT = 0

⇒ ∠QRS = ∠SPT ……….(3)

Additionally, we recognize that:

∠RQS = ∠RPS \ [Angles in the same segment are equal] ……..(4)

And since SP bisects ∠RPT, it follows that:

∠RPS = ∠SPT \ [PS bisects ∠RPT] …….(5)

Combining results from equations (3), (4), and (5), we conclude:

⇒ ∠QRS = ∠RQS.

Thus, the sides opposite these equal angles are also equal, leading to:

∴ SQ = SR

Hence, proved that SQ = SR.

Question 12

In the figure, O is the center of the circle, ∠AOE = 150°, ∠DAO = 51°. Calculate the sizes of the angles CEB and OCE.

Answer:

The angle formed at the center of a circle is twice the angle formed at the circumference by the same chord.

∴ Reflex ∠AOE = 2∠ADE

⇒ ∠ADE = $\dfrac{1}{2}$ Reflex ∠AOE

⇒ ∠ADE = $\dfrac{1}{2}$ (360° – 150°)

⇒ ∠ADE = $\dfrac{1}{2} \times 210°$ = 105°.

In the given figure:

∴ ∠DAB + ∠BED = 180° (The sum of opposite angles in a cyclic quadrilateral is 180°.)

⇒ ∠BED = 180° – ∠DAB = 180° – 51° = 129°.

Moreover:

∴ ∠CEB + ∠BED = 180° (As CED is a straight line)

⇒ ∠CEB = 180° – ∠BED = 180° – 129° = 51°.

Considering △ADC:

∴ ∠ADC + ∠ACD + ∠DAC = 180°

⇒ ∠ADE + ∠ACD + ∠DAO = 180° (From the figure, ∠ADC = ∠ADE and ∠DAC = ∠DAO)

⇒ 105° + ∠ACD + 51° = 180°

⇒ ∠ACD = 180° – 105° – 51° = 24°.

From the figure, we observe:

∠OCE = ∠ACD = 24°.

Hence, ∠CEB = 51° and ∠OCE = 24°.

Question 13

In the figure, given below, P and Q are the centers of two circles intersecting at B and C. ACD is a straight line. Calculate the numerical value of x.

Answer:

We are aware that the angle formed at the center of a circle is twice the angle formed at the circumference by the same chord.

∴ ∠APB = 2∠ACB

⇒ ∠ACB = $\dfrac{1}{2}$ ∠APB

⇒ ∠ACB = $\dfrac{1}{2} \times 150^\circ = 75^\circ$ .

Looking at the figure, we see:

∴ ∠ACB + ∠BCD = 180^\circ \text{ (since ACD is a straight line)}

⇒ 75^\circ + ∠BCD = 180^\circ

⇒ ∠BCD = 180^\circ – 75^\circ = 105^\circ.

Additionally, we have:

∴ Reflex ∠BQD = 2∠BCD \text{ (the angle at the center is double the angle at the circumference by the same chord.)}

⇒ (360^\circ – x) = 2 \times 105^\circ

⇒ x = 360^\circ – 210^\circ = 150^\circ.

Hence, x = 150^\circ.

Question 14

The figure shows two circles which intersect at A and B. The center of the smaller circle is O and lies on the circumference of the larger circle. Given ∠APB = a°.

Calculate, in terms of a°, the value of :

(i) obtuse ∠AOB,

(ii) ∠ACB,

(iii) ∠ADB.

Give reasons for your answers clearly.

Answer:

(i) Consider that the angle formed at the center of a circle is twice the angle formed at the circumference by the same chord. Therefore, the obtuse angle ∠AOB is given by 2∠APB, which equals 2a°.

Hence, obtuse ∠AOB = 2a°.

(ii) Notice that the quadrilateral OACB is cyclic. In a cyclic quadrilateral, the sum of the opposite angles is 180°. Thus, we have:

⇒ ∠AOB + ∠ACB = 180°

Substituting the known value of ∠AOB:

⇒ 2a° + ∠ACB = 180°

Solving for ∠ACB gives:

⇒ ∠ACB = 180° – 2a°.

Hence, ∠ACB = 180° – 2a°.

(iii)

By joining AD and BD, we recognize that angles in the same segment of a circle are equal. Therefore:

∴ ∠ADB = ∠ACB = 180° – 2a°.

Hence, ∠ADB = 180° – 2a°.

Question 15

In the given figure, O is the centre of the circle and ∠ABC = 55°. Calculate the values of x and y.

Answer:

Recall that the angle at the center of a circle is twice the angle at the circumference when subtended by the same chord. Thus, we have:

∠AOC = 2 × ∠ABC = 2 × 55° = 110°.

∴ x = 110°.

Since ABCD is a cyclic quadrilateral, the opposite angles must add up to 180°.

∴ ∠ADC + ∠ABC = 180°

⇒ y + 55° = 180°

⇒ y = 180° – 55° = 125°.

Hence, x = 110° and y = 125°.

Question 16

In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°

(i) Prove that AC is a diameter of the circle.

(ii) Find ∠ACB

Answer:

(i) Consider $\triangle ABD$ :

⇒ $\angle DAB + \angle ABD + \angle ADB = 180^\circ$ due to the angle sum property of triangles.

⇒ $65^\circ + 70^\circ + \angle ADB = 180^\circ$

⇒ $135^\circ + \angle ADB = 180^\circ$

⇒ $\angle ADB = 180^\circ - 135^\circ = 45^\circ$

Looking at the figure, observe:

⇒ $\angle ADC = \angle ADB + \angle BDC = 45^\circ + 45^\circ = 90^\circ$ .

Since $\angle ADC = 90^\circ$ , it confirms that $\angle ADC$ is a right angle, indicating that arc $ADC$ represents a semicircle.

∴ $AC$ is the diameter of the circle.

Thus, we have shown that $AC$ is indeed the diameter.

(ii) Recall that angles subtended by the same arc in a circle are equal.

∴ $\angle ACB = \angle ADB = 45^\circ$ .

Thus, ( \angle ACB = 45^\circ.

Question 17

In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and ∠EAB = 63°. Calculate :

(i) ∠EBA,

(ii) ∠BCD.

Answer:

(i) Since AB is the diameter, the angle subtended by it at any point on the circle is a right angle.

∴ ∠AEB = 90°.

In the triangle AEB, the sum of the angles is 180°.

⇒ ∠AEB + ∠EBA + ∠EAB = 180°

⇒ 90° + ∠EBA + 63° = 180°

Adding the known angles gives us:

⇒ 153° + ∠EBA = 180°

Solving for ∠EBA, we get:

⇒ ∠EBA = 180° – 153° = 27°.

Hence, ∠EBA = 27°.

(ii) Given that AB is parallel to ED,

∴ ∠DEB = ∠EBA = 27° because they are alternate angles.

Since BCDE forms a cyclic quadrilateral, the opposite angles add up to 180°.

∴ ∠DEB + ∠BCD = 180°

Substituting the known values:

⇒ 27° + ∠BCD = 180°

⇒ ∠BCD = 180° – 27° = 153°.

Hence, ∠BCD = 153°.

Question 18

In the given figure, AB is a diameter of the circle with center O. DO is parallel to CB and ∠DCB = 120°. Calculate :

(i) ∠DAB,

(ii) ∠DBA,

(iii) ∠DBC,

(iv) ∠ADC.

Also, show that the △AOD is an equilateral triangle.

Answer:

(i) Since ABCD forms a cyclic quadrilateral, we have:

∴ ∠DCB + ∠DAB = 180° (The sum of opposite angles in a cyclic quadrilateral is 180°.)

⇒ ∠DAB = 180° – ∠DCB

⇒ ∠DAB = 180° – 120° = 60°.

Hence, ∠DAB = 60°.

(ii) Recall that an angle inscribed in a semicircle is a right angle.

Thus, ∠ADB = 90°.

In △DAB, we apply the angle sum property:

⇒ ∠ADB + ∠DAB + ∠DBA = 180°

⇒ 90° + 60° + ∠DBA = 180°

⇒ 150° + ∠DBA = 180°

⇒ ∠DBA = 180° – 150° = 30°.

Hence, ∠DBA = 30°.

(iii) Since OD = OB (both are radii of the circle), it follows that:

∴ ∠ODB = ∠OBD (Angles opposite equal sides are equal)

From the figure, we see:

∠OBD = ∠DBA = 30°.

∴ ∠ODB = 30°.

Given that DO is parallel to BC:

∴ ∠DBC = ∠ODB = 30° (Alternate angles are equal.)

Hence, ∠DBC = 30°.

(iv) Referring to the figure:

∠ABC = ∠ABD + ∠DBC = 30° + 30° = 60°.

In the cyclic quadrilateral ABCD:

∴ ∠ADC + ∠ABC = 180° (The sum of opposite angles in a cyclic quadrilateral is 180°.)

⇒ ∠ADC = 180° – ∠ABC

⇒ ∠ADC = 180° – 60° = 120°.

Hence, ∠ADC = 120°.

In △AOD, since OA = OD (both are radii of the circle), we conclude:

∠AOD = ∠DAO (Angles opposite equal sides are equal)

From the figure, we know:

⇒ ∠DAO = ∠DAB = 60°.

∴ ∠AOD = ∠DAO = ∠ADO = 60°

Hence, it is shown that △AOD is an equilateral triangle.

Question 19

Calculate the angles x, y and z if :

\dfrac{x}{3} = \dfrac{y}{4} = \dfrac{z}{5}

Answer:

Consider the given equation $\dfrac{x}{3} = \dfrac{y}{4} = \dfrac{z}{5}$ and let this common ratio be $k$ .

∴ We have $x = 3k$ , $y = 4k$ , and $z = 5k$ .

From the diagram, notice that:

∠BCP is equal to ∠DCQ because they are vertically opposite angles.

In a triangle, the exterior angle is equal to the sum of the two opposite interior angles. Thus:

∠ABC = ∠BCP + ∠BPC = $x + y = 3k + 4k = 7k$ .

Similarly, ∠ADC = ∠DCQ + ∠DQC = $x + z = 3k + 5k = 8k$ .

Since ABCD is a cyclic quadrilateral, we know that the sum of the opposite angles is $180°$ .

⇒ ∠ABC + ∠ADC = $180°$

⇒ $8k + 7k = 180°$

⇒ $15k = 180°$

⇒ $k = \dfrac{180}{15} = 12°$ .

Now, substituting back to find $x$ , $y$ , and $z$ :

x = 3k = 3 \times 12° = 36°

y = 4k = 4 \times 12° = 48°

$z = 5k = 5 \times 12° = 60°$ .

Hence, $x = 36°$ , $y = 48°$ , and $z = 60°$ .

Question 20

In the given figure, AC is the diameter of the circle with center O. CD and BE are parallel. Angle ∠AOB = 80° and ∠ACE = 10°. Calculate :

(i) Angle BEC,

(ii) Angle BCD,

(iii) Angle CED.

Answer:

(i) Observing the figure, we see that:

∠BOC + ∠BOA = 180° since AOC forms a straight line.

⇒ ∠BOC + 80° = 180°

⇒ ∠BOC = 180° – 80° = 100°.

It’s known that the angle at the center is twice the angle at the circumference subtended by the same chord.

⇒ ∠BOC = 2∠BEC

⇒ ∠BEC = $\dfrac{1}{2}$ ∠BOC = $\dfrac{100°}{2}$ = 50°.

Hence, ∠BEC = 50°.

(ii) Considering that DC is parallel to EB,

∴ ∠DCE = ∠BEC = 50° because alternate angles are equal.

Also, the angle at the center is twice the angle at the circumference subtended by the same chord.

⇒ ∠AOB = 2∠ACB

⇒ ∠ACB = $\dfrac{1}{2}$ ∠AOB = $\dfrac{80°}{2}$ = 40°.

From the figure, we can write:

∠BCD = ∠ACB + ∠ACE + ∠DCE = 40° + 10° + 50° = 100°.

Hence, ∠BCD = 100°.

(iii) In a cyclic quadrilateral, the sum of opposite angles equals 180°.

⇒ ∠BED + ∠BCD = 180°

⇒ ∠BED = 180° – ∠BCD = 180° – 100° = 80°.

From the figure, it follows that:

⇒ ∠BED = ∠BEC + ∠CED

⇒ 80° = 50° + ∠CED

⇒ ∠CED = 80° – 50° = 30°.

Hence, ∠CED = 30°.

Question 21

In the given figure, AE is the diameter of the circle. Write down the numerical value of ∠ABC + ∠CDE. Give reasons for your answer.

Answer:

To solve this, let’s connect OA, OB, OC, and OD.

In the triangle △OAB, since OA and OB are radii of the same circle, they are equal. This implies that ∠1 = ∠2.

Similarly, in △OBC, OB equals OC (both radii), so ∠3 = ∠4.

In △OCD, OC equals OD, which means ∠5 = ∠6.

Finally, in △ODE, OD equals OE, leading to ∠7 = ∠8.

Now, consider the angle sum property for each triangle:

For △OAB:

∴ ∠1 + ∠2 + ∠a = 180° ……….(1)

For △OBC:

∴ ∠3 + ∠4 + ∠b = 180° ……….(2)

For △OCD:

∴ ∠5 + ∠6 + ∠c = 180° ……….(3)

For △ODE:

∴ ∠7 + ∠8 + ∠d = 180° ……….(4)

Adding equations (1), (2), (3), and (4) gives:

∴ ∠1 + ∠2 + ∠a + ∠3 + ∠4 + ∠b + ∠5 + ∠6 + ∠c + ∠7 + ∠8 + ∠d = 720°

Simplifying, we have:

∴ 2∠2 + 2∠3 + 2∠6 + 2∠7 + ∠a + ∠b + ∠c + ∠d = 720°

Since ∠a + ∠b + ∠c + ∠d = 180°, we substitute:

∴ 2[∠2 + ∠3] + 2[∠6 + ∠7] + 180° = 720°

This simplifies to:

∴ 2∠ABC + 2∠CDE = 540°

Dividing by 2, we find:

∴ ∠ABC + ∠CDE = 270°.

Hence, ∠ABC + ∠CDE = 270°.

Question 22

In the given figure, AOC is a diameter and AC is parallel to ED. If ∠CBE = 64°, calculate ∠DEC.

Answer:

Draw a line connecting points A and B.

Notice that ∠ABC is 90° because the angle subtended by a diameter in a semicircle is always a right angle.

Now, observe the figure:

∠ABE is calculated as follows:

∠ABE = ∠ABC – ∠CBE = 90° – 64° = 26°.

Since angles subtended by the same chord in the same segment are equal, we have:

∠ACE = ∠ABE = 26°.

Given that AC is parallel to ED, it follows that:

∠DEC = ∠ACE = 26° due to the property of alternate interior angles being equal.

Hence, ∠DEC = 26°.

Question 23

Use the given figure to find :

(i) ∠BAD

(ii) ∠DQB.

Answer:

(i) Considering $\triangle ADP$ :

∴ $\angle PAD + \angle ADP + \angle DPA = 180^\circ$ $\text{(Angle sum property of a triangle)}$

⇒ $\angle PAD + 85^\circ + 40^\circ = 180^\circ$

⇒ $\angle PAD + 125^\circ = 180^\circ$

⇒ $\angle PAD = 180^\circ - 125^\circ = 55^\circ$ .

From the diagram, we observe:

⇒ $\angle BAD = \angle PAD = 55^\circ$ .

Hence, $\angle BAD = 55^\circ$ .

(ii) Recall that the sum of opposite angles in a cyclic quadrilateral equals 180°.

⇒ $\angle ABC + \angle ADC = 180^\circ$

⇒ $\angle ABC + 85^\circ = 180^\circ$

⇒ $\angle ABC = 180^\circ - 85^\circ = 95^\circ$ .

Now, in $\triangle AQB$ :

∴ $\angle AQB + \angle QAB + \angle ABQ = 180^\circ$ $\text{(Angle sum property of a triangle)}$

⇒ $\angle AQB + \angle BAD + \angle ABC = 180^\circ$ $\text{From the diagram, } \angle QAB = \angle BAD \text{ and } \angle ABQ = \angle ABC$

⇒ $\angle AQB + 55^\circ + 95^\circ = 180^\circ$

⇒ $\angle AQB + 150^\circ = 180^\circ$

⇒ $\angle AQB = 180^\circ - 150^\circ = 30^\circ$ .

From the diagram, we see:

⇒ $\angle DQB = \angle AQB = 30^\circ$ .

Hence, $\angle DQB = 30^\circ$ .

Question 24

In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°, Calculate :

(i) ∠RPQ

(ii) ∠STP.

Answer:

To solve this, we start by connecting PR.

(i) Since PQ is a diameter, ∠PRQ must be 90° because the angle subtended by a diameter in a semicircle is always a right angle.

Now, consider △PQR:

∴ ∠RPQ + ∠PRQ + ∠PQR = 180°
⇒ ∠RPQ + 90° + 58° = 180°
⇒ ∠RPQ + 148° = 180°
⇒ ∠RPQ = 180° – 148° = 32°.

Thus, ∠RPQ = 32°.

(ii) Since SR is parallel to PQ, ∠PRS is equal to ∠RPQ due to the property of alternate angles.

∴ ∠PRS = ∠RPQ = 32°

Now, in cyclic quadrilateral PRST, the sum of the opposite angles is 180°:

⇒ ∠STP + ∠PRS = 180°
⇒ ∠STP = 180° – ∠PRS = 180° – 32° = 148°.

Thus, ∠STP = 148°.

Question 25

AB is the diameter of the circle with center O. OD is parallel to BC and ∠AOD = 60°. Calculate the numerical values of :

(i) ∠ABD,

(ii) ∠DBC,

(iii) ∠ADC.

Answer:

Connect BD.

(i) Observe the figure:

⇒ ∠BDA = 90° [Since the angle inscribed in a semicircle is always a right angle.]

In the triangle △OAD:

OA = OD [Both are radii of the circle]

∠OAD = ∠ODA = x (assume) [Angles opposite equal sides are equal]

In △OAD:

⇒ ∠OAD + ∠ODA + ∠AOD = 180° [Triangle angle sum property]

⇒ x + x + 60° = 180°

⇒ 2x = 180° – 60°

⇒ 2x = 120°

⇒ x = $\dfrac{120°}{2}$ = 60°.

From the figure:

∠ODB = ∠BDA – ∠ADO = 90° – 60° = 30°.

Given that OD is parallel to BC:

∠DBC = ∠ODB = 30° [Alternate interior angles are equal]

Hence, ∠DBC = 30°.

(ii) Recall that:

The angle in a semicircle is a right angle.

∴ ∠BDA = 90°.

Since:

∠OAD = ∠ODA = ∠AOD = 60°.

From the figure:

⇒ ∠BDA = ∠ODA + ∠ODB

⇒ 90° = 60° + ∠ODB

⇒ ∠ODB = 90° – 60° = 30°.

Given:

OD is parallel to BC

∴ ∠DBC = ∠ODB = 30° [Alternate interior angles are equal].

Hence, ∠DBC = 30°.

(iii) From the figure:

∠ABC = ∠ABD + ∠DBC = 30° + 30° = 60°.

Since, the sum of opposite angles in a cyclic quadrilateral is 180°.

∴ ∠ABC + ∠ADC = 180°

⇒ 60° + ∠ADC = 180°

⇒ ∠ADC = 120°.

Hence, ∠ADC = 120°.

Question 26

In the given figure, the center O of the small circle lies on the circumference of the bigger circle. If ∠APB = 75° and ∠BCD = 40°, find :

(i) ∠AOB,

(ii) ∠ACB,

(iii) ∠ABD,

(iv) ∠ADB.

Answer:

(i) Begin by connecting points A and B, as well as A and D.

Notice that the angle an arc makes at the center is twice the angle it makes at any point on the other part of the circle’s circumference.

∴ ∠AOB = 2∠APB = 2 × 75° = 150°.

Hence, ∠AOB = 150°.

(ii) Recall that the sum of opposite angles in a cyclic quadrilateral is 180°.

∴ ∠ACB + ∠AOB = 180°

⇒ ∠ACB + 150° = 180°

⇒ ∠ACB = 180° – 150° = 30°.

Hence, ∠ACB = 30°.

(iii) Referring to the figure, we have:

∠ACD = ∠ACB + ∠BCD = 30° + 40° = 70°.

Again, using the property of cyclic quadrilaterals where opposite angles sum to 180°:

In cyclic quadrilateral ABDC,

∴ ∠ABD + ∠ACD = 180°

⇒ ∠ABD + 70° = 180°

⇒ ∠ABD = 180° – 70° = 110°.

Hence, ∠ABD = 110°.

(iv) Applying the cyclic quadrilateral property once more:

In cyclic quadrilateral AOBD,

∴ ∠ADB + ∠AOB = 180°

⇒ ∠ADB + 150° = 180°

⇒ ∠ADB = 180° – 150° = 30°.

Hence, ∠ADB = 30°.

Question 27

In the given figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°. Find :

(i) ∠BCD

(ii) ∠ACB

Hence, show that AC is a diameter.

Answer:

(i) Consider the property of a cyclic quadrilateral: the sum of its opposite angles equals 180°.

For cyclic quadrilateral ABCD, we have:

∴ ∠BCD + ∠BAD = 180°

⇒ ∠BCD + 65° = 180°

⇒ ∠BCD = 180° – 65° = 115°.

Hence, ∠BCD = 115°.

(ii) Now, look at △ABD:

According to the angle sum property of a triangle:

⇒ ∠ADB + ∠BAD + ∠DBA = 180°

⇒ ∠ADB + 65° + 70° = 180°

⇒ ∠ADB + 135° = 180°

⇒ ∠ADB = 180° – 135° = 45°.

Notice that angles in the same segment of a circle are equal, so:

∴ ∠ACB = ∠ADB = 45°.

Hence, ∠ADB = 45°.

From the diagram, observe:

∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°.

Since an angle subtended by a diameter in a semicircle is a right angle, it follows that:

Hence, proved that AC is a diameter.

Question 28

In a cyclic quadrilateral ABCD, ∠A : ∠C = 3 : 1 and ∠B : ∠D = 1 : 5; find each angle of the quadrilateral.

Answer:

Consider ∠A as 3x and ∠C as x.

Since ABCD is a cyclic quadrilateral, the sum of opposite angles is 180°.

∴ ∠A + ∠C = 180°

⇒ 3x + x = 180°

⇒ 4x = 180°

⇒ x = $\dfrac{180°}{4}$ = 45°.

Thus, ∠A = 3x = 3 × 45° = 135° and ∠C = x = 45°.

Next, let ∠B be y and ∠D be 5y.

Again, because ABCD is cyclic, ∠B + ∠D = 180°.

⇒ y + 5y = 180°

⇒ 6y = 180°

⇒ y = $\dfrac{180°}{6}$ = 30°.

Therefore, ∠B = y = 30° and ∠D = 5y = 5 × 30° = 150°.

Hence, ∠A = 135°, ∠B = 30°, ∠C = 45°, and ∠D = 150°.

Exercise 17(C)

Question 1(a)

In the given figure, O is the center of the circle and chord AB : chord CD = 3 : 5. If angle AOB = 60°, angle COD is equal to :

60°
120°
90°
100°

Answer:

We know that the ratio of the lengths of chords AB to CD is 3:5.

∴ The ratio of the angles subtended by these chords at the center, ∠AOB to ∠COD, will also be 3:5.

Given that ∠AOB is 60°, we can set up the proportion:

⇒ 60° : ∠COD = 3 : 5

This implies:

⇒ $\dfrac{60°}{∠COD} = \dfrac{3}{5}$

To find ∠COD, solve for it by cross-multiplying:

⇒ ∠COD = $\dfrac{60° \times 5}{3} = \dfrac{300°}{3}$ = 100°.

Hence, Option 4 is the correct option.

Question 1(b)

In the given figure, O is the center of the circle and angle OAB = 55°, then angle ACB is equal to :

55°
35°
70°
30°

Answer:

Consider the triangle $\triangle OAB$ . Since $OA$ and $OB$ are radii of the circle, they are equal. This implies that the angles opposite these sides must also be equal. Therefore, $\angle OBA = \angle OAB = 55^\circ$ .

Using the angle sum property of a triangle, we have:

\angle OBA + \angle OAB + \angle AOB = 180^\circ

Substituting the known angles:

55^\circ + 55^\circ + \angle AOB = 180^\circ

\angle AOB + 110^\circ = 180^\circ

Solving for $\angle AOB$ , we find:

\angle AOB = 180^\circ - 110^\circ = 70^\circ

Recall that the angle subtended by an arc at the center of a circle is twice the angle subtended at any point on the remaining part of the circumference. Thus:

\angle AOB = 2 \times \angle ACB

Solving for $\angle ACB$ , we get:

\angle ACB = \frac{\angle AOB}{2} = \frac{70^\circ}{2} = 35^\circ

Hence, Option 2 is the correct option.

Question 1(c)

In the given figure, O is the center of a circle. AB is the side of a square and BC is side of a regular hexagon. Also arc AD = arc CD. Angle DOC is equal to :

150°
105°
130°
210°

Answer:

Connect OD to the center.

Since AB is a side of the square, it follows that:

∴ ∠AOB = $\dfrac{360°}{4}$ = 90°.

Given BC is a side of a regular hexagon, hence:

∴ ∠BOC = $\dfrac{360°}{6}$ = 60°.

Remember, equal arcs create equal angles at the circle’s center. Given arc AD = arc CD, we have:

∴ ∠AOD = ∠COD = x (assume)

From the diagram, we see:

⇒ ∠AOD + ∠COD + ∠AOB + ∠BOC = 360°

⇒ x + x + 90° + 60° = 360°

⇒ 2x + 150° = 360°

⇒ 2x = 360° – 150°

⇒ 2x = 210°

⇒ x = $\dfrac{210°}{2}$ = 105°.

Hence, Option 2 is the correct option.

Question 1(d)

In the given figure, O is the center of the circle, AB is side of a regular pentagon, then angle ACB is equal to :

36°
72°
50°
40°

Answer:

To solve this problem, let’s connect the points O, A, and B with line segments OA and OB.

Since AB is a side of a regular pentagon, we know:

∴ ∠AOB = $\dfrac{360°}{5}$ = 72°.

Recall the property of a circle: the angle subtended by an arc at the center is twice the angle subtended at any point on the remaining circumference.

∴ ∠AOB = 2∠ACB

This implies:

∠ACB = $\dfrac{∠AOB}{2} = \dfrac{72°}{2}$ = 36°.

Hence, Option 1 is the correct option.

Question 1(e)

In the given figure, O is the center of the circle, chords AB, CD and EF are equal whereas chords BC, DE and FA are separately equal. The angle AOC is equal to :

80°
100°
90°
120°

Answer:

We know that chords AB, CD, and EF are equal in length. Consequently, the angles subtended by these chords at the center, ∠AOB, ∠COD, and ∠EOF, are equal. Let’s denote each of these angles as x.

Similarly, chords BC, DE, and FA are also equal. Therefore, the angles subtended by these chords, ∠BOC, ∠DOE, and ∠AOF, are equal as well. Let’s denote each of these angles as y.

Now, consider the sum of all these angles around the center O:

⇒ ∠AOB + ∠COD + ∠EOF + ∠BOC + ∠DOE + ∠AOF = 360°

Substituting the expressions for these angles, we have:

⇒ x + x + x + y + y + y = 360°

⇒ 3x + 3y = 360°

Dividing the entire equation by 3 gives:

⇒ x + y = \dfrac{360°}{3}

⇒ x + y = 120°

From the diagram, we observe that ∠AOC is composed of ∠AOB and ∠BOC. Thus, we find:

∠AOC = ∠AOB + ∠BOC = x + y = 120°.

Hence, Option 4 is the correct option.

Question 2

In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal. Prove it.

Answer:

Consider a cyclic trapezium ABCD where AB is parallel to DC and AC and BD are the diagonals.

Notice that chord AD creates ∠ABD, and chord BC creates ∠BDC at the circle’s edge.

Since ∠ABD = ∠BDC (because AB is parallel to DC with BD as the transversal),

∴ Chord AD equals chord BC, as equal chords create equal angles at the circle’s edge.

From the diagram, observe:

⇒ ∠B + ∠D = 180° (The sum of opposite angles in a cyclic quadrilateral is 180°)

Additionally,

⇒ ∠B + ∠C = 180° (The sum of co-interior angles is 180° due to AB being parallel to CD)

Thus, we have:

∴ ∠B + ∠C = ∠B + ∠D

⇒ ∠C = ∠D

Now, examining triangles ∆ADC and ∆BCD:

⇒ DC = DC (This is common)

⇒ ∠C = ∠D (As shown above)

⇒ AD = BC (As demonstrated earlier)

Therefore, using the SAS congruence criterion:

∆ADC ≅ ∆BCD

∴ AC = BD (By C.P.C.T.)

Hence, proved above AC = BD and AD = BC.

Question 3

In the following figure, AD is the diameter of the circle with centre O. Chords AB, BC and CD are equal. If ∠DEF = 110°, calculate :

(i) ∠AEF, (ii) ∠FAB.

Answer:

To solve this, start by connecting AE, OB, and OC.

(i) Since AD is the diameter, we know that:

∠AED = 90°

This is because an angle subtended by a diameter in a semicircle is always a right angle.

Given that ∠DEF = 110°, we can find ∠AEF by subtracting:

∠AEF = ∠DEF – ∠AED = 110° – 90° = 20°.

Thus, ∠AEF = 20°.

(ii) We are told chords AB, BC, and CD are equal, which implies:

∠AOB = ∠BOC = ∠COD

This is due to the property that equal chords in a circle subtend equal angles at the center.

Considering the straight line AOD:

∠AOB + ∠BOC + ∠COD = 180°

Therefore:

∠AOB = ∠BOC = ∠COD = $rac{180°}{3}$ = 60°

In triangle OAB, since OA = OB (radii of the circle), it follows that:

∠OAB = ∠OBA

Using the angle sum property for triangle OAB:

∠OAB + ∠OBA + ∠AOB = 180°

Substituting the known values:

∠OAB + ∠OBA + 60° = 180°

⇒ ∠OAB + ∠OBA = 120°

Since ∠OAB = ∠OBA, we have:

∠OAB = ∠OBA = $rac{120°}{2}$ = 60°

Now, in the cyclic quadrilateral ADEF, the opposite angles satisfy:

∠DEF + ∠DAF = 180°

Thus, finding ∠DAF:

∠DAF = 180° – ∠DEF = 180° – 110° = 70°

From the diagram, we can express ∠FAB as:

∠FAB = ∠DAF + ∠OAB = 70° + 60° = 130°

Hence, ∠FAB = 130°.

Question 4

If two sides of a cyclic-quadrilateral are parallel; prove that :

(i) its other two sides are equal.

(ii) its diagonals are equal.

Answer:

Consider a cyclic quadrilateral ABCD where AB is parallel to DC. The diagonals are AC and BD.

Given that AB || DC, we have:

∠DCA = ∠CAB, because alternate angles are equal when two lines are parallel.

Notice that chord AD creates ∠DCA and chord BC creates ∠CAB at the circle’s circumference.

Since ∠DCA = ∠CAB, and angles subtended by equal chords are equal, it follows that chord AD = chord BC or AD = BC.

Thus, AD = BC is proven.

(ii) From the diagram, we know:

⇒ ∠A + ∠C = 180° because opposite angles in a cyclic quadrilateral sum to 180°.

Also, since AB || CD:

⇒ ∠B + ∠C = 180° [Sum of co-interior angles = 180°]

Thus, we have:

∴ ∠B + ∠C = ∠A + ∠C

⇒ ∠B = ∠A

Now, consider triangles ∆ABC and ∆ADB:

⇒ AB = AB [This is the common side]

⇒ ∠B = ∠A [As proven earlier]

⇒ BC = AD [As proven earlier]

By the SAS (Side-Angle-Side) criterion of congruence, we conclude that:

∆ACB ≅ ∆ADB

∴ AC = BD [By C.P.C.T.]

Thus, AC = BD is proven.

Question 5

The given figure show a circle with centre O. Also, PQ = QR = RS and ∠PTS = 75°.

Calculate:

(i) ∠POS,

(ii) ∠QOR,

(iii) ∠PQR.

Answer:

To solve this problem, we first connect the points O to P, Q, R, and S.

Given that PQ = QR = RS, it follows that ∠POQ = ∠QOR = ∠ROS = x. This is because equal chords in a circle subtend equal angles at the centre.

The arc PQRS creates ∠POS at the centre and ∠PTS at the circle’s circumference. Since the angle at the centre is twice the angle at the circumference, we have:

∠POS = 2 × ∠PTS = 2 × 75° = 150°.

Thus,

∠POQ + ∠QOR + ∠ROS = 150°.

This implies:

x + x + x = 150°

3x = 150°

x = $\dfrac{150°}{3}$ = 50°.

Now, consider ∆OPQ:

OP = OQ, being radii of the circle.

Thus, ∠OPQ = ∠OQP = y.

In ∆OPQ:

∠OPQ + ∠OQP + ∠POQ = 180°

∠OPQ + ∠OQP + 50° = 180°

∠OPQ + ∠OQP = 180° – 50°

∠OPQ + ∠OQP = 130°

2y = 130°

y = $\dfrac{130°}{2}$ = 65°

Thus, ∠OPQ = ∠OQP = y = 65°.

Now, consider ∆OQR:

OQ = OR, as they are radii of the circle.

Thus, ∠OQR = ∠ORQ = z.

In ∆OQR:

∠OQR + ∠ORQ + ∠QOR = 180°

z + z + 50° = 180°

2z = 180° – 50°

2z = 130°

z = $\dfrac{130°}{2}$ = 65°

Thus, ∠OQR = ∠ORQ = z = 65°.

(i) Hence, ∠POS = 150°.

(ii) Hence, ∠QOR = 50°.

(iii) Observing the figure, we find:

∠PQR = ∠PQO + ∠OQR = 65° + 65° = 130°.

Hence, ∠PQR = 130°.

Question 6

In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. Calculate the sizes of :

(i) ∠AOB,

(ii) ∠ACB,

(iii) ∠ABC.

Answer:

To solve this, let’s connect OC.

(i) In a regular hexagon inscribed in a circle, each side creates a central angle of 60°.

⇒ ∠AOB = 60°.

Thus, ∠AOB = 60°.

(ii) According to the circle’s properties, the central angle is twice the angle at the circumference.

∴ ∠AOB = 2∠ACB

⇒ ∠ACB = $\dfrac{1}{2}$ ∠AOB = $\dfrac{1}{2} \times 60°$ = 30°.

So, ∠ACB = 30°.

(iii) Given AC is a side of a regular octagon,

∠AOC = $\dfrac{360°}{8}$ = 45°.

Here, the arc AC creates ∠AOC at the centre and ∠ABC at the circle’s edge.

∴ ∠AOC = 2∠ABC

∴ ∠ABC = $\dfrac{1}{2}$ x ∠AOC = $\dfrac{1}{2}$ x 45° = 22.5°

Therefore, ∠ABC = 22.5°.

Question 7

In a regular pentagon ABCDE, inscribed in a circle; find the ratio between angle EDA and angle ADC.

Answer:

Consider a regular pentagon ABCDE inscribed in a circle as depicted in the diagram.

In a circle, the angle at the center is double the angle at the circumference subtended by the same arc.

∴ ∠AOE = 2∠ADE

This implies:

⇒ ∠ADE = $\dfrac{1}{2}$ ∠AOE

Since ∠AOE is subtended by the side AE of a regular pentagon inscribed in a circle, we have:

∴ ∠AOE = $\dfrac{360°}{5}$ = 72°

Thus, we find:

⇒ ∠ADE = $\dfrac{1}{2}$ ∠AOE = $\dfrac{1}{2} \times 72°$ = 36°.

Recall that each interior angle of a regular pentagon measures 108°.

From the diagram, we can see:

⇒ ∠ADC = ∠EDC – ∠ADE = 108° – 36° = 72°.

Therefore, the ratio of ∠ADE to ∠ADC is:

∴ ∠ADE : ∠ADC = 36° : 72° = 1 : 2.

Hence, the ratio between angle EDA and angle ADC = 1 : 2.

Question 8

In the given figure, AB = BC = CD and ∠ABC = 132°. Calculate :

(i) ∠AEB,

(ii) ∠AED,

(iii) ∠COD.

Answer:

(i) First, connect EB and EC.

In cyclic quadrilateral ABCE, we have:

⇒ ∠ABC + ∠AEC = 180°

This is because the sum of opposite angles in a cyclic quadrilateral is always 180°.

⇒ 132° + ∠AEC = 180°

⇒ ∠AEC = 180° – 132° = 48°.

Given that AB = BC, we apply the theorem stating that the angle subtended by an arc at the center is twice the angle it subtends at the circumference.

∴ ∠AEB = $\dfrac{1}{2}$ ∠AEC

= $\dfrac{1}{2} \times 48°$ = 24°.

Thus, ∠AEB = 24°.

(ii) According to the property that equal chords subtend equal angles at the circumference:

∠AEB = ∠BEC = ∠CED = 24°.

Therefore, ∠AED = ∠AEB + ∠BEC + ∠CED = 24° + 24° + 24° = 72°.

Hence, ∠AED = 72°.

(iii) Using the same theorem that an arc’s angle at the center is double that at the circumference:

∴ ∠COD = 2∠CED = 2 × 24° = 48°.

Thus, ∠COD = 48°.

Question 9

In the figure, O is the centre of the circle and the length of arc AB is twice the length of arc BC. If angle AOB = 108°, find :

(i) ∠CAB,

(ii) ∠ADB.

Answer:

To solve this, let’s connect points AD and BD.

(i) We are given that:

⇒ arc AB = 2 arc BC

This implies:

⇒ ∠AOB = 2∠BOC

Thus, we find ∠BOC as follows:

⇒ ∠BOC = $\dfrac{1}{2}$ ∠AOB = $\dfrac{1}{2} \times 108°$ = 54°.

Recall that the angle subtended by an arc at the centre of a circle is double the angle it subtends at any point on the remaining circumference.

∴ ∠BOC = 2∠CAB

Therefore:

⇒ ∠CAB = $\dfrac{1}{2}$ ∠BOC = $\dfrac{1}{2} \times 54°$ = 27°.

Hence, ∠CAB = 27°.

(ii) Arc AB creates ∠AOB at the centre and ∠ACB at the circle’s circumference.

Thus:

∴ ∠AOB = 2∠ACB

So we have:

∠ACB = $\dfrac{1}{2}$ ∠AOB = $\dfrac{1}{2} \times 108°$ = 54°.

Considering cyclic quadrilateral ADBC:

⇒ ∠ADB + ∠ACB = 180° [Since the sum of opposite angles in a cyclic quadrilateral is 180°]

Thus:

⇒ ∠ADB + 54° = 180°

⇒ ∠ADB = 180° – 54° = 126°.

Hence, ∠ADB = 126°.

Question 10

The figure shows a circle with centre O. AB is the side of regular pentagon and AC is the side of regular hexagon.

Find the angles of triangle ABC.

Answer:

To tackle this problem, let’s connect the points $O$ , $A$ , $B$ , and $C$ with lines $OA$ , $OB$ , and $OC$ .

Notice that since $AB$ is a side of a regular pentagon, the angle at the center, $\angle AOB$ , is given by:

\angle AOB = \frac{360^\circ}{5} = 72^\circ.

Similarly, because $AC$ is a side of a regular hexagon, the angle $\angle AOC$ at the center is:

\angle AOC = \frac{360^\circ}{6} = 60^\circ.

According to the figure, we have:

\angle AOB + \angle AOC + \text{reflex } \angle BOC = 360^\circ.

Substituting the known values:

72^\circ + 60^\circ + \text{reflex } \angle BOC = 360^\circ.

This simplifies to:

\text{reflex } \angle BOC + 132^\circ = 360^\circ,

\text{reflex } \angle BOC = 360^\circ - 132^\circ = 228^\circ.

Recall that the angle subtended by an arc at the center is twice the angle it subtends at any point on the circle’s circumference. Therefore, for arc $BC$ :

\text{reflex } \angle BOC = 2 \times \angle BAC.

Thus:

\angle BAC = \frac{1}{2} \times 228^\circ = 114^\circ.

For arc $AC$ , which subtends $\angle AOC$ at the center and $\angle ABC$ on the circle:

\angle AOC = 2 \times \angle ABC.

Hence:

\angle ABC = \frac{1}{2} \times 60^\circ = 30^\circ.

Finally, for arc $AB$ , which subtends $\angle AOB$ at the center and $\angle ACB$ on the circle:

\angle AOB = 2 \times \angle ACB.

Therefore:

\angle ACB = \frac{1}{2} \times 72^\circ = 36^\circ.

Hence, the angles of triangle $ABC$ are $\angle ABC = 30^\circ$ , $\angle ACB = 36^\circ$ , and $\angle BAC = 114^\circ$ .

Question 11

In the given figure, BD is a side of regular hexagon, DC is a side of a regular pentagon and AD is a diameter. Calculate :

(i) ∠ADC,

(ii) ∠BDA,

(iii) ∠ABC,

(iv) ∠AEC.

Answer:

Consider point O as the center of the circle.

Connect the points BC, BO, CO, and EO.

Since BD is a side of a regular hexagon,

∴ ∠BOD = $\dfrac{360°}{6}$ = 60°.

Since DC is a side of a regular pentagon,

∴ ∠COD = $\dfrac{360°}{5}$ = 72°.

In △BOD,

OB = OD [Both are radii of the circle]

∴ ∠OBD = ∠ODB = x (assume) [Angles opposite to equal sides are equal]

⇒ ∠OBD + ∠ODB + ∠BOD = 180°

⇒ x + x + 60° = 180°

⇒ 2x = 180° – 60°

⇒ 2x = 120°

⇒ x = $\dfrac{120°}{2}$ = 60°.

(i) In △OCD,

⇒ OD = OC [Both are radii of the circle]

⇒ ∠ODC = ∠OCD = y (assume) [Angles opposite to equal sides are equal]

⇒ ∠OCD + ∠ODC + ∠COD = 180°

⇒ y + y + 72° = 180°

⇒ 2y = 180° – 72°

⇒ 2y = 108°

⇒ y = $\dfrac{108°}{2}$ = 54°.

From the diagram,

∠ADC = ∠ODC = y = 54°.

Hence, ∠ADC = 54°.

(ii) Observing the diagram,

∠BDA = ∠BDO = 60°.

Hence, ∠BDA = 60°.

(iii) Recall that the angle at the center is twice the angle at the circumference.

Arc AC forms ∠AOC at the center and ∠ABC at the remaining circumference.

∴ ∠AOC = 2∠ABC

⇒ ∠ABC = $\dfrac{1}{2}$ ∠AOC

⇒ ∠ABC = $\dfrac{1}{2}$ [∠AOD – ∠COD]

⇒ ∠ABC = $\dfrac{1}{2}$ [180° – 72°]

⇒ ∠ABC = $\dfrac{1}{2} \times 108°$

⇒ ∠ABC = 54°.

Hence, ∠ABC = 54°.

(iv) Considering the cyclic quadrilateral AECD,

⇒ ∠AEC + ∠ADC = 180°

⇒ ∠AEC + 54° = 180°

⇒ ∠AEC = 180° – 54°

⇒ ∠AEC = 126°.

Hence, ∠AEC = 126°.

Test Yourself

Question 1(a)

In the given figure x°, y°, z° and p° are exterior angles of cyclic quadrilateral ABCD, then x° + y° + z° + p° is :

180°
270°
360°
720°

Answer:

Recall that for any cyclic quadrilateral, the exterior angle is equal to the opposite interior angle. In the cyclic quadrilateral ABCD, the angles are related as follows: ∠C = x°, ∠B = p°, ∠D = y°, and ∠A = z°. Thus, we can express the sum of the exterior angles as:

∴ x° + y° + z° + p° = ∠C + ∠D + ∠A + ∠B

It’s known that the total of the interior angles in a cyclic quadrilateral is 360°.

∴ ∠A + ∠B + ∠C + ∠D = 360°.

Hence, Option 3 is the correct option.

Question 1(b)

In the given figure, O is center of the circle and OABC is a rhombus, then :

x° + y° = 180°
x° = y° = 90°
x° + 2y° = 360°
x° = y° = 45°

Answer:

Connect OB to form triangle OAB.

Observe the diagram:

OB = OA (Both are radii of the circle) ……..(1)

Recall that all sides of a rhombus are equal.

∴ OA = AB ………….(2)

Combining equations (1) and (2), we have:

⇒ OA = OB = AB

Thus, OAB forms an equilateral triangle.

In a rhombus, diagonals bisect the angles at the vertices.

In △OAB:

∠AOB = $\dfrac{x}{2}$

∠OBA = $\dfrac{y}{2}$

Given that every angle in an equilateral triangle measures 60°.

∴ ∠AOB = 60°

⇒ $\dfrac{x}{2} = 60°$

⇒ x = 120°.

Similarly, ∠OBA = 60°

⇒ $\dfrac{y}{2} = 60°$

⇒ y = 120°.

Now, substitute x and y into the equation x° + 2y° = 360°:

⇒ 120° + 2(120°)

⇒ 120° + 240°

⇒ 360°.

Since L.H.S. equals R.H.S., Option 3 is the correct option.

Question 1(c)

Arcs AB and BC are of lengths in the ratio 11 : 4 and O is center of the circle. If angle BOC = 32°, the angle AOB is :

64°
88°
128°
132°

Answer:

We know that the lengths of arcs AB and BC are given in the ratio 11 : 4. This implies that the corresponding angles subtended at the center, ∠AOB and ∠BOC, must also be in the same ratio, 11 : 4.

Given that ∠BOC is 32°, we can set up the relationship:

∠AOB : 32° = 11 : 4

This means:

\dfrac{∠AOB}{32°} = \dfrac{11}{4}

Solving for ∠AOB, we multiply both sides by 32°:

∠AOB = $\dfrac{11}{4} \times 32°$

Calculating this gives us:

∠AOB = 88°.

Hence, Option 2 is the correct option.

Question 1(d)

In the given figure, AB is the side of regular pentagon and BC is the side of regular hexagon. Angle BAC is :

132°
66°
90°
120°

Answer:

Consider that AB represents a side of a regular pentagon. Therefore, the central angle ∠AOB is calculated as:

∴ ∠AOB = $\dfrac{360°}{5}$ = 72°.

Similarly, since BC is a side of a regular hexagon, the central angle ∠BOC is:

∴ ∠BOC = $\dfrac{360°}{6}$ = 60°.

From the diagram, it is clear that:

∠AOC = ∠AOB + ∠BOC = 72° + 60° = 132°.

According to the property of circles, the angle subtended by an arc at the center of the circle is twice the angle it subtends at any point on the remaining part of the circle. Thus, we have:

⇒ ∠AOC = 2∠APC

This gives us:

⇒ ∠APC = $\dfrac{∠AOC}{2} = \dfrac{132°}{2}$ = 66°.

Hence, Option 2 is the correct option.

Question 1(e)

In the given figure, O is center of the circle. Chord BC = chord CD and angle A = 80°. Angle BOC is :

120°
80°
100°
160°

Answer:

We have the chords BC and CD being equal in length.

∴ ∠BOC = ∠COD = x (let’s assume)

Observing the figure,

∠BOD = ∠BOC + ∠COD = x + x = 2x

According to the property of circles, the angle subtended by an arc at the center is twice the angle it subtends at any point on the circle’s remaining part.

⇒ ∠BOD = 2∠BAD

⇒ 2x = 2 × 80°

⇒ x = 80°.

Hence, Option 2 is the correct option.

Question 1(f)

In the given circle, ∠BAD = 95°, ∠ABD = 40° and ∠BDC = 45°.

Assertion (A) : To show that AC is a diameter, the angle ADC or angle ABC need to be proved to be 90°.

Reason (R) : In △ADB,

∠ADB = 180° – 95° – 40° = 45°

∴ Angle ADC = 45° + 45° = 90°

A is true, R is false.
A is false, R is true.
Both A and R are true and R is correct reason for A.
Both A and R are true and R is incorrect reason for A.

Answer:

Consider the property that an angle inscribed in a semicircle is always a right angle. Therefore, if AC is indeed the diameter, then both ∠ADC and ∠ABC should measure 90°.

Thus, the assertion (A) holds true.

Examining △ADB, we apply the angle sum property of triangles:

∠ADB + ∠DBA + ∠BAD = 180°

Substituting the given angles:

∠ADB + 40° + 95° = 180°

Simplifying gives:

∠ADB + 135° = 180°

This results in:

∠ADB = 180° – 135° = 45°.

Now, observe that:

∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°.

Thus, the reason (R) is also true.

Hence, Option 3 is the correct option.

Question 1(g)

ABCD is a cyclic quadrilateral, BD and AC are its diameters. Also, ∠DBC = 50°.

Assertion (A) : ∠BAC = 40°.

Reason (R) : ∠BAC = ∠BDC = 180° – (50° + 90°) = 40°.

(a) A is true, R is false.
(b) A is false, R is true.
(c) Both A and R are true and R is correct reason for A.
(d) Both A and R are true and R is incorrect reason for A.

Answer: (c) Both A and R are true and R is correct reason for A.

Since both BD and AC are diameters, it follows that:

∠ABC and ∠BCD are right angles because angles in a semicircle measure 90°.

In the triangle DBC, applying the angle sum property:

∴ ∠DBC + ∠BCD + ∠BDC = 180°

⇒ 50° + 90° + ∠BDC = 180°

⇒ 140° + ∠BDC = 180°

⇒ ∠BDC = 180° – 140°

⇒ ∠BDC = 40°

Since angles subtended by the same arc in a circle are equal:

⇒ ∠BAC = ∠BDC

⇒ ∠BAC = 40°

Thus, both the assertion and the reason are correct, and the reason correctly supports the assertion.

Hence, option 3 is the correct option.

Question 1(h)

Points A, C, B and D are concyclic, AB is diameter and ∠ABC = 60°.

Assertion (A) : ∠BAC = 60°.

Reason (R) : AB is diameter so ∠ACB = 90° and ∠ABC + ∠BAC = 90°.

(a) A is true, R is false.
(b) A is false, R is true.
(c) Both A and R are true and R is correct reason for A.
(d) Both A and R are true and R is incorrect reason for A.

Answer: (b) A is false, R is true.

Draw a line segment AC.

Notice that since AB is the diameter, we have:

∠ACB = 90° (because the angle subtended by a diameter on the circle is always a right angle).

In △ ABC, applying the angle sum property, we get:

∴ ∠ABC + ∠ACB + ∠BAC = 180° ………………..(1)

Substituting the known values:

⇒ 60° + 90° + ∠BAC = 180°

⇒ 150° + ∠BAC = 180°

⇒ ∠BAC = 180° – 150°

⇒ ∠BAC = 30°

Thus, the assertion (A) is false.

From equation (1), we can also deduce:

⇒ ∠ABC + 90° + ∠BAC = 180°

⇒ ∠ABC + ∠BAC = 180° – 90°

⇒ ∠ABC + ∠BAC = 90°

Therefore, the reason (R) is true.

Hence, option 2 is the correct option.

Question 1(i)

AB is diameter of the circle and ∠ACD = 38°.

Assertion (A) : x = 38°.

Reason (R) : ∠ACB = 90°, x = ∠DCB = 90° – 38° = 52°.

(a) A is true, R is false.
(b) A is false, R is true.
(c) Both A and R are true and R is correct reason for A.
(d) Both A and R are true and R is incorrect reason for A.

Answer: (b) A is false, R is true.

Let’s connect DB and CB. Given that AB is the circle’s diameter, we know that the angle subtended by a diameter is a right angle.

∴ ∠ACB = 90°

According to the problem, ∠ACD = 38°. Therefore, we have:

⇒ ∠ACD + ∠DCB = 90°

Substituting the given angle:

⇒ 38° + ∠DCB = 90°

Solving for ∠DCB gives:

⇒ ∠DCB = 90° – 38°

⇒ ∠DCB = 52°

Since angles subtended by the same chord in the same segment of a circle are equal, we find:

⇒ ∠BAD (x) = ∠BCD = 52°

Thus, the assertion (A) is incorrect, while the reason (R) is correct.

Hence, option 2 is the correct option.

Question 1(j)

Chords AC and BD intersect each other at point P.

Assertion (A) : PA x PC = PB x PD.

Reason (R) : Δ APD ∼ Δ BPC

\Rightarrow \dfrac{PA}{PB} = \dfrac{PD}{PC}

(a) A is true, R is false.
(b) A is false, R is true.
(c) Both A and R are true and R is correct reason for A.
(d) Both A and R are true and R is incorrect reason for A.

Answer: (c) Both A and R are true and R is correct reason for A.

Consider triangles Δ APD and Δ BPC. Notice that ∠APD is equal to ∠BPC because vertically opposite angles are equal. Additionally, ∠ADP equals ∠BCP since angles in the same segment of a circle are equal.

∴ Δ APD is similar to Δ BPC by the Angle-Angle (A.A.) similarity criterion.

Because the triangles are similar, their corresponding sides are in proportion:

$\Rightarrow \dfrac{AP}{PB} = \dfrac{PD}{PC}$ ……..(1)

This confirms that the reason (R) is true.

By rearranging equation (1), we find:

⇒ AP \times PC = PD \times PB

Thus, the assertion (A) is also true, and the reason (R) correctly explains the assertion.

Option 3 is the correct option.

Question 1(k)

A circle with center at point O and ∠AOC = 160°.

Statement (1) : Angle x = 100° and angle y = 80°.

Statement (2) : The angle, which an arc of a circle subtends at the center of the circle is double the angle which it subtends at any point on the remaining part of the circumference.

(a) Both statements are true.
(b) Both statements are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.

Answer: (d) Statement 1 is false, and statement 2 is true.

Notice that the angle subtended by an arc at the center of a circle is twice the angle it subtends at any point on the remaining part of the circle. This confirms that statement 2 is correct.

∴ ∠AOC = 2x

⇒ 160° = 2x

⇒ x = $\dfrac{160°}{2}$ = 80°

Also, remember that in a cyclic quadrilateral, the sum of either pair of opposite angles is 180°.

∴ x + y = 180°

⇒ 80° + y = 180°

⇒ y = 180° – 80° = 100°

Thus, statement 1 is incorrect.

Hence, option 4 is the correct option.

Question 1(l)

AC is diameter, AE is parallel to BC and ∠BAC = 50°.

Statement (1) : ∠EDC + 50° = 180°.

Statement (2) : ∠EDC + ∠EAC = 180°.

(a) Both statements are true.
(b) Both statements are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.

Answer: (d) Statement 1 is false, and statement 2 is true.

Given that AC is the diameter, we know that angles subtended by a diameter in a semicircle are right angles.

∴ ∠ABC = 90°

Since AE is parallel to BC and AB acts as a transversal, the co-interior angles formed must sum up to 180°.

⇒ ∠ABC + ∠BAE = 180°

⇒ 90° + ∠BAE = 180°

⇒ ∠BAE = 180° – 90°

⇒ ∠BAE = 90°

This implies that ∠BAC + ∠EAC = 90°

⇒ 50° + ∠EAC = 90°

⇒ ∠EAC = 90° – 50°

⇒ ∠EAC = 40°

In the cyclic quadrilateral AEDC, the sum of opposite angles is always 180°.

⇒ ∠EDC + ∠EAC = 180°

⇒ ∠EDC + 40° = 180°

Thus, statement 1 is incorrect, while statement 2 holds true.

Hence, option 4 is the correct option.

Question 1(m)

O is the center of the circle, OB = BC and ∠BOC = 20°.

Statement (1) : x = 2 x 20° = 40°

Statement (2) : ∠BOC = 20°.

x = ∠OAB + 20° = ∠OBA + 20° = 40° + 20° = 60°

(a) Both statements are true.
(b) Both statements are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.

Answer: (d) Statement 1 is false, and statement 2 is true.

Consider the given conditions:

∴ OB = OC, which implies that the triangle OBC is isosceles.

∴ ∠BOC = ∠BCO = 20° (since angles opposite to equal sides are equal).

For △OBC, applying the angle sum property:

∠OBC + ∠BCO + ∠BOC = 180°

⇒ ∠OBC + 20° + 20° = 180°

⇒ ∠OBC + 40° = 180°

⇒ ∠OBC = 180° – 40°

⇒ ∠OBC = 140°

Since ∠OBC and ∠OBA form a linear pair:

∠OBC + ∠OBA = 180°

⇒ 140° + ∠OBA = 180°

⇒ ∠OBA = 180° – 140°

⇒ ∠OBA = 40°

Given OB = OA (as both are radii of the circle), we have:

∠OBA = ∠OAB = 40° (angles opposite to equal sides are equal).

Using the exterior angle property in triangle OAC, the exterior angle (∠EOA) equals the sum of the two opposite interior angles:

x = ∠OAB + 20°

x = ∠OBA + 20° = 40° + 20° = 60°

Thus, statement 1 is incorrect, while statement 2 is correct.

Hence, option 4 is the correct option.

Question 1(n)

O is the center of the circle and ∠AOC = 120°.

Statement (1) : ∠ABC = 120°

Statement (2) : ∠ABC + ∠ADC = 180° ⇒ ∠ABC + 60° = 180°.

(a) Both statements are true.
(b) Both statements are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.

Answer: (a) Both statements are true.

Notice that the angle subtended by an arc at the center of a circle is twice the angle it subtends at any point on the remaining part of the circle’s circumference.

∴ ∠AOC = 2 × ∠ADC

Given ∠AOC = 120°, we have:

⇒ 120° = 2 × ∠ADC

⇒ ∠ADC = $\dfrac{120°}{2}$ = 60°

Now, since ABCD is a cyclic quadrilateral, the sum of the opposite angles is 180°.

⇒ ∠ADC + ∠ABC = 180°

Substituting the value of ∠ADC, we get:

⇒ 60° + ∠ABC = 180°

⇒ ∠ABC = 180° – 60°

⇒ ∠ABC = 120°

Thus, both statements are indeed true.

Hence, option 1 is the correct option.

Question 2

In the given circle with diameter AB, find the value of x.

Answer:

Notice that angles subtended by the same arc in a circle are equal.

∴ ∠ABD = ∠ACD = 30°.

Looking at the figure, we see:

∠ADB = 90° because an angle formed in a semicircle is always a right angle.

In △ADB, we apply the angle sum property of triangles:

⇒ ∠ABD + ∠ADB + ∠BAD = 180°

Substituting the known values:

⇒ 30° + 90° + x = 180°

Adding the angles gives:

⇒ 120° + x = 180°

Solving for x, we find:

⇒ x = 180° – 120° = 60°.

Hence, the value of x = 60°.

Question 3

In the given figure, ABC is a triangle in which ∠BAC = 30°. Show that BC is equal to the radius of the circumcircle of the triangle ABC, whose centre is O.

Answer:

Connect points O to B and O to C.

We know the principle that the angle subtended at the center of a circle is twice the angle subtended at any point on the remaining circumference.

∴ ∠BOC = 2 × ∠BAC = 2 × 30° = 60°.

Consider triangle △OBC:

⇒ OB = OC, since these are radii of the same circle.

⇒ Let ∠OBC = ∠OCB = x, as angles opposite equal sides are equal.

Now, the sum of angles in △OBC is 180°:

⇒ ∠BOC + ∠OBC + ∠OCB = 180°

⇒ 60° + x + x = 180°

⇒ 2x + 60° = 180°

⇒ 2x = 180° – 60°

⇒ 2x = 120°

⇒ x = $\dfrac{120°}{2}$ = 60°.

Thus, ∠OBC = ∠OCB = ∠BOC = 60°.

This confirms that △OBC is an equilateral triangle.

∴ OB = OC = BC.

Therefore, BC is indeed equal to the radius of the circumcircle of triangle ABC.

Question 4

Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Answer:

Consider △ABC where AB = AC and a circle is drawn with AB as its diameter, intersecting BC at D.

From the diagram, we observe:

∠ADB = 90° because an angle subtended by a diameter in a semicircle is a right angle.

Now, notice:

⇒ ∠ADB + ∠ADC = 180°, as they form a linear pair.

⇒ ∠ADC = 180° – ∠ADB

⇒ ∠ADC = 180° – 90° = 90°.

In triangles △ABD and △ACD, we have:

⇒ AB = AC, which is given.

⇒ ∠ADB = ∠ADC = 90°.

⇒ AD is common to both triangles.

Thus, by the RHS (Right angle-Hypotenuse-Side) congruency criterion, △ABD ≅ △ACD.

Therefore, BD = CD due to C.P.C.T. (Corresponding Parts of Congruent Triangles).

Hence, point D is the midpoint of BC.

Hence, proved that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Question 5

In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.

Answer:

Let’s connect the points OE and AB in the diagram.

Notice that arc EC forms ∠EOC at the centre and ∠EBC on the circle’s circumference.

According to the circle’s properties, the angle at the centre is twice the size of the angle at the circumference.

∴ ∠EOC = 2 × ∠EBC = 2 × 65° = 130°.

Now, consider triangle OEC:

Since OE and OC are radii of the circle, they are equal.

∴ ∠OEC = ∠OCE (Angles opposite equal sides are equal.)

Applying the angle sum property in ∆OCE:

⇒ ∠OEC + ∠OCE + ∠EOC = 180°

⇒ 2∠OCE + 130° = 180°

⇒ 2∠OCE = 180° – 130°

⇒ 2∠OCE = 50°

⇒ ∠OCE = $\frac{50°}{2}$ = 25°.

Given that AC is parallel to ED, we have:

∴ ∠DEC = ∠OCE (Alternate angles are equal)

⇒ ∠DEC = 25°.

Hence, ∠DEC = 25°.

Question 6

In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate :

(i) ∠BDC

(ii) ∠BEC

(iii) ∠BAC

Answer:

(i) Since BD is the diameter, the angle subtended by it on the circle is a right angle. Thus, ∠BCD = 90°.

Given that ∠DBC = 58°, we can apply the angle sum property of a triangle in ∆BDC:

∠DBC + ∠BCD + ∠BDC = 180°

Substituting the known values:

58° + 90° + ∠BDC = 180°

⇒ 148° + ∠BDC = 180°

⇒ ∠BDC = 180° – 148° = 32°.

Hence, ∠BDC = 32°.

(ii) In the cyclic quadrilateral BECD, the opposite angles are supplementary. Therefore:

∠BEC + ∠BDC = 180°

Substituting the value of ∠BDC:

∠BEC + 32° = 180°

⇒ ∠BEC = 180° – 32° = 148°

Hence, ∠BEC = 148°.

(iii) Considering the cyclic quadrilateral ABEC, the opposite angles also sum up to 180°:

∠BAC + ∠BEC = 180°

Substituting the value of ∠BEC:

∠BAC + 148° = 180°

⇒ ∠BAC = 180° – 148° = 32°.

Hence, ∠BAC = 32°.

Question 7

D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.

Answer:

Consider joining the line segment DE.

In the isosceles triangle ∆ABC, we are given that the sides AB and AC are equal, denoted by AB = AC = x. This implies that the angles opposite these equal sides are also equal, so ∠B = ∠C.

Now, observe the smaller triangle ∆ADE, where AD = AE = y. This equality indicates that the angles opposite these sides are equal, thus ∠ADE = ∠AED.

In ∆ABC, the ratios of corresponding sides are equal: $\dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{y}{x}$ . This ratio confirms that DE is parallel to BC by the converse of the Basic Proportionality Theorem (BPT).

Given that DE || BC, it follows that:

⇒ ∠ADE = ∠B (since corresponding angles are equal)

⇒ (180° – ∠EDB) = ∠B

⇒ ∠B + ∠EDB = 180°

Since we have established that ∠B = ∠C, it follows that:

⇒ ∠C + ∠EDB = 180°

This shows that opposite angles are supplementary.

Similarly, we find that:

⇒ ∠B + ∠CED = 180°

The sum of opposite angles in a cyclic quadrilateral equals 180°.

Hence, proved that B, C, E, and D are concyclic.

Question 8

In the given figure, ABCD is a cyclic quadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ADC = 92°, ∠FAE = 20°; determine ∠BCD. Given reason in support of your answer.

Answer:

Given that ABCD is a cyclic quadrilateral, we know the sum of the opposite angles must be 180°. Thus, ∠B + ∠D = 180°.

Since ∠ADC = 92°, we have:

⇒ ∠B + 92° = 180°

⇒ ∠B = 180° – 92° = 88°

With AF parallel to CB, alternate angles are equal, so ∠FAB = ∠B = 88°.

However, it is given that ∠FAE = 20°.

From the figure, we can find ∠BAE as follows:

∠BAE = ∠BAF + ∠FAE = 88° + 20° = 108°.

Now, to find ∠BAD, we use:

∠BAD = 180° – ∠BAE = 180° – 108° = 72°.

For the cyclic quadrilateral ABCD, the opposite angles also satisfy:

∠BCD + ∠BAD = 180°

⇒ ∠BCD + 72° = 180°

⇒ ∠BCD = 180° – 72° = 108°.

Hence, ∠BCD = 108°.

Question 9

If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66° and ∠ABC = 80°. Calculate :

(i) ∠DBC,

(ii) ∠IBC,

(iii) ∠BIC.

Answer:

To solve this, we first connect DB, DC, IB, and IC.

(i) Observing the diagram:

∠DAC is given by $\dfrac{1}{2}$ ∠BAC since I is the incenter.

∠DAC = $\dfrac{1}{2}$ x 66° = 33°.

Angles in the same segment are equal, thus:

∴ ∠DBC = ∠DAC = 33°.

Therefore, ∠DBC = 33°.

(ii) Since I is the incenter of ∆ABC, IB divides ∠ABC into two equal parts.

∴ ∠IBC = $\dfrac{1}{2}$ ∠ABC

= $\dfrac{1}{2} \times 80°$ = 40°.

Thus, ∠IBC = 40°.

(iii) In ∆ABC, we have:

⇒ ∠ACB + ∠ABC + ∠BAC = 180° by the angle sum property.

⇒ ∠ACB = 180° – ∠ABC – ∠BAC

⇒ ∠ACB = 180° – 80° – 66°

⇒ ∠ACB = 180° – 146°

⇒ ∠ACB = 34°.

Since I is the incenter, IC divides ∠C into two equal parts.

∴ ∠ICB = $\dfrac{1}{2}$ ∠ACB = $\dfrac{1}{2} \times 34°$ = 17°.

In ∆IBC, applying the angle sum property:

⇒ ∠IBC + ∠ICB + ∠BIC = 180°

⇒ 40° + 17° + ∠BIC = 180°

⇒ 57° + ∠BIC = 180°

⇒ ∠BIC = 180° – 57° = 123°.

Thus, ∠BIC = 123°.

Question 10

In the given Figure, AB = AD = DC = PB and ∠DBC = x°. Determine, in terms of x :

(i) ∠ABD, (ii) ∠APB.

Hence or otherwise, prove that AP is parallel to DB.

Answer:

(i) Draw lines AC and BD.

Notice that $\angle DAC = \angle DBC = x^\circ$ because angles in the same segment of a circle are equal.

Also, $\angle DCA = \angle DAC = x^\circ$ since they are opposite angles to equal sides.

Therefore, $\angle ABD = \angle DCA = x^\circ$ as they are angles in the same segment.

Thus, $\angle ABD = x^\circ$ .

(ii) Consider triangle $\triangle ABP$ .

The exterior angle $\angle ABC = \angle BAP + \angle APB$ .

Given that $\angle BAP = \angle APB$ because they are angles opposite to equal sides, we have:

2x^\circ = \angle APB + \angle APB = 2\angle APB

2\angle APB = 2x^\circ

\angle APB = x^\circ

Therefore, $\angle APB = x^\circ$ .

Since $\angle APB = \angle DBC = x^\circ$ , these are corresponding angles.

This proves that AP is parallel to DB.

Question 11

In the given figure; ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠CQE are supplementary.

Answer:

Connect point E to point B.

Recall that in a cyclic quadrilateral, the sum of opposite angles equals 180°.

Consider the cyclic quadrilateral ABEP:

∴ ∠APE + ∠ABE = 180° …..(1) [Opposite angles in a cyclic quadrilateral are supplementary]

Now, look at the cyclic quadrilateral BCQE:

∴ ∠CQE + ∠CBE = 180° …..(2)

By adding equations (1) and (2), we obtain:

⇒ ∠APE + ∠ABE + ∠CQE + ∠CBE = 180° + 180°

⇒ ∠APE + ∠ABE + ∠CQE + ∠CBE = 360° …..(3)

From the diagram, notice:

∠ABE + ∠CBE = 180° [They form a linear pair]

Substituting this into equation (3), we have:

∠APE + ∠CQE + 180° = 360°

⇒ ∠APE + ∠CQE = 360° – 180°

⇒ ∠APE + ∠CQE = 180°

Thus, it is established that ∠APE and ∠CQE are supplementary.

Question 12

In the given figure, AB is the diameter of the circle with centre O.

If ∠ADC = 32°, find angle BOC.

Answer:

Recall that the angle formed at the centre of a circle is twice the angle formed at any point on the remaining circumference by the same arc. Here, arc AC creates ∠AOC at the centre and ∠ADC on the circle’s boundary.

⇒ ∠AOC = 2 × ∠ADC

⇒ ∠AOC = 2 × 32° = 64°

Now, observe that ∠AOC and ∠BOC are a linear pair, meaning they are supplementary and together add up to 180°.

∴ ∠AOC + ∠BOC = 180°

⇒ 64° + ∠BOC = 180°

⇒ ∠BOC = 180° – 64° = 116°.

Hence, ∠BOC = 116°.

Question 13

In a cyclic quadrilateral PQRS, angle PQR = 135°. Sides SP and RQ produced meet at point A whereas sides PQ and SR produced meet at point B.

If ∠A : ∠B = 2 : 1, find angles A and B.

Answer:

Consider ∠A as 2x and ∠B as x.

In the cyclic quadrilateral PQRS, we have:

∴ ∠PSR + ∠PQR = 180°
⇒ ∠PSR + 135° = 180°
⇒ ∠PSR = 180° – 135° = 45°.

From the diagram, observe:

∴ ∠PQR + ∠PQA = 180° (since they form a linear pair)
⇒ ∠PQA + 135° = 180°
⇒ ∠PQA = 180° – 135° = 45°.

Now, in triangle PBS:

∴ ∠BSP + ∠BPS + ∠PBS = 180°
⇒ ∠PSR + ∠BPS + ∠B = 180°
⇒ ∠BPS + 45° + x = 180°
⇒ ∠BPS = 180° – 45° – x
⇒ ∠BPS = 135° – x \quad \text{…(1)}

Recall that an exterior angle of a triangle equals the sum of the two opposite interior angles. Thus, in triangle PQA:

∴ ∠BPS = ∠PQA + ∠A
⇒ ∠BPS = 45° + 2x \quad \text{…(2)}

By equating the two expressions for ∠BPS from equations (1) and (2), we have:

⇒ 135° – x = 45° + 2x
⇒ 2x + x = 135° – 45°
⇒ 3x = 90°
⇒ x = \dfrac{90°}{3} = 30°.

Thus, ∠A = 2x = 2 \times 30° = 60°
And ∠B = x = 30°.

Hence, ∠A = 60° and ∠B = 30°.

Question 14

In the following figure, ABCD is a cyclic quadrilateral in which AD is parallel to BC. If the bisector of angle A meets BC at point E and the given circle at point F, prove that :

(i) EF = FC

(ii) BF = DF

Answer:

Since $AF$ is the angle bisector of ∠A, we have ∠BAF = ∠DAF.

Thus, ∠DAE = ∠BAE ……..(1)

Now, since ∠DAE = ∠AEB ……..(2) [Alternate angles are equal],

From equations (1) and (2), it follows that ∠BAE = ∠AEB.

(i) Consider △ABE,

⇒ ∠BAE + ∠AEB + ∠ABE = 180°

⇒ ∠AEB + ∠AEB + ∠ABE = 180°

⇒ 2∠AEB + ∠ABE = 180°

⇒ ∠ABE = 180° – 2∠AEB

Since ABCD is a cyclic quadrilateral, the sum of opposite angles in a cyclic quadrilateral is 180°.

∴ ∠ABC + ∠ADC = 180°

⇒ ∠ABE + ∠ADC = 180°

⇒ 180° – 2∠AEB + ∠ADC = 180°

⇒ ∠ADC = 180° – 180° + 2∠AEB

⇒ ∠ADC = 2∠AEB ……..(3)

Given that ADCF is also a cyclic quadrilateral,

∴ ∠AFC + ∠ADC = 180°

⇒ ∠AFC = 180° – ∠ADC

⇒ ∠AFC = 180° – 2∠AEB [Using (3)]

In △ECF,

⇒ ∠EFC + ∠ECF + ∠FEC = 180°

⇒ ∠AFC + ∠ECF + ∠FEC = 180° [From the figure, ∠EFC = ∠AFC]

⇒ ∠AFC + ∠ECF + ∠AEB = 180° [Since ∠FEC = ∠AEB (Vertically opposite angles are equal)]

⇒ ∠ECF = 180° – (∠AFC + ∠AEB)

⇒ ∠ECF = 180° – (180° – 2∠AEB + ∠AEB)

⇒ ∠ECF = ∠AEB

⇒ ∠ECF = ∠FEC

∴ EF = FC [As sides opposite to equal angles are equal.]

Hence, proved that EF = FC.

(ii) Since $AF$ is the angle bisector of ∠A,

∴ ∠BAF = ∠DAF

∴ arc BF = arc DF [As equal arcs subtend equal angles]

∴ BF = DF [Equal arcs have equal chords]

Hence, proved that BF = DF.

Question 15

ABCD is a cyclic quadrilateral. Sides AB and DC produced meet at point E; whereas sides BC and AD produced meet at point F. If ∠DCF : ∠F : ∠E = 3 : 5 : 4, find the angles of the cyclic quadrilateral ABCD.

Answer:

Assume ∠DCF = 3x, ∠F = 5x, and ∠E = 4x.

Notice that in a cyclic quadrilateral, the sum of opposite angles is always 180°.

∴ ∠A + ∠DCB = 180°

⇒ ∠A = 180° – ∠DCB

Since ∠DCF and ∠DCB are a linear pair, we have:

⇒ ∠A = 180° – (180° – ∠DCF)

⇒ ∠A = ∠DCF = 3x.

Consider △CDF,

∠CDA is the exterior angle, thus:

∠CDA = ∠F + ∠DCF

= 5x + 3x = 8x

Now, in △BCE,

The exterior angle ∠ABC is equal to the sum of the two opposite interior angles:

⇒ ∠ABC = ∠BCE + ∠E

Since ∠BCE = ∠DCF (as vertically opposite angles are equal), we have:

⇒ ∠ABC = 3x + 4x = 7x.

For the cyclic quadrilateral ABCD, the sum of opposite angles is:

⇒ ∠ABC + ∠CDA = 180°

⇒ 7x + 8x = 180°

⇒ 15x = 180°

Solving for x, we get:

⇒ x = $\dfrac{180°}{15}$ = 12°.

Thus, the angles of the quadrilateral are:

∠A = 3x = 3 \times 12° = 36°,

∠B = 7x = 7 \times 12° = 84°,

∠C = 180° – ∠A = 180° – 36° = 144°,

∠D = 8x = 8 \times 12° = 96°.

Hence, ∠A = 36°, ∠B = 84°, ∠C = 144°, ∠D = 96°.

Question 16

In the given figure, AB is the diameter of a circle with center O. If chord AC = chord AD, prove that :

(i) arc BC = arc DB

(ii) AB is the bisector of ∠CAD.

Further, if the length of arc AC is twice the length of arc BC, find :

(a) ∠BAC

(b) ∠ABC

Answer:

To solve this, let’s connect BC and BD.

In triangles △ABC and △ABD, observe the following:

⇒ AC = AD [This is given]

⇒ AB = AB [This is common to both triangles]

⇒ ∠ACB = ∠ADB [Each is 90°, as angles in a semicircle are right angles]

∴ △ABC ≅ △ABD [By the RHS (Right angle-Hypotenuse-Side) congruency criterion]

(i) From the congruency of △ABC and △ABD:

∴ BC = BD [Corresponding Parts of Congruent Triangles (C.P.C.T.)]

∴ Arc BC = Arc BD [Since equal chords subtend equal arcs]

Thus, arc BC is equal to arc BD.

(ii) Again, from the congruency of △ABC and △ABD:

∴ ∠BAC = ∠BAD [By C.P.C.T.]

∴ AB bisects ∠CAD.

Thus, AB is the bisector of ∠CAD.

For the additional conditions:

(a) Given that arc AC is twice arc BC, we have:

∴ ∠ABC = 2∠BAC

Also, note:

⇒ ∠ABC + ∠BAC = 90° [Since △ABC is right-angled at C]

⇒ 2∠BAC + ∠BAC = 90°

⇒ 3∠BAC = 90°

⇒ ∠BAC = $\dfrac{90°}{3}$ = 30°.

Therefore, ∠BAC = 30°.

(b) Now, calculate ∠ABC:

∠ABC = 2∠BAC = 2 \times 30° = 60°.

Therefore, ∠ABC = 60°.

Question 17

In cyclic quadrilateral ABCD; AD = BC, ∠BAC = 30° and ∠CBD = 70°; find :

(i) ∠BCD

(ii) ∠BCA

(iii) ∠ABC

(iv) ∠ADC

Answer:

(i) Let’s start by observing that in the cyclic quadrilateral ABCD, since AD = BC, the angles in the same segment are equal. Therefore, ∠DAC = ∠CBD = 70°. Now, calculate ∠BAD as follows:

∠BAD = ∠BAC + ∠DAC = 30° + 70° = 100°.

In a cyclic quadrilateral, the sum of the opposite angles is 180°.

∴ ∠BAD + ∠BCD = 180°
⇒ 100° + ∠BCD = 180°
⇒ ∠BCD = 180° – 100° = 80°.

Hence, ∠BCD = 80°.

(ii) Given AD = BC, equal chords subtend equal angles, thus ∠ACD = ∠BDC. Also, angles in the same segment are equal, so ∠ACB = ∠ADB.

Adding these, we have:

∠ACD + ∠ACB = ∠BDC + ∠ADB
⇒ ∠BCD = ∠ADC = 80°.

In triangle BCD, the sum of angles is 180°:

∠CBD + ∠BCD + ∠BDC = 180°
⇒ 70° + 80° + ∠BDC = 180°
⇒ 150° + ∠BDC = 180°
⇒ ∠BDC = 180° – 150° = 30°.

Since equal chords subtend equal angles, ∠ACD = ∠BDC = 30°.

Now, calculate ∠BCA:

∠BCA = ∠BCD – ∠ACD = 80° – 30° = 50°.

Hence, ∠BCA = 50°.

(iii) In a cyclic quadrilateral, opposite angles add up to 180°:

∠ADC + ∠ABC = 180°
⇒ 80° + ∠ABC = 180°
⇒ ∠ABC = 180° – 80° = 100°.

Hence, ∠ABC = 100°.

(iv) From part (ii), we have already established that:

Hence, ∠ADC = 80°.

Question 18

In the given figure, ∠ACE = 43° and ∠CAF = 62°; find the values of a, b and c.

Answer:

Consider the triangle △AEC. According to the angle sum property, we have:

∠ACE + ∠CAE + ∠AEC = 180°

Substituting the given values, we find:

43° + 62° + ∠AEC = 180°

Solving for ∠AEC, we get:

∠AEC = 180° – 105° = 75°.

Now, observe the cyclic quadrilateral in the figure. The sum of opposite angles in a cyclic quadrilateral is 180°:

∠ABD + ∠AED = 180°

From the figure, we know ∠AED = ∠AEC, so:

a + 75° = 180°

Thus, solving for a gives:

a = 180° – 75° = 105°.

Next, for the vertically opposite angles, we have:

∠BDC = c

And for the linear pair, we know:

∠DBC = 180° – a

Substituting the value of a:

= 180° – 105°
= 75°.

In triangle △DBC, applying the angle sum property:

∠DBC + ∠BCD + ∠BDC = 180°

Substituting the known values:

75° + 43° + c = 180°

This simplifies to:

118° + c = 180°

Therefore, solving for c gives:

c = 180° – 118° = 62°.

Finally, consider triangle △BAF. Again using the angle sum property:

∠ABF + ∠BAF + ∠AFB = 180°

Substituting the known angles:

a + 62° + b = 180°

With a = 105°, we have:

105° + 62° + b = 180°

Solving for b results in:

b + 167° = 180°

Hence, b = 180° – 167° = 13°.

Therefore, the values are a = 105°, b = 13°, and c = 62°.

Question 19

In the given figure, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°.

Find :

(i) ∠CAD

(ii) ∠CBD

(iii) ∠ADC

Answer:

(i) Recall that in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle. Here, ∠BAD is the exterior angle, which equals ∠BCE = 80°.

From the figure, we have:

∠CAD = ∠BAD – ∠BAC = 80° – 25° = 55°.

Hence, ∠CAD = 55°.

(ii) Notice that angles in the same segment of a circle are equal. Therefore, ∠CBD is equal to ∠CAD.

∴ ∠CBD = 55°.

Hence, ∠CBD = 55°.

(iii) Again, angles in the same segment are equal, so ∠BDC = ∠BAC = 25°.

Since AB is parallel to DC and BD acts as a transversal, alternate angles are equal. Therefore, ∠ABD = ∠BDC = 25°.

From the figure:

∠ABC = ∠ABD + ∠CBD = 25° + 55° = 80°.

In cyclic quadrilateral ABCD, the sum of the opposite angles is 180°.

⇒ ∠ABC + ∠ADC = 180°

⇒ 80° + ∠ADC = 180°

⇒ ∠ADC = 180° – 80° = 100°.

Hence, ∠ADC = 100°.

Question 20

In the figure, given below, CP bisects angle ACB. Show that DP bisects angle ADB.

Answer:

Given that CP bisects ∠ACB, it follows that ∠ACP = ∠BCP. $\because$ CP is the angle bisector.

Now, using the property that angles subtended by the same segment are equal, we have:

∠ACP = ∠ADP \quad \text{…(2)}

Similarly, ∠BCP = ∠BDP \quad \text{…(3)}

From equations (1) and (2), we deduce:

∠BCP = ∠ADP \quad \text{…(4)}

Combining equations (3) and (4), we find:

∠ADP = ∠BDP.

∴ DP divides ∠ADB into two equal parts, confirming that DP is indeed the bisector of ∠ADB.

Hence, proved that DP is the bisector of ∠ADB.

Question 21

In the figure, given below, AD = BC, ∠BAC = 30° and ∠CBD = 70°. Find :

(i) ∠BCD

(ii) ∠BCA

(iii) ∠ABC

(iv) ∠ADB

Answer:

Given that angles in the same segment of a circle are equal, we have:

∴ ∠CAD = ∠CBD = 70°.

Looking at the figure:

∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°.

(i) The sum of opposite angles in a cyclic quadrilateral is 180°.

In the cyclic quadrilateral ABCD:

⇒ ∠BCD + ∠BAD = 180°

⇒ ∠BCD + 100° = 180°

⇒ ∠BCD = 180° – 100° = 80°.

Hence, ∠BCD = 80°.

(ii) Since AD = BC, ABCD forms an isosceles trapezium and AB is parallel to DC.

∠DCA = ∠BAC = 30° (Alternate angles)

From the figure:

∠BCA = ∠BCD – ∠DCA = 80° – 30° = 50°.

Hence, ∠BCA = 50°.

(iii) Recognizing that angles in the same segment are equal:

∠ABD = ∠DCA = 30°

From the figure:

∠ABC = ∠ABD + ∠CBD = 30° + 70° = 100°.

Hence, ∠ABC = 100°.

(iv) Again, considering angles in the same segment are equal:

∠ADB = ∠BCA = 50°.

Hence, ∠ADB = 50°.

Question 22

In the given figure, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°. Find :

(i) ∠OBD

(ii) ∠AOB

(iii) ∠BED

Answer:

(i) We know that AD is parallel to BC.

∴ OD is also parallel to BC and BD acts as a transversal.

Using the property of alternate angles, ∠ODB is equal to ∠CBD, which is 32°.

In the triangle OBD, since OB and OD are radii of the same circle, they are equal.

Thus, ∠OBD is equal to ∠ODB, both being 32°.

Hence, ∠OBD = 32°.

(ii) As AD is parallel to BC, AO is also parallel to BC, and OB serves as a transversal.

This makes ∠AOB equal to ∠OBC due to alternate angles.

From the figure, we find:

∠OBC = ∠OBD + ∠DBC = 32° + 32° = 64°.

∴ ∠AOB = 64°.

Hence, ∠AOB = 64°.

(iii) In triangle OAB, since OA and OB are radii of the same circle, they are equal.

Let ∠OAB = ∠OBA = x.

According to the angle sum property of a triangle:

x + x + 64° = 180°

⇒ 2x = 180° – 64°

⇒ 2x = 116°

⇒ x = $\dfrac{116°}{2}$ = 58°

Thus, ∠OAB = 58°.

From the figure, ∠DAB is equal to ∠OAB, which is 58°.

Since angles in the same segment are equal, ∠BED is also equal to ∠DAB.

Therefore, ∠BED = 58°.

Hence, ∠BED = 58°.

Question 23

In the figure given, O is the centre of the circle. ∠DAE = 70°. Find giving suitable reasons, the measure of

(i) ∠BCD

(ii) ∠BOD

(iii) ∠OBD

Answer:

(i) Given that ∠DAE = 70°, notice that ∠DAE and ∠BAD form a linear pair. This implies:

∠DAE + ∠BAD = 180°
⇒ 70° + ∠BAD = 180°
⇒ ∠BAD = 180° – 70° = 110°.

Since the sum of opposite angles in a cyclic quadrilateral equals 180°, we have:

∠BCD + ∠BAD = 180°
⇒ ∠BCD + 110° = 180°
⇒ ∠BCD = 180° – 110° = 70°.

Thus, ∠BCD = 70°.

(ii) The angle subtended by an arc at the center is twice the angle it subtends on the circle’s circumference. Therefore:

∠BOD = 2∠BCD = 2 × 70° = 140°.

Thus, ∠BOD = 140°.

(iii) In △OBD, since OB = OD (both being radii of the circle), it follows that ∠OBD = ∠ODB = x.

Thus, in △OBD:

∠OBD + ∠ODB + ∠BOD = 180°
⇒ x + x + 140° = 180°
⇒ 2x = 180° – 140°
⇒ 2x = 40°
⇒ x = $\dfrac{40°}{2}$ = 20°.

∴ ∠OBD = 20°.

Therefore, ∠OBD = 20°.

Frequently Asked Questions

A secant is a line that intersects a circle at two distinct points. A tangent is a line that touches the circle at exactly one point, known as the point of tangency. Essentially, a tangent is a special case of a secant where the two intersection points coincide.

A cyclic quadrilateral is a four-sided figure whose vertices all lie on a single circle. Its key properties are that the sum of opposite angles is always 180 degrees (e.g., ∠A + ∠C = 180°), and the exterior angle at any vertex is equal to the interior opposite angle.

To prove this theorem, you consider a point on the tangent other than the point of contact. The distance from the center to this external point will always be greater than the radius, making the radius the shortest distance from the center to the tangent line. The shortest distance from a point to a line is the perpendicular distance.

This page provides complete, step-by-step solutions for all 110 questions from the Circles chapter in the Selina Concise Mathematics textbook. This includes all problems from Exercise 17(A), Exercise 17(B), Exercise 17(C), and the final Test Yourself section.

ICSE Board

Exercise 17(A)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3(i)

Question 3(ii)

Question 3(iii)

Question 3(iv)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Exercise 17(B)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6(a)

Question 6(b)

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24

Question 25

Question 26

Question 27

Question 28

Exercise 17(C)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Test Yourself

Question 1(a)