Concise Mathematics Selina Solutions Class 10 ICSE Chapter 18 Tangents and Intersecting Chords

This page provides the most reliable ICSE Class 10 Tangents and Intersecting Chords Solutions from the Selina Concise Mathematics textbook. In this chapter, we move beyond basic circle properties to explore the dynamic relationship between circles and lines. You will learn about key concepts such as tangents, secants, and the properties that arise when chords intersect inside or outside a circle. We will apply important theorems like the Alternate Segment Theorem and the Tangent-Secant Theorem to solve a variety of geometrical problems. Mastering these concepts is crucial as they form the foundation for more advanced topics in coordinate geometry and are frequently tested in board examinations.

If you are stuck on a specific proof or calculation in Chapter 18, you have come to the right place. This page contains detailed, step-by-step solutions for all 76 questions found in Exercise 18(A), Exercise 18(B), and the Test Yourself section. Each solution is crafted to follow the exact method and format that the ICSE board expects, ensuring you learn the correct way to present your work for full marks. Here, you will find clear explanations for every problem, helping you verify your own answers and understand the underlying theorems.

Exercise 18(A)

Question 1(a)

In the given figure, PA, PB and QR are tangents to a circle. If perimeter of the △PQR = 18 cm, the length of tangent PA is :

1. 18 cm
2. 27 cm
3. 9 cm
4. none of these

Consider that tangent QR intersects the circle at point D.

Given:

The perimeter of △PQR is 18 cm.

⇒ PQ + QD + RD + PR = 18 \hspace{0.2cm} \text{…(1)}

Recall that tangents drawn from an external point to a circle are equal in length.

∴ Let PA = PB = x, RD = RA, and QD = QB.

By substituting these values into equation (1), we have:

⇒ PQ + QB + RA + PR = 18

⇒ PB + PA = 18

⇒ x + x = 18

⇒ 2x = 18

⇒ x = 9 \text{ cm.}

∴ The length of tangent PA is 9 cm.

Hence, Option 3 is the correct option.

Question 1(b)

In the given figure, APB is tangent to the inner circle and also a chord of outer circle. Both the circles are concentric. If OA = 10 cm and OP = 6 cm, the length of AB is :

1. 16 cm
2. 10 cm
3. 14 cm
4. 20 cm

The tangent to a circle at any point and the radius drawn to that point are always perpendicular.

∴ OP ⊥ AP

In the right-angled triangle OAP, we apply the Pythagorean theorem:

⇒ OA^2 = OP^2 + AP^2

Substituting the given values:

⇒ 10^2 = 6^2 + AP^2

⇒ 100 = 36 + AP^2

⇒ AP^2 = 100 – 36

⇒ AP^2 = 64

⇒ AP = \sqrt{64} = 8 cm.

Since AB is a chord of the larger circle centered at O, the perpendicular from the center to the chord bisects it.

∴ PB = AP = 8 cm.

Thus, the length of AB is the sum of AP and PB:

AB = AP + PB = 8 + 8 = 16 cm.

Hence, Option 1 is the correct option.

Question 1(c)

A, B and C are three circles which touch each other as shown. Using the information, given in the diagram, we find the length AB as :

1. 6 cm
2. 17 cm
3. (289 – 9 – 2) cm
4. 11 cm

Consider the given figure.

For the line segment AC, we have:

Similarly, for the line segment BC, we find:

Now, using the Pythagoras theorem in the right-angled triangle ACB, we can write:

Substituting the values, we get:

Calculating further:

Taking the square root on both sides, we find:

Hence, Option 2 is the correct option.

Question 1(d)

BC is a tangent to the circle with center O. OD is radius of the circle. If ∠DOC = 100°, ∠B is equal to :

1. 50°
2. 60°
3. 40°
4. 70°

Observe the diagram given.

Since OD and OA are both radii of the circle, they are equal in length.

In the triangle △OAD, the angles ∠ODA and ∠OAD are equal. Let’s call each of these angles ‘x’. This is because angles opposite to equal sides in a triangle are equal.

Now, considering the property of triangles where an exterior angle is equal to the sum of the two opposite interior angles, we have:

∴ ∠DOC = ∠ODA + ∠OAD

⇒ 100° = 2x

Solving for x, we have:

⇒ x = \dfrac{100°}{2}

⇒ x = 50°.

Thus, ∠OAD = 50°.

Recall that a tangent to a circle and the radius at the point of tangency are perpendicular. Therefore:

∴ ∠BCA = 90°.

In triangle △ABC, using the angle sum property, we know:

⇒ ∠ABC + ∠BCA + ∠CAB = 180°

From the diagram, ∠CAB is equal to ∠OAD.

⇒ ∠ABC + ∠BCA + ∠OAD = 180°

Substituting the known values:

⇒ ∠ABC + 90° + 50° = 180°

⇒ ∠ABC + 140° = 180°

⇒ ∠ABC = 180° – 140° = 40°.

Hence, Option 3 is the correct option.

Question 1(e)

PA and PB are tangents to a circle with center O. If angle BPA = 70°, the angle ACB is :

1. 70°
2. 105°
3. 140°
4. 55°

Connect the points OA and OB.

Notice that the tangent at any point on a circle is perpendicular to the radius at that point. This means ∠OAP = 90° and ∠OBP = 90°.

Consider quadrilateral OAPB. The sum of the interior angles of a quadrilateral is 360°.

∴ ∠OAP + ∠APB + ∠PBO + ∠BOA = 360°

Substitute the known angles:

⇒ 90° + 70° + 90° + ∠BOA = 360°

Simplifying gives:

⇒ ∠BOA + 250° = 360°

⇒ ∠BOA = 360° – 250° = 110°

Next, recall 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.

∴ ∠AOB = 2∠ACB

This gives us:

⇒ ∠ACB = \dfrac{1}{2} ∠AOB

Substitute the value of ∠AOB:

⇒ ∠ACB = \dfrac{1}{2} \times 110° = 55°.

Hence, Option 4 is the correct option.

Question 2

The radius of a circle is 8 cm. Calculate the length of a tangent drawn to this circle from a point at a distance of 10 cm from its centre.

Consider a circle with the center labeled as O and a radius measuring 8 cm. Suppose P is an external point from which a tangent is drawn that touches the circle at point T.

∴ OP = 10 cm and OT = 8 cm

Recall that the tangent to a circle at any point is perpendicular to the radius at that point.

In the right-angled triangle ∆OTP, we apply the Pythagorean Theorem:

⇒ OP^2 = OT^2 + PT^2

Substituting the given values:

⇒ 10^2 = 8^2 + PT^2

⇒ PT^2 = 100 – 64 = 36

⇒ PT = 6 cm.

Thus, the length of the tangent is 6 cm.

Question 3

In the given figure, O is the centre of the circle and AB is a tangent to the circle at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.

Assume the radius of the circle is r cm.

Thus, the length of AO is given by ( AC + OC = (7.5 + r) ) cm.

Consider the right triangle \triangle AOB. According to the Pythagorean Theorem:

Substitute the known values:

Expanding the equation, we have:

Notice that r^2 cancels out from both sides, leading to:

Simplifying further:

Solving for r, we get:

Therefore, the radius of the circle is 11.25 cm.

Question 4

Two circles of radii 5 cm and 3 cm are concentric. Calculate the length of a chord of the outer circle which touches the inner.

Consider that the tangent to a circle at any point is perpendicular to the radius drawn to the point of tangency.

In the right-angled triangle \triangle OST, we apply the Pythagorean theorem:

∴ OS^2 = OT^2 + ST^2

⇒ 5^2 = 3^2 + ST^2

⇒ ST^2 = 25 - 9

⇒ ST^2 = 16

⇒ ST = \sqrt{16}

⇒ ST = 4 cm.

Similarly, examining the right-angled triangle \triangle OQT:

∴ OQ^2 = OT^2 + QT^2

⇒ 5^2 = 3^2 + QT^2

⇒ QT^2 = 25 - 9

⇒ QT^2 = 16

⇒ QT = \sqrt{16}

⇒ QT = 4 cm.

From the diagram, the entire chord QS is the sum of ST and QT:

∴ QS = ST + QT = 4 + 4 = 8 cm.

Thus, the length of the chord of the outer circle that touches the inner circle is 8 cm.

Question 5

Three circles touch each other externally. A triangle is formed when the centers of these circles are joined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.

Consider three circles, each with centers labeled as A, B, and C. When we connect these centers, a triangle ABC is formed.

Given:
– AB = 6 cm
– AC = 8 cm
– BC = 9 cm

Let the radii of the circles centered at A, B, and C be r_1, r_2, and r_3 respectively.

The relationships between the sides of the triangle and the radii are:

• AB = r_1 + r_2 = 6 …………(1)
• AC = r_1 + r_3 = 8 …………(2)
• BC = r_2 + r_3 = 9 ………….(3)

By adding these equations, we find:

This simplifies to:

Dividing by 2, we obtain:

r_1 + r_2 + r_3 = 11.5 \text{ cm} ………..(4)

Now, substitute r_2 + r_3 = 9 from equation (3) into equation (4):

Solving for r_1, we get:

r_1 = 11.5 - 9
r_1 = 2.5 \text{ cm}

Next, use r_1 + r_2 = 6 from equation (1) in equation (4):

Solving for r_3, we find:

r_3 = 11.5 - 6
r_3 = 5.5 \text{ cm}

Finally, substitute r_1 + r_3 = 8 from equation (2) into equation (4):

Solving for r_2, we have:

r_2 = 11.5 - 8
r_2 = 3.5 \text{ cm}

Hence, the radii of the circles are 2.5 cm, 3.5 cm, and 5.5 cm.

Question 6

If the sides of a quadrilateral ABCD touch a circle, prove that AB + CD = BC + AD.

Consider a circle that is tangent to the sides AB, BC, CD, and DA of quadrilateral ABCD at points P, Q, R, and S, respectively.

We have a fundamental property of tangents: when two tangents are drawn from a single external point to a circle, their lengths are equal.

Thus, from point A, the tangents AP and AS are equal:

AP = AS \text{......... (1)}

Similarly, from point B, the tangents BP and BQ are equal:

BP = BQ \text{......... (2)}

From point C, the tangents CR and CQ are equal:

CR = CQ \text{......... (3)}

And from point D, the tangents DR and DS are equal:

DR = DS \text{......... (4)}

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

⇒ AP + BP + CR + DR = AS + DS + BQ + CQ

Referring to the diagram, we note:

• AP + BP = AB
• CR + DR = CD
• AS + DS = AD
• BQ + CQ = BC

Substituting these into our equation, we find:

⇒ AB + CD = AD + BC

Thus, we have shown that AB + CD = AD + BC.

Question 7

If the sides of a parallelogram touch a circle, prove that the parallelogram is a rhombus.

Consider a circle that is tangent to the sides AB, BC, CD, and DA of parallelogram ABCD at points P, Q, R, and S, respectively.

Recall that if two tangents are drawn from a single external point to a circle, those tangents will be of equal length.

Therefore, from point A, the tangents AP and AS are equal.

Thus, AP = AS ………. (1)

Similarly, for the other points, we have:

BP = BQ ………. (2)

CR = CQ ………. (3)

DR = DS ………. (4)

Adding equations (1), (2), (3), and (4), we obtain:

AP + BP + CR + DR = AS + BQ + CQ + DS

From the diagram, notice that:

⇒ AP + BP = AB,

BQ + CQ = BC,

CR + DR = CD,

AS + DS = AD.

Thus, we have:

∴ AB + CD = AD + BC ……….(5)

Since the opposite sides of a parallelogram are equal, we know:

AB = CD and BC = AD

Substituting these into equation (5), we find:

⇒ AB + AB = BC + BC

⇒ 2AB = 2BC

⇒ AB = BC

Therefore, all sides are equal: ∴ AB = BC = CD = DA.

Hence, proved that ABCD is a rhombus.

Question 8

From the given figure, prove that :

AP + BQ + CR = BP + CQ + AR.

Also, show that :

AP + BQ + CR = \dfrac{1}{2} x Perimeter of triangle ABC.

Consider that when two tangents are drawn from an external point to a circle, they are of equal length.

From point B, the tangents BQ and BP are equal:

BQ = BP …….. (1)

From point A, the tangents AP and AR are equal:

AP = AR …….. (2)

From point C, the tangents CR and CQ are equal:

CR = CQ …….. (3)

By adding equations (1), (2), and (3), we obtain:

AP + BQ + CR = BP + CQ + AR ……… (4)

Thus, we have shown that AP + BQ + CR = BP + CQ + AR.

Next, by adding AP + BQ + CR to both sides of equation (4), we derive:

2(AP + BQ + CR) = AP + BP + CQ + BQ + AR + CR ……….(5)

From the figure, observe that:

AP + BP = AB, BQ + CQ = BC, and AR + CR = AC.

Substituting these into equation (5), we have:

⇒ 2(AP + BQ + CR) = AB + BC + CA

⇒ AP + BQ + CR = \dfrac{1}{2}(AB + BC + CA).

⇒ AP + BQ + CR = \dfrac{1}{2} (Perimeter of △ABC). [∵ Perimeter = AB + BC + CA]

Therefore, it is proved that AP + BQ + CR = \dfrac{1}{2} (Perimeter of △ABC).

Question 9

In the given figure, if AB = AC then prove that BQ = CQ.

Consider that from point A, the lines AP and AR are tangents to the circle. It is a known property that when two tangents are drawn from an external point to a circle, they are of equal length. Therefore, AP = AR ……….(1)

Similarly, from point B, the lines BP and BQ serve as tangents to the circle. Thus, BP = BQ ……….(2)

Likewise, from point C, the lines CQ and CR are tangents to the circle, leading to CQ = CR …………(3)

By adding equations (1), (2), and (3), we obtain:

⇒ AP + BP + CQ = AR + BQ + CR

Rearranging terms, we have:

⇒ (AP + BP) + CQ = (AR + CR) + BQ

This simplifies to:

⇒ AB + CQ = AC + BQ

According to the problem, AB = AC.

∴ BQ = CQ.

Hence, proved that BQ = CQ.

Question 10

Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centers if :

(i) they touch each other externally,

(ii) they touch each other internally.

Consider two circles with centers at O and O' and radii 6.3 cm and 3.6 cm respectively.

(i) External Touching:

When the circles touch each other externally at point P, the line segment OO' represents the distance between their centers. Here, the trick is to recognize that the distance OO' is simply the sum of the radii of the two circles.

So, OO' = O'P + OP = 6.3 + 3.6 = 9.9 cm.

∴ The centers are 9.9 cm apart when the circles touch externally.

(ii) Internal Touching:

In the case where the circles touch internally at point P, the distance between their centers, OO', is the difference between the larger radius and the smaller radius.

Thus, OO' = O'P - OP = 6.3 - 3.6 = 2.7 cm.

∴ When the circles touch internally, the centers are 2.7 cm apart.

Question 11

From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that:

(i) ∠AOP = ∠BOP

(ii) OP is the ⊥ bisector of chord AB.

The diagram illustrates a circle with center O and tangents PA and PB originating from a point P outside the circle.

(i) Consider triangles ∆AOP and ∆BOP:

⇒ AP = BP because tangents drawn from the same external point P are congruent.

⇒ OA = OB since both are radii of the circle.

⇒ OP = OP as it is common to both triangles.

∴ By the Side-Side-Side (SSS) congruence criterion, ΔAOP ≅ ΔBOP.

∴ By Corresponding Parts of Congruent Triangles (C.P.C.T.), ∠AOP = ∠BOP.

Thus, we have shown that ∠AOP = ∠BOP.

(ii) Now, examine triangles ∆OAM and ∆OBM:

OA = OB, as they are both radii of the circle.

∠AOM = ∠BOM because we have already proved ∠AOP = ∠BOP.

OM = OM, as it is common to both triangles.

∴ By the Side-Angle-Side (SAS) congruence criterion, ΔOAM ≅ ΔOBM.

By C.P.C.T.,

⇒ AM = MB and ∠OMA = ∠OMB = x.

Observing the figure,

⇒ ∠OMA + ∠OMB = 180° as AB is a straight line.

⇒ x + x = 180°

⇒ 2x = 180°

⇒ x = \dfrac{180°}{2}

⇒ x = 90°.

Thus, ∠OMA = ∠OMB = 90°.

Therefore, we have demonstrated that OP is the ⊥ bisector of chord AB.

Question 12

In the given Figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that :

(i) tangent at point P bisects AB.

(ii) angle APB = 90°.

(i) Consider the property that if two tangents are drawn from an external point to a circle, they are of equal length.

From the figure, observe that:

TA and TP are tangents to the circle with center O.

∴ TA = TP ………..(1)

Similarly, TB and TP are tangents to the circle with center O’.

∴ TB = TP ………..(2)

By comparing (1) and (2), we conclude:

TA = TB.

Thus, it is established that the tangent at point P bisects AB.

(ii) Now, consider △ATP:

Since TA = TP (as shown earlier),

∴ ∠TAP = ∠TPA ………(1) [Because angles opposite equal sides are equal.]

Next, in △BTP:

TB = TP (as shown earlier),

∴ ∠TBP = ∠TPB ……….(2) [Because angles opposite equal sides are equal.]

Adding equations (1) and (2), we find:

∠TAP + ∠TBP = ∠TPA + ∠TPB

∠TAP + ∠TBP = ∠APB ………..(3)

In △ABP, using the angle sum property:

∠APB + ∠BAP + ∠ABP = 180°

From the figure, notice that ∠TAP = ∠BAP and ∠TBP = ∠ABP.

⇒ ∠APB + ∠TAP + ∠TBP = 180°

⇒ ∠APB + ∠APB = 180°

⇒ 2∠APB = 180°

⇒ ∠APB = 90°.

Thus, it is proven that ∠APB = 90°.

Question 13

Tangents AP and AQ are drawn to a circle, with center O, from an exterior point A. Prove that :

∠PAQ = 2∠OPQ

Recall that a tangent to a circle is perpendicular to the radius at the point of contact.

In quadrilateral OPAQ, we have:

∠OPA = ∠OQA = 90°.

The sum of the angles in a quadrilateral is 360°, so:

∠OPA + ∠OQA + ∠POQ + ∠PAQ = 360°

Substituting the known angles, we get:

90° + 90° + ∠POQ + ∠PAQ = 360°

This simplifies to:

∠POQ + ∠PAQ = 360° – 180°

Thus, we have:

∠POQ + ∠PAQ = 180° ……….(1)

Now, consider △OPQ. Since OP = OQ (both are radii of the circle), it follows that:

∴ ∠OPQ = ∠OQP

Using the angle sum property of a triangle:

∠OPQ + ∠OQP + ∠POQ = 180°

Since ∠OPQ = ∠OQP, we have:

∠OPQ + ∠OPQ + ∠POQ = 180°

This simplifies to:

2∠OPQ + ∠POQ = 180° ………(2)

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

∠POQ + ∠PAQ = 2∠OPQ + ∠POQ

Subtracting ∠POQ from both sides gives:

∠PAQ = 2∠OPQ.

Hence, proved that ∠PAQ = 2∠OPQ.

Question 14

ABC is a right angled triangle with AB = 12 cm and AC = 13 cm. A circle, with center O, has been inscribed inside the triangle. Calculate the value of x, the radius of the inscribed circle.

Let AB touches the circle at L, AC at N and BC at M.

Notice that LBMO forms a square, which implies that LB = BM = OM = OL = x.

The segment AL is calculated as follows: AL = AB – LB = (12 – x) cm.

Since tangents drawn from an external point to a circle are equal, AL = AN = (12 – x) cm.

Given that triangle ABC is a right-angled triangle, we apply the Pythagorean theorem:

∴ AC^2 = AB^2 + BC^2

⇒ 13^2 = 12^2 + BC^2

⇒ BC^2 = 13^2 - 12^2

⇒ BC^2 = 169 - 144

⇒ BC^2 = 25

⇒ BC = \sqrt{25}

⇒ BC = 5 cm.

From the diagram, MC is calculated as: MC = BC – BM = (5 – x) cm.

Similarly, since tangents from an external point are equal, CN = CM = (5 – x) cm.

We know that:

⇒ AC = AN + CN

⇒ 13 = (12 – x) + (5 – x)

⇒ 13 = 17 – 2x

⇒ 2x = 17 – 13

⇒ 2x = 4

⇒ x = \dfrac{4}{2}

⇒ x = 2 cm.

Hence, x = 2.

Question 15

In a triangle ABC, the incircle (center O) touches BC, CA and AB at points P, Q and R respectively. Calculate :

(i) ∠QOR

(ii) ∠QPR;

given that ∠A = 60°.

In the triangle \triangle ABC, the incircle with center O touches sides BC, CA, and AB at points P, Q, and R, respectively.

(i) Consider the tangent properties of a circle. The tangent at any point on the circle is perpendicular to the radius at that point.

∴ \angle ORA = \angle OQA = 90^\circ.

Now, look at quadrilateral AROQ:

The sum of the interior angles of a quadrilateral is 360^\circ.

∴ \angle ORA + \angle OQA + \angle QOR + \angle A = 360^\circ

⇒ 90^\circ + 90^\circ + \angle QOR + 60^\circ = 360^\circ

⇒ 240^\circ + \angle QOR = 360^\circ

⇒ \angle QOR = 360^\circ - 240^\circ

⇒ \angle QOR = 120^\circ.

Thus, \angle QOR = 120^\circ.

(ii) Refer to the figure:

Arc RQ creates \angle ROQ at the center and \angle QPR at the circumference of the circle.

∴ \angle QPR = \dfrac{1}{2}\angle QOR

⇒ \angle QPR = \dfrac{1}{2} \times 120^\circ = 60^\circ.

Therefore, \angle QPR = 60^\circ.

Question 16

In the given figure, PT touches the circle with center O at point R. Diameter SQ is produced to meet the tangent TR at P.

Given ∠SPR = x° and ∠QRP = y°;

prove that :

(i) ∠ORS = y°

(ii) write an expression connecting x and y.

(i) Observing the figure, notice that:

⇒ ∠QRP = ∠OSR = y° because angles in alternate segments are equal.

Since OS and OR are both radii of the circle, they are equal.

∴ Angles opposite these equal sides are also equal.

Thus, ∠ORS = ∠OSR = y°.

Therefore, ∠ORS = y° is proven.

(ii) From the figure, we know:

∠ORP = 90° because a tangent at a point is perpendicular to the radius drawn to that point.

⇒ ∠ORQ = ∠ORP – ∠QRP = 90° – y° ………..(1)

Since OQ and OR are radii of the circle, they are equal.

∴ Angles opposite these equal sides are equal, so:

∠OQR = ∠ORQ = 90° – y°

Now, consider △PQR:

⇒ ∠OQR = ∠QPR + ∠QRP because an exterior angle of a triangle equals the sum of the two opposite interior angles.

⇒ 90° – y° = x° + y°

⇒ x° + 2y° = 90°.

Thus, the expression connecting x and y is x + 2y = 90°.

Question 17

PT is a tangent to the circle at T. If ∠ABC = 70° and ∠ACB = 50°; calculate :

(i) ∠CBT

(ii) ∠BAT

(iii) ∠APT

To solve this, first connect AT and BT.

(i) TC serves as the circle’s diameter.

Notice that the angle formed in a semicircle is a right angle.

∴ ∠CBT = 90°.

Hence, ∠CBT = 90°.

(ii) Consider the cyclic quadrilateral ATBC.

⇒ ∠CBT + ∠CAT = 180° (since the sum of opposite angles in a cyclic quadrilateral is 180°)

⇒ 90° + ∠CAT = 180°

⇒ ∠CAT = 180° – 90°

⇒ ∠CAT = 90°.

Now, in △ABC,

⇒ ∠CBA + ∠CAB + ∠ACB = 180° [According to the angle sum property of a triangle]

⇒ 70° + ∠CAB + 50° = 180°

⇒ ∠CAB + 120° = 180°

⇒ ∠CAB = 180° – 120°

⇒ ∠CAB = 60°.

From the diagram,

∠BAT = ∠CAT – ∠CAB = 90° – 60° = 30°.

Hence, ∠BAT = 30°.

(iii) From the diagram,

∠BTX = ∠BAT = 30° [Angles in the same segment are equal]

∠PBT = ∠CBT – ∠CBA = 90° – 70° = 20°.

⇒ ∠PTB = 180° – ∠BTX = 180° – 30° = 150°.

In △PBT,

⇒ ∠PBT + ∠PTB + ∠APT = 180° [Using the angle sum property of a triangle]

⇒ 20° + 150° + ∠APT = 180°

⇒ ∠APT + 170° = 180°

⇒ ∠APT = 180° – 170°

⇒ ∠APT = 10°.

Hence, ∠APT = 10°.

Question 18

In the given figure, O is the center of the circumcircle ABC. Tangents A and C intersect at P. Given angle AOB = 140° and angle APC = 80°; find the angle BAC.

To solve this, first connect OC.

Since PA and PC are tangents, we have:

∴ OA ⊥ PA and OC ⊥ PC

Consider quadrilateral APCO:

⇒ ∠APC + ∠AOC = 180°

⇒ 80° + ∠AOC = 180°

⇒ ∠AOC = 180° – 80°

⇒ ∠AOC = 100°

From the diagram, calculate:

∠BOC = 360° – (∠AOB + ∠AOC)

= 360° – (140° + 100°)

= 360° – 240° = 120°.

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

Arc BC creates ∠BOC at the centre and ∠BAC on the circle’s circumference.

∴ ∠BAC = \dfrac{1}{2}∠BOC = \dfrac{1}{2} \times 120° = 60°.

Hence, ∠BAC = 60°.

Question 19

In the given figure, PQ is a tangent to the circle at A. AB and AD are bisectors of ∠CAQ and ∠PAC. If ∠BAQ = 30°, prove that : BD is diameter of the circle.

Examining the figure, we note:

∠CAB = ∠BAQ = 30° because AB is the bisector of ∠CAQ.

∴ ∠CAQ = 2 × ∠BAQ = 60°.

Continuing with the figure, observe:

∠CAQ + ∠PAC = 180° as they form a linear pair.

Thus, 60° + ∠PAC = 180°

⇒ ∠PAC = 180° – 60°

⇒ ∠PAC = 120°.

Now, since AD bisects ∠PAC:

∠PAC = 2 × ∠CAD

⇒ 120° = 2 × ∠CAD

⇒ ∠CAD = \dfrac{120°}{2}

⇒ ∠CAD = 60°.

From the figure, calculate:

∠DAB = ∠CAD + ∠CAB = 60° + 30° = 90°.

Therefore, BD subtends a right angle (90°) at the circle, confirming that angle in a semicircle is a right angle.

Hence, BD is the diameter of the circle.

Exercise 18(B)

Question 1(a)

Chords AB and CD of a circle intersect each other at point O such that OA : OC = 4 : 7. Then OB : OD is equal to :

1. 4 : 7
2. 5 : 4
3. 7 : 4
4. 4 : 5

We’re given that the ratio of segments OA to OC is 4 : 7.

Recall the property of intersecting chords: when two chords intersect within or outside a circle, the products of the lengths of the segments of each chord are equal.

From the diagram, this means:

∴ OA × OB = OC × OD

This implies:

⇒ \dfrac{OD}{OB} = \dfrac{OA}{OC}

Substituting the given ratio:

⇒ \dfrac{OD}{OB} = \dfrac{4}{7}

Therefore, the reciprocal gives us:

⇒ \dfrac{OB}{OD} = \dfrac{7}{4}

Thus, the ratio OB : OD is 7 : 4.

Hence, Option 3 is the correct option.

Question 1(b)

If ∠PAC : ∠PCA = 4 : 5, ∠P is :

1. 40°
2. 60°
3. 105°
4. 45°

In a trapezium, the sum of the interior angles on the same side is 180°. For trapezium ABDC, we have:

∴ ∠B + ∠BAC = 180°

Given that ∠B is 60°, it follows:

⇒ 60° + ∠BAC = 180°

⇒ ∠BAC = 180° – 60° = 120°.

Considering the figure, we can see:

∠PAC + ∠BAC = 180°

⇒ ∠PAC + 120° = 180°

⇒ ∠PAC = 180° – 120° = 60°.

With the given ratio:

∠PAC : ∠PCA = 4 : 5

This implies:

Substituting for ∠PAC:

Solving for ∠PCA:

In triangle △PCA, using the angle sum property:

∠APC + ∠PAC + ∠PCA = 180°

⇒ ∠APC + 60° + 75° = 180°

⇒ ∠APC + 135° = 180°

⇒ ∠APC = 180° – 135° = 45°.

Hence, Option 4 is the correct option.

Question 1(c)

AC is a tangent to the given circle which touches the circle at point B. If ∠EBC = 45°; angle EDB is equal to :

1. 45°
2. 90°
3. 125°
4. 135°

Observe the diagram provided.

∴ ∠ABE + ∠EBC = 180° due to the linear pair postulate.

Given that ∠EBC = 45°, we substitute to find:

⇒ ∠ABE + 45° = 180°

⇒ ∠ABE = 180° – 45° = 135°.

Recall the theorem: The angle between a tangent and a chord through the point of contact is congruent to the angle in the alternate segment.

∴ ∠EDB = ∠ABE = 135°.

Hence, Option 4 is the correct option.

Question 1(d)

In the given circle, PA is tangent and PBC is secant, PA = 8 cm and PB = 4 cm. The length of BC is :

1. 8 cm
2. 12 cm
3. 16 cm
4. 2 cm

We have a tangent PA and a secant PBC. According to the theorem for tangents and intersecting chords, the square of the tangent length from the point of contact is equal to the product of the lengths of the secant segments.

∴ PB \times PC = PA^2

⇒ 4 \times PC = 8^2

⇒ 4 \times PC = 64

⇒ PC = \dfrac{64}{4} = 16 \text{ cm}.

Now, looking at the figure:

BC = PC – PB = 16 – 4 = 12 \text{ cm}.

Hence, Option 2 is the correct option.

Question 1(e)

In the given figure, O is the center of the circle, PA is tangent and PBC is secant. If ∠ABC = 60°; ∠P is :

1. 30°
2. 60°
3. 120°
4. 90°

Consider △ABC. Here, we have ∠BAC = 90° since an angle inscribed in a semicircle is a right angle.

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

⇒ ∠ABC + ∠BAC + ∠ACB = 180°

Substituting the known angles:

⇒ 60° + 90° + ∠ACB = 180°

This simplifies to:

⇒ 150° + ∠ACB = 180°

⇒ ∠ACB = 180° – 150° = 30°.

Recall that the angle between a tangent and a chord at the point of contact equals the angle in the alternate segment.

⇒ ∠BAP = ∠ACB = 30°.

Now, observe from the figure:

⇒ ∠PBA + ∠ABC = 180° (as they form a linear pair)

⇒ ∠PBA + 60° = 180°

⇒ ∠PBA = 180° – 60° = 120°.

In △PBA, apply the angle sum property:

⇒ ∠PBA + ∠BAP + ∠APB = 180°

Substitute the values:

⇒ 120° + 30° + ∠APB = 180°

⇒ 150° + ∠APB = 180°

⇒ ∠APB = 180° – 150° = 30°.

∴ ∠P = 30°.

Hence, Option 1 is the correct option.

Question 2

In the given figure, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find :

(i) AB

(ii) the length of tangent PT.

(i) Consider the figure provided.

The segment PC can be calculated as the sum of PD and CD:

According to the intersecting chords theorem, if two chords intersect at a point outside the circle, the product of the lengths of the segments of one chord is equal to the product of the segments of the other chord.

In this situation, since AB and CD intersect externally at point P, we have:

Substituting the known values gives:

Solving for AP:

Therefore, the diameter AB is:

Hence, AB = 12 cm.

(ii) For a tangent and a chord intersecting externally, the square of the tangent’s length from the point of contact to the intersection point is equal to the product of the lengths of the segments of the chord.

Thus, we have:

Substituting the values, we get:

Solving for PT:

Hence, PT = 8 cm.

Question 3

In the following figure, PQ is the tangent to the circle at A, DB is a diameter and O is the centre of the circle. If ∠ADB = 30° and ∠CBD = 60°, calculate:

(i) ∠QAB,

(ii) ∠PAD,

(iii) ∠CDB.

(i) We know that ∠QAB is equal to ∠ADB because angles in alternate segments are equal.

∴ ∠QAB = 30°.

Hence, the value of ∠QAB = 30°.

(ii) Consider △DAO. Since OA and OD are radii of the circle, they are equal.

Thus, ∠OAD = ∠ODA = 30° because angles opposite equal sides in a triangle are equal.

The tangent at any point on a circle is perpendicular to the radius at the point of contact.

∴ ∠OAP = 90°.

From the figure, we find:

⇒ ∠PAD = ∠OAP – ∠OAD = 90° – 30° = 60°.

Hence, ∠PAD = 60°.

(iii) In △BCD, notice that ∠BCD is 90° since an angle in a semicircle is always a right angle.

Given, ∠CBD = 60°.

Using the angle sum property of a triangle, ∠CDB + ∠CBD + ∠BCD = 180°.

⇒ ∠CDB + 60° + 90° = 180°.

⇒ ∠CDB = 180° – 150° = 30°.

Hence, ∠CDB = 30°.

Question 4

If PQ is a tangent to the circle at R; calculate :

(i) ∠PRS,

(ii) ∠ROT.

Given, O is the center of the circle and angle TRQ = 30°.

(i) Since ST passes through the center O, it is the diameter of the circle. According to the property of circles, the angle formed in a semi-circle is always 90°.

∴ ∠SRT = 90°.

Given that PQ is a straight line, we have:

∠PRS + ∠SRT + ∠TRQ = 180°

Substituting the known angles:

⇒ ∠PRS + 90° + 30° = 180°

⇒ ∠PRS + 120° = 180°

⇒ ∠PRS = 180° – 120°

⇒ ∠PRS = 60°.

Hence, ∠PRS = 60°.

(ii) The angle formed between a tangent and a chord through the point of contact is equal to the angle in the alternate segment on the circle. Thus,

∠TSR = ∠TRQ = 30°.

Furthermore, the angle subtended by a segment at the center of the circle is twice the angle subtended at the circumference.

∠ROT = 2∠TSR = 2 × 30° = 60°.

Hence, ∠ROT = 60°.

Question 5

Two circles with centers O and O’ are drawn to intersect each other at points A and B. Center O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with center O’ at A. Prove that OA bisects angle BAC.

Let’s connect points O'A and O'B.

Notice that the angle formed between a tangent and a chord at the point of contact is equal to the angle in the alternate segment. Here, CD is the tangent, and AO is the chord.

∴ ∠OAC = ∠OBA ……… (1)

In the triangle ∆OAB:

Since OA = OB (both being radii of the circle centered at O), we can say:

∠OAB = ∠OBA ………. (2) [Angles opposite to equal sides]

Combining results from (1) and (2), we conclude:

∠OAC = ∠OAB

Thus, it is established that OA bisects ∠BAC.

Question 6

In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If ∠BCG = 108° and O is the center of the circle, find :

(i) angle BCT

(ii) angle DOC

To solve this problem, let’s first connect the points OC, OD, and BD.

Given that ∠BCG equals 108°, we can use the linear pair relationship:

∴ ∠BCG + ∠BCD = 180°

Substituting the given value:

⇒ 108° + ∠BCD = 180°

Solving for ∠BCD, we have:

⇒ ∠BCD = 180° – 108°

⇒ ∠BCD = 72°.

Since BC = CD, the triangle BDC is isosceles, which means:

∠BDC = ∠DBC = x (let’s assume)

Applying the angle sum property in ΔBDC:

⇒ ∠DBC + ∠BDC + ∠BCD = 180°

⇒ x + x + 72° = 180°

⇒ 2x + 72° = 180°

Solving for x:

⇒ 2x = 180° – 72°

⇒ 2x = 108°

⇒ x = \dfrac{108°}{2}

⇒ x = 54°.

Now, consider the tangent property:

∠BCT = ∠BDC (since angles in alternate segments are equal)

∴ ∠BCT = 54°.

Hence, ∠BCT = 54°.

For part (ii), remember that the angle subtended by a segment at the center is twice the angle subtended at any point on the circle’s circumference:

∴ ∠DOC = 2∠DBC

Substituting the value of ∠DBC:

⇒ ∠DOC = 2(54°) = 108°.

Hence, ∠DOC = 108°.

Question 7

In the figure; PA is a tangent to the circle, PBC is secant and AD bisects angle BAC. Show that triangle PAD is an isosceles triangle. Also, show that :

∠CAD = \dfrac{1}{2} [∠PBA – ∠PAB]

Consider the rule that the angle formed between a tangent and a chord at the point of contact is equal to the angle in the alternate segment.

In the given figure:

• PA is a tangent, and AB is a chord.

∴ ∠PAB = ∠C [Angles in alternate segments are equal] ………..(1)

It’s given that AD bisects ∠BAC.

∴ ∠BAD = ∠DAC ………..(2)

Recall the property that an exterior angle is equal to the sum of the two opposite interior angles.

⇒ ∠ADP = ∠C + ∠DAC

⇒ ∠ADP = ∠PAB + ∠BAD [Using equations (1) and (2)]

⇒ ∠ADP = ∠PAD

Since the angles opposite equal sides are equal, it follows that:

∴ PA = PD

This implies that triangle PAD is isosceles.

Now, in △ABC:

⇒ ∠PBA = ∠C + ∠BAC [An exterior angle equals the sum of the opposite interior angles]

⇒ ∠BAC = ∠PBA – ∠C

⇒ ∠BAC = ∠PBA – ∠PAB [Since ∠C = ∠PAB]

Since AD bisects ∠BAC:

⇒ 2∠CAD = ∠PBA – ∠PAB

⇒ ∠CAD = \dfrac{1}{2}(∠PBA – ∠PAB).

Thus, it is proved that ∠CAD = \dfrac{1}{2}(∠PBA – ∠PAB).

Question 8

Two circles intersect each other at point A and B. Their common tangent touches the circles at points P and Q as shown in the figure. Show that the angles PAQ and PBQ are supplementary.

Consider the line segment AB.

Recall that the angle formed between a tangent and a chord at the point of contact is equal to the angle in the alternate segment of the circle. Therefore:

In the given figure, PQ is the tangent and AB is the chord.

∴ ∠QPA = ∠PBA [Angles in alternate segment are equal] ………(1)

Similarly, we have:

∴ ∠PQA = ∠QBA [Angles in alternate segment are equal] ……….(2)

Adding equations (1) and (2), we obtain:

⇒ ∠QPA + ∠PQA = ∠PBA + ∠QBA ……….(3)

Now, observe:

⇒ ∠PBA + ∠QBA = ∠PBQ ………..(4)

In the triangle PAQ:

⇒ ∠QPA + ∠PQA + ∠PAQ = 180° [Angle sum property of triangle]

⇒ ∠QPA + ∠PQA = 180° – ∠PAQ

Substituting from equation (3), we have:

⇒ ∠PBA + ∠QBA = 180° – ∠PAQ [From (3)] ……….(5)

Combining equations (4) and (5), we deduce:

⇒ ∠PBQ = 180° – ∠PAQ

Therefore, it follows that:

⇒ ∠PBQ + ∠PAQ = 180°.

Hence, proved that PAQ and PBQ are supplementary.

Question 9

In the figure, chords AE and BC intersect each other at point D.

(i) If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm; find DE.

(ii) If AD = BD, show that : AE = BC.

(i) Consider joining AB in the diagram.

Notice that ∠ADB and ∠CDE are both 90°, as vertically opposite angles are equal.

In the right-angled triangle ADB, we apply the Pythagorean theorem:

∴ AB² = AD² + BD²

⇒ 5² = AD² + 4²

⇒ 25 = AD² + 16

⇒ AD² = 25 – 16

⇒ AD² = 9

⇒ AD = \sqrt{9}

⇒ AD = 3 cm.

According to the intersecting chords theorem, when two chords intersect, the product of the segments of one chord equals the product of the segments of the other.

From the figure, chords AE and CB intersect at D, so:

∴ AD × DE = CD × BD

⇒ 3 × DE = 4 × 9

⇒ DE = \frac{36}{3}

⇒ DE = 12 cm.

Thus, DE = 12 cm.

(ii) Given that AD = BD, denote this common length as x.

We know:

∴ AD × DE = CD × BD

⇒ (x)DE = (x)CD

⇒ DE = CD ……….(2)

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

⇒ AD + DE = BD + CD

∴ AE = BC.

Hence, it is shown that AE = BC.

Question 10

In the adjoining figure, O is the center of the circle and AB is a tangent to it at point B. ∠BDC = 65°. Find ∠BAO.

Let’s analyze the figure. Notice that the angles ∠ADE and ∠BDE form a linear pair.

∴ ∠ADE + ∠BDE = 180°

Given that ∠BDE = 65°, we have:

⇒ ∠ADE + 65° = 180°
⇒ ∠ADE = 180° – 65°
⇒ ∠ADE = 115° …………..(1)

Now, observe that ∠DBO is 90° because DB is tangent to the circle at B and BC is the diameter.

In △BDC, using the angle sum property:

⇒ ∠BDC + ∠DBC + ∠DCB = 180°
⇒ 65° + 90° + ∠DCB = 180°
⇒ ∠DCB = 180° – 155° = 25°.

Since OE and OC are radii of the same circle, they are equal, making △OCE isosceles.

⇒ ∠OCE = ∠OEC
⇒ ∠OEC = ∠DCB = 25° [because ∠OCE = ∠DCB]

In △ADE, applying the angle sum property again:

⇒ ∠ADE + ∠DEA + ∠DAE = 180°
⇒ 115° + 25° + ∠DAE = 180°

Given that ∠DEA = ∠OEC (as they are vertically opposite), we find:

⇒ ∠DAE = 180° – 140° = 40°.

Finally, from the figure, we see:

∠BAO = ∠DAE = 40°.

Hence, ∠BAO = 40°.

Test Yourself

Question 1(a)

AP is a tangent to the given circle. If AB = 8 cm and BC = 10 cm, then AP is :

1. 8 cm
2. 16 cm
3. 12 cm
4. 24 cm

Consider the property involving a tangent and a chord intersecting outside a circle. When this occurs, the product of the lengths of the chord’s segments equals the square of the tangent’s length from the contact point to the intersection point.

In the given problem, calculate the length of segment AC as follows:

AC = AB + BC = 8 + 10 = 18 cm.

Using the property mentioned, we have:

Substituting the known values:

This simplifies to:

Taking the square root of both sides gives:

AP = \sqrt{144} = 12 cm.

Hence, Option 3 is the correct option.

Question 1(b)

In the given figure, O is center of the circle and PQ is a tangent. If angle OAB = x; the measure of angle ABP; in terms of x, is :

1. x
2. 180° – 2x
3. 90° + x
4. 90° – x

Consider △OAB in the diagram. Since OA and OB are both radii of the circle, they are equal in length. This implies that the angles opposite these equal sides must also be equal. Thus, ∠OBA = ∠OAB = x.

Recall that a tangent to a circle is perpendicular to the radius at the point of contact. Therefore, OB is perpendicular to PQ, making ∠PBO = 90°.

Now, observe that ∠ABP can be calculated as the difference between ∠PBO and ∠OBA. Thus, ∠ABP = 90° – x.

Hence, Option 4 is the correct option.

Question 1(c)

In the given figure, AB is tangent to the circle with center O. If OCB is a straight line segment, the angle BAC is :

1. 40°
2. 55°
3. 35°
4. 20°

The tangent at any point on a circle is perpendicular to the radius at that point. Thus, OA \perp AB, which gives us \angle OAB = 90^\circ.

Let \angle BAC = x.

Now, observe \triangle OAC:

• \angle A = \angle OAB - \angle BAC = 90^\circ - x.
• Since OA = OC (both are radii of the circle), the angles opposite these equal sides are also equal.

Thus, \angle C = \angle A = 90^\circ - x.

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

Substituting the known values:

Simplifying:

This implies:

Now, consider \triangle OAB:

Using the angle sum property again:

Substituting the values:

Simplifying gives:

Solving for x:

Thus, \angle BAC = 35^\circ.

Hence, Option 3 is the correct option.

Question 1(d)

In the given figure O is center, PQ is tangent at point A. BD is diameter and ∠AOD = 84° then angle QAD is :

1. 32°
2. 84°
3. 48°
4. 42°

Consider the triangle \triangle OAD. Since OA and OD are radii of the same circle, they are equal.

Notice that angles opposite to equal sides in a triangle are equal. Let \angle A = \angle D = x.

Applying the angle sum property in \triangle OAD:

Substitute the known values:

Simplifying gives:

Thus, \angle OAD = \angle A = 48^\circ.

Recall that a tangent at any point on a circle is perpendicular to the radius at that point. Therefore, \angle OAQ = 90^\circ.

Now, calculate \angle DAQ:

\angle DAQ = \angle OAQ - \angle OAD = 90^\circ - 48^\circ = 42^\circ.

Hence, Option 4 is the correct option.

Question 1(e)

Two mutually perpendicular tangents are drawn to a circle with radius R\sqrt{2} units. The shortest distance between the two points of contact is :

• (a) R units
• (b) \dfrac{1}{2}R units
• (c) R\sqrt{2} units
• (d) 2R units

Consider two perpendicular tangents drawn from an external point A to the circle, touching it at points B and C.

The radius of the circle is given as R\sqrt{2} units.

From the diagram:

• The distance AC, which is the same as the radius OB, is R\sqrt{2}.
• Similarly, the distance AB, which is the same as the radius OC, is also R\sqrt{2}.

Now, let’s analyze the right triangle ABC:

• By the Pythagorean theorem, we have:

• Substituting the given values:

• Simplifying, we find:

BC^2 = 2R^2 + 2R^2
BC^2 = 4R^2

• Taking the square root of both sides gives:

BC = \sqrt{4R^2}
BC = 2R

Therefore, the shortest distance between the points of contact B and C is 2R units.

Hence, Option 4 is the correct option.

Question 1(f)

For the three circles with centers A, B and C and radii 5 cm, 2 cm and 6 cm respectively.

Assertion (A) : To find the perimeter of the triangle ABC, add the radii of given three circles.

Reason (R) : The required perimeter is the product of sum of radii by 2.

• (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.

Consider the points where the circles intersect, labeled as D, E, and F.

Referring to the figure, the perimeter of triangle ABC can be calculated as follows:

Perimeter of triangle ABC = AB + BC + CA

= (AD + BD) + (BE + CE) + (CF + FA)

= 5 + 2 + 2 + 6 + 6 + 5

= 26 cm.

When we simply add the radii of the three circles, we get:

5 + 2 + 6 = 13 cm, which does not match the perimeter.

However, if we multiply the sum of the radii by 2, we have:

Sum of radii × 2 = 13 × 2 = 26 cm, which indeed equals the perimeter.

∴ Assertion is false, but the Reason is true.

Question 1(g)

AB is diameter of the circle. PA is tangent and ∠AOC = 60°.

Assertion(A): x + 30° = 90°.

Reason(R): PA is tangent

⇒ ∠BAP = 90°

∴ x + 30° = 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.

To solve this, remember that the angle formed by an arc at the circle’s center is twice the angle formed at any point on the circle’s circumference.

∴ ∠AOC = 2 × ∠ABC

⇒ 60° = 2 × ∠ABC

⇒ ∠ABC = \dfrac{60°}{2} = 30°

A tangent to a circle is always perpendicular to the radius at the point of contact.

∴ AP ⊥ OA

⇒ ∠OAP = 90°

Since ∠OAP = ∠BAP, we have ∠BAP = 90°

In ΔABP, using the angle sum property:

∴ ∠ABP + ∠APB + ∠BAP = 180°

⇒ 30° + x + 90° = 180°

⇒ 30° + x = 180° – 90°

⇒ 30° + x = 90°.

Thus, both the assertion and the reason are true, and the reason correctly explains the assertion.

Hence, option 3 is the correct option.

Question 1(h)

Chords AD and BC one produce meet at exterior point P.

Assertion(A): PD x AD = PC x BC.

Reason(R): In triangles PAB and PCD.

∠PAB = ∠PCD ⇒ ΔPAB ∼ ΔPCD

• (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.

Consider triangles \Delta PAB and \Delta PCD.

Notice that \angle PBA = \angle PDC because exterior angles of a cyclic quadrilateral are equal to the interior opposite angles. Similarly, \angle PAB = \angle PCD for the same reason.

Thus, by the Angle-Angle (A.A.) similarity criterion, \Delta PAB \sim \Delta PCD.

Since corresponding sides of similar triangles are proportional, we have:

Therefore, the Reason (R) is indeed true.

However, from the proportionality of sides, it follows that:

This reveals that the Assertion (A) is false because it states PD \times AD = PC \times BC, which is incorrect.

∴ Assertion (A) is false, while Reason (R) is true.

Hence, option 2 is the correct option.

Question 1(i)

Two circles touch each other externally at point P. OA and OB are the tangent of the two circles (as shown) and OA = 10 cm.

Statement (1): OB = 10 cm.

Statement (2): On joining O and P, tangent OP = tangent OA and tangent OP = tangent OB

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

Consider joining the point O to point P.

Recall that when two tangents are drawn from a common external point to a circle, the tangents are equal in length.

In the figure, point O is the external point from which tangents OA and OP are drawn to the circle with center Q’. Therefore, we have:

OA = OP \quad \text{…(1)}

Similarly, from the same point O, tangents OB and OP are drawn to the circle with center Q. Thus, we have:

OB = OP \quad \text{…(2)}

From equations (1) and (2), it follows that:

⇒ OA = OB

Given that OA = 10 cm, it implies:

⇒ OB = 10 cm

∴ Both statements are indeed true.

Hence, option 1 is the correct option.

Question 1(j)

O is centre of the circle, PB and PC are tangents and ∠BPC = 50°.

Statement (1): ∠BAC = ∠P = 50°

Statement (2): ∠BOC + 50° = 180°

⇒ ∠BOC = 130°

∴ ∠BAC = 65°

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

In a circle, the tangent at any point is always perpendicular to the radius at that point of contact.

∴ OB ⊥ BP and OC ⊥ CP

⇒ ∠OBP = 90° and ∠OCP = 90°

Consider the quadrilateral OCPB. The sum of its interior angles is 360°.

∴ ∠OBP + ∠BPC + ∠OCP + ∠BOC = 360°

Substituting the known angles:

⇒ 90° + 50° + 90° + ∠BOC = 360°

⇒ 230° + ∠BOC = 360°

⇒ ∠BOC = 360° – 230°

⇒ ∠BOC = 130°

Now, recall the property of circles that the angle subtended by an arc at the center is twice the angle subtended on the circle’s circumference.

∴ ∠BOC = 2 × ∠BAC

⇒ 130° = 2 × ∠BAC

⇒ ∠BAC = \frac{130°}{2} = 65°.

Thus, Statement 1 is incorrect, whereas Statement 2 holds true.

Hence, option 4 is the correct option.

Question 2

Prove that, of any two chords of a circle, the greater chord is nearer to the centre.

Consider a circle with centre O and radius r. Let AB and CD be two chords in this circle. Assume OM < ON, where OM ⊥ AB and ON ⊥ CD.

We aim to demonstrate: AB > CD.

Recall that the perpendicular from the centre of a circle to a chord bisects the chord.

∴ AM = \dfrac{AB}{2} and CN = \dfrac{CD}{2}

Connect OA and OC.

In the right-angled triangle OAM:

⇒ OA^2 = AM^2 + OM^2

⇒ AM^2 = OA^2 - OM^2 ….. (1)

In the right-angled triangle ONC:

⇒ OC^2 = CN^2 + ON^2

⇒ CN^2 = OC^2 - ON^2 ….. (2)

Since OM < ON:

⇒ OM^2 < ON^2

⇒ -OM^2 > -ON^2

⇒ OA^2 - OM^2 > OC^2 - ON^2 [∵ OA = OC]

⇒ AM^2 > CN^2 [From (1) and (2)]

⇒ \Big(\dfrac{\text{AB}}{2}\Big)^2 > \Big(\dfrac{\text{CD}}{2}\Big)^2

⇒ \dfrac{1}{4}AB^2 > \dfrac{1}{4}CD^2

⇒ AB^2 > CD^2

⇒ AB > CD

Hence proved that, of any two chords of a circle, the greater chord is nearer to the centre.

Question 3

ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = 110° and angle BAC = 50°. Find angle DAC and angle DCA.

In the cyclic quadrilateral ABCD, where AD is parallel to BC, we are given that ∠ADC = 110° and ∠BAC = 50°.

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

∴ ∠B + ∠D = 180°

⇒ ∠B + 110° = 180°

⇒ ∠B = 180° – 110°

⇒ ∠B = 70°.

Consider the triangle ABC:

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

⇒ ∠BAC + ∠ABC + ∠ACB = 180°

⇒ 50° + 70° + ∠ACB = 180°

⇒ ∠ACB = 180° – 120° = 60°

Since AD is parallel to BC, alternate angles are equal.

∴ ∠DAC = ∠ACB = 60°

Now, in triangle ADC:

The angle sum property gives us:

⇒ ∠DAC + ∠ADC + ∠DCA = 180°

⇒ 60° + 110° + ∠DCA = 180°

⇒ ∠DCA = 180° – 170° = 10°

Hence, ∠DAC = 60° and ∠DCA = 10°.

Question 4

In the given figure, C and D are points on the semi-circle described on AB as diameter.

Given angle BAD = 70° and angle DBC = 30°, calculate angle BDC.

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

∴ ∠BCD + ∠BAD = 180°

⇒ ∠BCD + 70° = 180°

⇒ ∠BCD = 180° – 70° = 110°

Now, consider triangle BCD. By the angle sum property, the sum of angles in a triangle is 180°.

⇒ ∠CBD + ∠BCD + ∠BDC = 180°

⇒ 30° + 110° + ∠BDC = 180°

⇒ ∠BDC = 180° – 140° = 40°.

Hence, ∠BDC = 40°.

Question 5

In cyclic quadrilateral ABCD, ∠A = 3∠C and ∠D = 5∠B. Find the measure of each angle of the quadrilateral.

In cyclic quadrilateral ABCD, it’s known that opposite angles are supplementary, meaning their sum is 180°. Given that ∠A = 3∠C, we have:

Substituting for ∠A, we get:

This simplifies to:

⇒ ∠C = \dfrac{180°}{4}

⇒ ∠C = 45°.

Now, substituting back to find ∠A:

⇒ ∠A = 3∠C = 3 \times 45° = 135°.

For the angles ∠B and ∠D, we know:

Given ∠D = 5∠B, we substitute:

This simplifies to:

⇒ ∠B = \dfrac{180°}{6}

⇒ ∠B = 30°.

Finally, we find ∠D:

⇒ ∠D = 5∠B = 5 \times 30° = 150°.

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

Question 6

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

Consider a circle drawn with one of the equal sides, AB, of the isosceles triangle ABC as its diameter, as illustrated in the diagram.

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

Since ∠ADB and ∠ADC form a linear pair:

⇒ ∠ADB + ∠ADC = 180°

Substituting the known value:

⇒ 90° + ∠ADC = 180°

Solving for ∠ADC gives:

⇒ ∠ADC = 180° – 90°

⇒ ∠ADC = 90°.

In triangles ∆ABD and ∆ACD, observe the following:

⇒ ∠ADB = ∠ADC, both are 90°.

⇒ AB = AC, as given.

⇒ AD is common to both triangles.

Therefore, by the RHS congruence criterion, ∆ABD ≅ ∆ACD.

From the congruence, corresponding parts are equal, so:

BD = DC

Hence, the circle bisects base BC at D.

Question 7

Bisectors of vertex angles A, B and C of a triangle ABC intersect its circumcircle at points D, E and F respectively. Prove that angle EDF = 90° – \dfrac{1}{2}∠A.

Consider the bisector BE of angle ∠B. This implies:

⇒ ∠ABE = \dfrac{∠B}{2}

Looking at the diagram, notice that:

⇒ ∠ADE = ∠ABE \text{(since angles subtended by the same arc are equal)}

Thus, we have:

⇒ ∠ADE = \dfrac{∠B}{2} \text{...(1)}

Similarly, since FC bisects ∠C, we have:

⇒ ∠ACF = \dfrac{∠C}{2}

By the same arc property, ∠ACF = ∠ADF, therefore:

⇒ ∠ADF = \dfrac{∠C}{2} \text{...(2)}

Now, from the diagram, we can write:

⇒ ∠D = ∠ADE + ∠ADF

Substituting from equations (1) and (2), we get:

⇒ ∠D = \dfrac{∠B + ∠C}{2} \text{...(3)}

In triangle ABC, the sum of angles is:

⇒ ∠A + ∠B + ∠C = 180°

This implies:

⇒ ∠B + ∠C = 180° – ∠A

Substituting this into equation (3), we find:

⇒ ∠D = \dfrac{180° - ∠A}{2}

⇒ ∠D = 90° – \dfrac{1}{2}∠A

Finally, from the diagram, we observe:

∠EDF = ∠D.

Thus, it is proved that ∠EDF = 90° - \dfrac{1}{2}∠A.

Question 8

In the figure, AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠C = 20°, find angle AOD.

To solve this, start by drawing a line from O to B.

We know from the problem that:

BC = OD = radius of the circle

⇒ BC = OB.

Since BC and OB are equal, the angles opposite these sides must also be equal.

⇒ ∠BOC = ∠BCO = 20°.

For any triangle, the exterior angle is equal to the sum of the two opposite interior angles.

∴ ∠ABO = ∠BCO + ∠BOC = 20° + 20° = 40° …………(1)

Consider triangle OAB:

OA = OB [because they are radii of the same circle]

This implies that the angles opposite these equal sides are also equal.

∠OAB = ∠ABO = 40° [from equation (1)]

Using the angle sum property of triangles:

⇒ ∠AOB + ∠OAB + ∠OBA = 180°

⇒ ∠AOB + 40° + 40° = 180°

⇒ ∠AOB + 80° = 180°

⇒ ∠AOB = 180° – 80°

⇒ ∠AOB = 100°

Since DOC forms a straight line, we have:

⇒ ∠AOD + ∠AOB + ∠BOC = 180°

⇒ ∠AOD + 100° + 20° = 180°

⇒ ∠AOD = 180° – 120° = 60°.

Hence, ∠AOD = 60°.

Question 9

P is the midpoint of an arc APB of a circle. Prove that the tangent drawn at P will be parallel to the chord AB.

We understand that the angle formed between a tangent and a chord at the point of contact is equal to the angle in the opposite segment.

Refer to the figure provided:

Given that TPS is a tangent, and PA is a chord of the circle, we have:

∠BPT = ∠PAB [Angles in alternate segments are equal] ……….(1)

Additionally, since P is the midpoint of arc APB, we know:

∠PBA = ∠PAB [As PA = PB] ……..(2)

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

∠BPT = ∠PBA

These angles are alternate angles, which implies:

∴ TPS || AB

Hence, proved that the tangent drawn at P will be parallel to the chord AB.

Question 10

In the given figure, ABCD is a cyclic quadrilateral, PQ is tangent to the circle at point C and BD is its diameter. If ∠DCQ = 40° and ∠ABD = 60°, find:

(i) ∠DBC

(ii) ∠BCP

(iii) ∠ADB

(i) According to the theorem of tangents and alternate segments, the angle formed between a tangent and a chord at the point of contact equals the angle in the opposite segment. In this case, since PQ is tangent and CD is the chord, we have:

∴ ∠DBC = ∠DCQ.

Given ∠DCQ = 40°, it follows that:

∠DBC = 40°.

Hence, ∠DBC = 40°.

(ii) Consider △DCB:

The sum of angles in a triangle is 180°:

∠DBC + ∠DCB + ∠CDB = 180°.

Since ∠DCB = 90° (angle in a semicircle is a right angle), substituting the known values gives:

40° + 90° + ∠CDB = 180°.

Solving for ∠CDB, we find:

∠CDB = 180° – 130° = 50°.

By the property of alternate segment angles, we conclude:

∠BCP = ∠CDB = 50°.

Hence, ∠BCP = 50°.

(iii) In △ABD:

Since BD is the diameter, ∠BAD = 90° (angle in a semicircle is a right angle).

Given ∠ABD = 60°, we apply the angle sum property:

∠ADB + ∠BAD + ∠ABD = 180°.

Substituting the known angles:

∠ADB + 90° + 60° = 180°.

This simplifies to:

∠ADB = 180° – 150° = 30°.

Hence, ∠ADB = 30°.

Question 11

The given figure shows a circle with centre O and BCD is a tangent to it at C. Show that : ∠ACD + ∠BAC = 90°.

Connect OC.

We know that the radius drawn to a tangent at the point of contact is perpendicular to the tangent. Here, BCD serves as the tangent and OC is the radius.

Thus, OC ⊥ BD, which gives us:

∠OCD = 90°

∴ ∠OCA + ∠ACD = 90° …………. (1)

Considering ∆OCA,

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

∴ ∠OCA = ∠OAC [Angles opposite equal sides in a triangle are equal]

Substituting this into equation (1), we obtain:

⇒ ∠OAC + ∠ACD = 90°

⇒ ∠BAC + ∠ACD = 90° [From the figure, ∠BAC = ∠OAC]

Therefore, it is demonstrated that ∠ACD + ∠BAC = 90°.

Question 12

ABC is a right triangle with angle B = 90°. A circle with BC as diameter meets hypotenuse AC at point D. Prove that:

(i) AC x AD = AB^2

(ii) BD^2 = AD x DC.

(i) In the right triangle \triangle ABC, we know \angle B = 90^\circ and BC serves as the diameter of the circle. This implies that AB acts as a tangent to the circle at point B.

According to the tangent-segment theorem, when a tangent and a chord intersect externally, the square of the tangent’s length from the point of contact to the intersection point equals the product of the chord’s segments.

∴ AB^2 = AD \times AC.

Thus, it is proven that AB^2 = AD \times AC.

(ii) Observe from the diagram that \angle BDC = 90^\circ because an angle inscribed in a semicircle is a right angle.

Also, from the figure, we have:

⇒ \angle ADB + \angle BDC = 180^\circ [Linear pair]

⇒ \angle ADB + 90^\circ = 180^\circ

⇒ \angle ADB = 180^\circ - 90^\circ

⇒ \angle ADB = 90^\circ

In \triangle ADB:

⇒ \angle ADB + \angle A + \angle ABD = 180^\circ [Angle sum property of a triangle]

⇒ 90^\circ + \angle A + \angle ABD = 180^\circ

⇒ \angle A + \angle ABD = 90^\circ ……………(1)

In \triangle ABC, \angle ABC = 90^\circ.

⇒ \angle ABC + \angle A + \angle ACB = 180^\circ [Angle sum property of a triangle]

⇒ 90^\circ + \angle A + \angle ACB = 180^\circ

⇒ \angle A + \angle ACB = 90^\circ ……………(2)

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

⇒ \angle A + \angle ABD = \angle A + \angle ACB

⇒ \angle ABD = \angle ACB.

From the diagram, it follows that:

⇒ \angle ACB = \angle BCD

∴ \angle ABD = \angle BCD

Now, consider \triangle ABD and \triangle CBD:

Both have \angle BDA = \angle BDC = 90^\circ.

And \angle ABD = \angle BCD.

Thus, by the AA criterion, \triangle ABD \sim \triangle CBD.

For similar triangles, the ratios of corresponding sides are equal.

∴ BD^2 = AD \times DC.

Therefore, it is established that BD^2 = AD \times DC.

Question 13

In the given figure, AC = AE.

Show that :

(i) CP = EP

(ii) BP = DP

(i) Consider the triangles \triangle ADC and \triangle ABE.

⇒ \angle ACD = \angle AEB because angles in the same segment are equal.

⇒ AC = AE as given.

⇒ \angle A = \angle A since it’s a common angle.

Thus, \triangle ADC \cong \triangle ABE by the ASA congruence criterion.

By C.P.C.T (Corresponding Parts of Congruent Triangles), we get:

⇒ AD = AB …………..(1)

We know:

⇒ AE = AC ………….(2)

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

⇒ AE - AD = AC - AB

⇒ DE = BC

Now, consider \triangle BPC and \triangle DPE.

⇒ \angle C = \angle E due to angles in the same segment being equal.

⇒ BC = DE as proven above.

⇒ \angle CBP = \angle PDE because angles in the same segment are equal.

Thus, \triangle BPC \cong \triangle DPE by the ASA criterion.

By C.P.C.T, we conclude:

⇒ CP = EP

Hence, proved that CP = EP.

(ii) As established above,

∴ BP = DP by C.P.C.T

Hence, proved that BP = DP.

Question 14

ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC = 120°.

Calculate :

(i) ∠BEC

(ii) ∠BED

Cyclic pentagon ABCDE with its circumcircle with centre at point O is shown in the figure below:

(i) Given that AB = BC = CD and ∠ABC = 120°, it follows that ∠BCD = ∠ABC = 120° as well.

The lines OB and OC bisect ∠ABC and ∠BCD, respectively. Thus, ∠OBC = ∠BCO = 60°.

In the triangle BOC, applying the angle sum property:

∴ ∠OBC + ∠BCO + ∠BOC = 180°

⇒ 60° + 60° + ∠BOC = 180°

⇒ ∠BOC = 180° – 120° = 60°.

The arc BC creates ∠BOC at the center and ∠BEC at the circle’s circumference.

Recall that an arc’s angle at the center is twice the angle it subtends at any point on the circle.

∴ ∠BEC = \dfrac{1}{2}∠BOC = \dfrac{1}{2} × 60° = 30°.

Thus, ∠BEC = 30°.

(ii) In the cyclic quadrilateral BCDE, the opposite angles sum to 180°:

∴ ∠BED + ∠BCD = 180°

⇒ ∠BED + 120° = 180°

⇒ ∠BED = 180° – 120°

⇒ ∠BED = 60°.

Thus, ∠BED = 60°.

Question 15

In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If ∠ACO = 30°, find:

(i) ∠BCO

(ii) ∠AOB

(iii) ∠APB

(i) Consider triangles \triangle OAC and \triangle OBC. Notice:

⇒ OC = OC [Common side]

⇒ OA = OB [Both are radii of the circle]

⇒ CA = CB [Tangents drawn from an external point C to a circle are equal]

Thus, \triangle OAC \cong \triangle OBC by the SSS (Side-Side-Side) congruence rule.

∴ \angle ACO = \angle BCO = 30^\circ [Corresponding Parts of Congruent Triangles]

Hence, \angle BCO = 30^\circ.

(ii) From the diagram, we can write:

\angle ACB = \angle ACO + \angle BCO = 30^\circ + 30^\circ = 60^\circ.

In a cyclic quadrilateral, the sum of opposite angles is 180^\circ.

⇒ \angle AOB + \angle ACB = 180^\circ

⇒ \angle AOB = 180^\circ - 60^\circ = 120^\circ.

Hence, \angle AOB = 120^\circ.

(iii) The arc AB creates \angle AOB at the center and \angle APB at the circumference.

We know:

The angle at the center is twice the angle at the circumference for the same arc.

⇒ \angle APB = \dfrac{1}{2}\angle AOB = \dfrac{1}{2} \times 120^\circ = 60^\circ.

Hence, \angle APB = 60^\circ.

Question 16

The given figure shows a semi-circle with center O and diameter PQ. If PA = AB and ∠BCQ = 140°; find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.

To solve this, let’s connect point B to point P.

In the cyclic quadrilateral PBCQ, we have:

∠BPQ + ∠BCQ = 180° because the opposite angles of a cyclic quadrilateral sum up to 180°.

⇒ ∠BPQ + 140° = 180°

⇒ ∠BPQ = 180° – 140°

⇒ ∠BPQ = 40° …………(1)

Now, consider △PBQ:

Since ∠PBQ = 90° (as the angle in a semi-circle is a right angle), we can write:

⇒ ∠PBQ + ∠BPQ + ∠PQB = 180° by the angle sum property of triangles.

⇒ 90° + 40° + ∠PQB = 180°

⇒ 130° + ∠PQB = 180°

⇒ ∠PQB = 180° – 130°

⇒ ∠PQB = 50°.

Next, in the cyclic quadrilateral PQBA:

⇒ ∠PQB + ∠PAB = 180° because opposite angles of a cyclic quadrilateral sum to 180°.

⇒ 50° + ∠PAB = 180°

⇒ ∠PAB = 180° – 50°

⇒ ∠PAB = 130°.

Now, in △PAB:

⇒ ∠PAB + ∠PBA + ∠BPA = 180° by the angle sum property of triangles.

⇒ 130° + ∠PBA + ∠BPA = 180°

⇒ ∠PBA + ∠BPA = 180° – 130°

⇒ ∠PBA + ∠BPA = 50° …………….(2)

Given that PA = PB, the angles opposite to equal sides are equal.

∴ ∠PBA = ∠BPA = x (let)

Substituting in equation (2), we get:

⇒ x + x = 50°

⇒ 2x = 50°

⇒ x = dfrac{50°}{2}

⇒ x = 25°.

From the diagram, we see:

∠AQB = ∠APB = 25° because angles in the same segment are equal.

Now, ∠APQ = ∠APB + ∠BPQ = 25° + 40° = 65°.

We know that the angle subtended by an arc at the center is twice the angle subtended at any other point on the circumference.

Thus, arc AQ subtends ∠AOQ at the center and ∠APQ at the circle’s circumference.

∠AOQ = 2∠APQ = 2 × 65° = 130°.

In △AOQ, since OA = OQ (radii of the same circle), the angles opposite these equal sides are equal.

Let ∠OAQ = ∠OQA = y.

⇒ ∠OAQ + ∠OQA + ∠AOQ = 180° by the angle sum property of triangles.

⇒ y + y + 130° = 180°

⇒ 2y = 180° – 130°

⇒ 2y = 50°

⇒ y = dfrac{50°}{2}

⇒ y = 25°.

∴ ∠OAQ = 25°.

Since ∠OAQ = ∠AQB = 25°, these are alternate angles.

Therefore, AO is parallel to BQ.

Hence, ∠AQB = 25° and ∠PAB = 130°.

Question 17

The given figure shows a circle with center O such that chord RS is parallel to chord QT, angle PRT = 20° and angle POQ = 100°. Calculate :

(i) angle QTR

(ii) angle QRP

(iii) angle QRS

(iv) angle STR

(i) Observing the figure, notice that ∠POQ and ∠QOR form a linear pair.

∴ ∠POQ + ∠QOR = 180°.

Given ∠POQ = 100°, we find:

⇒ 100° + ∠QOR = 180°

⇒ ∠QOR = 180° – 100°

⇒ ∠QOR = 80°.

Recall that the angle an arc makes at the center is twice what it makes on the circle.

Arc RQ makes ∠QOR at the center and ∠QTR on the circle.

∴ ∠QOR = 2∠QTR

⇒ ∠QTR = \dfrac{1}{2}∠QOR = \dfrac{1}{2} \times 80° = 40°.

Thus, ∠QTR = 40°.

(ii) Similarly, the arc QP makes ∠QOP at the center and ∠QRP on the circle.

∴ ∠QOP = 2∠QRP

⇒ ∠QRP = \dfrac{1}{2}∠QOP = \dfrac{1}{2} \times 100° = 50°.

Thus, ∠QRP = 50°.

(iii) Given that RS is parallel to QT, we have:

∴ ∠SRT = ∠QTR = 40° (Alternate angles are equal)

Looking at the figure, we add:

∠QRS = ∠QRP + ∠PRT + ∠SRT = 50° + 20° + 40° = 110°.

Thus, ∠QRS = 110°.

(iv) Since RSTQ forms a cyclic quadrilateral, the opposite angles sum to 180°.

∴ ∠QRS + ∠QTS = 180°

And since ∠QTS = ∠QTR + ∠STR:

⇒ ∠QRS + ∠QTR + ∠STR = 180°

⇒ 110° + 40° + ∠STR = 180°

⇒ ∠STR = 180° – 150°

⇒ ∠STR = 30°.

Thus, ∠STR = 30°.

Question 18

TA and TB are tangents to a circle with center O from an external point T. OT intersects the circle at point P. Prove that AP bisects the angle TAB.

We have tangents TA and TB drawn from an external point T to a circle whose center is O.

Let’s examine triangles ΔOAT and ΔOBT.

• OA and OB are equal as they are radii of the circle.
• OT is a common side to both triangles.
• TA equals TB because tangents drawn from an external point to a circle are of equal length.

∴ ΔOAT ≅ ΔOBT by the SSS congruence criterion.

From this congruence, we know ∠ATO = ∠BTO by C.P.C.T. (Corresponding Parts of Congruent Triangles).

⇒ From the figure, ∠ATO is the same as ∠ATP and ∠BTO is the same as ∠BTP, so ∠ATP = ∠BTP.

Now consider triangles ΔAPT and ΔBPT:

• AT equals BT, as tangents from a common external point are equal.
• PT is a shared side.
• ∠ATP equals ∠BTP, as shown earlier.

∴ ΔAPT ≅ ΔBPT by the SAS congruence criterion.

From this, ∠PAT = ∠PBT by C.P.C.T. and also AP = BP by C.P.C.T.

In triangle ΔPAB:

• ∠PAB = ∠PBA because angles opposite to equal sides are equal.
• ∠PAT = ∠PBA since angles in alternate segments are equal.

∴ ∠PAB = ∠PAT.

Thus, AP bisects ∠TAB.

Hence, proved that AP bisects ∠TAB.

Question 19

Chords AB and CD of a circle when extended meet at point X. Given AB = 4 cm, BX = 6 cm and XD = 5 cm, calculate the length of CD.

Consider the chords AB and CD of a circle, which when extended, meet at a point X. In such scenarios, a key geometric property states that the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.

∴ XB \times XA = XD \times XC

Given: AB = 4 cm, BX = 6 cm, and XD = 5 cm, we substitute these into the formula:

⇒ 6 \times (6 + 4) = 5 \times (5 + CD)

⇒ 6 \times 10 = 25 + 5CD

⇒ 60 = 25 + 5CD

Rearranging gives:

⇒ 5CD = 60 – 25

⇒ 5CD = 35

Dividing both sides by 5, we find:

⇒ CD = \dfrac{35}{5}

⇒ CD = 7 \text{ cm}.

Hence, CD = 7 cm.

Question 20

In the following figure, a circle is inscribed in the quadrilateral ABCD. If BC = 38 cm, QB = 27 cm, DC = 25 cm and that AD is perpendicular to DC, find the radius of the circle.

Consider the given figure.

∴ BR = BQ = 27 cm, as the tangents drawn from an external point to a circle are equal in length.

Now, calculate CR:

⇒ CR = BC – BR = 38 – 27 = 11 cm.

Since CR = CS, we have:

⇒ CR = CS = 11 cm, again due to the property of tangents from an external point.

Next, find DS:

⇒ DS = DC – CS = 25 – 11 = 14 cm.

In the quadrilateral DSOP, the sum of the interior angles is:

⇒ ∠SDP + ∠DPO + ∠OSD + ∠POS = 360°

Given that AD is perpendicular to DC, we have three right angles:

⇒ 90° + 90° + 90° + ∠POS = 360°

Solving for ∠POS:

⇒ ∠POS = 360° – 270° = 90°.

All angles in the quadrilateral DPOS are right angles, and OS = OP because they are radii of the same circle.

This confirms that DPOS is a square.

Thus, OP = DS = 14 cm.

Hence, radius of circle = 14 cm.

Question 21

In the given figure, XY is the diameter of the circle and PQ is a tangent to the circle at Y.

If ∠AXB = 50° and ∠ABX = 70°, find ∠BAY and ∠APY.

Consider △AXB. By applying the angle sum property of a triangle, we have:

⇒ ∠AXB + ∠XAB + ∠ABX = 180°

Substituting the given values:

⇒ 50° + XAB + 70° = 180°

Solving for ∠XAB gives:

⇒ ∠XAB = 180° – 120° = 60°.

From the figure, notice that ∠XAY is 90° since the angle in a semicircle is a right angle.

Thus, ∠BAY = ∠XAY – ∠XAB = 90° – 60° = 30°.

Furthermore, ∠BXY = ∠BAY = 30° because angles in the same segment are equal.

Now, consider the exterior angle property:

An exterior angle equals the sum of two opposite interior angles:

⇒ ∠ACX = ∠BXC + ∠CBX

In the figure, we observe that ∠BXC = ∠BXY and ∠CBX = ∠ABX. Therefore:

⇒ ∠ACX = ∠BXY + ∠ABX

⇒ ∠ACX = 30° + 70° = 100°.

Since the diameter is perpendicular to the tangent, we know:

⇒ ∠XYP = 90°

Using the exterior angle property in the triangle, we have:

⇒ ∠ACX = ∠APY + ∠CYP

Solving for ∠APY gives:

⇒ ∠APY = ∠ACX – ∠CYP = 100° – 90° = 10°.

Hence, ∠APY = 10° and ∠BAY = 30°.

Question 22

In the given figure, QAP is the tangent at point A and PBD is a straight line.

If ∠ACB = 36° and ∠APB = 42°, find :

(i) ∠BAP

(ii) ∠ABD

(iii) ∠QAD

(iv) ∠BCD

(i) Notice that the angle formed between a tangent and a chord at the point of contact equals the angle in the opposite segment of the circle. Thus, ∠BAP is equal to ∠ACB, which is 36°.

Hence, ∠BAP = 36°.

(ii) In any triangle, an exterior angle is the sum of the two non-adjacent interior angles. So, in △APB, ∠ABD is the sum of ∠APB and ∠BAP. Therefore, ∠ABD = 42° + 36° = 78°.

Hence, ∠ABD = 78°.

(iii) From the diagram, ∠ADB is equal to ∠ACB, as angles in the same segment are equal. Thus, in △PAD, the exterior angle ∠QAD equals the sum of ∠APB and ∠ADB, which gives us ∠QAD = 42° + 36° = 78°.

Hence, ∠QAD = 78°.

(iv) The angle between a tangent and a chord through the point of contact equals the angle in the alternate segment, so ∠ACD equals ∠QAD, which is 78°. From the figure, ∠BCD is the sum of ∠ACB and ∠ACD, thus ∠BCD = 36° + 78° = 114°.

Hence, ∠BCD = 114°.

Question 23

In the given figure, AB is the diameter. The tangent at C meets AB produced at Q.

If ∠CAB = 34°, find :

(i) ∠CBA

(ii) ∠CQB

(i) Observing the figure, we know that ∠ACB = 90° because the angle subtended by a diameter in a semicircle is always a right angle.

In △ACB, we apply the angle sum property:

∴ ∠ACB + ∠CAB + ∠CBA = 180°

Substituting the known values:

⇒ 90° + 34° + ∠CBA = 180°

Solving for ∠CBA:

⇒ ∠CBA = 180° – 124° = 56°.

Hence, ∠CBA = 56°.

(ii) Referring to the figure again, we see that ∠QCB is equal to ∠CAB, which is 34°, due to the property of angles in alternate segments being equal.

Now, considering the linear pair formed by ∠CBQ and ∠CBA:

∴ ∠CBQ + ∠CBA = 180°

Inserting the known angle:

⇒ ∠CBQ + 56° = 180°

Therefore, ∠CBQ = 124°.

Looking at △CBQ, we use the angle sum property:

⇒ ∠CBQ + ∠QCB + ∠CQB = 180°

Plugging in the angles we have:

⇒ 124° + 34° + ∠CQB = 180°

Solving for ∠CQB:

⇒ ∠CQB = 180° – 158° = 22°.

Hence, ∠CQB = 22°.

Question 24

In the given figure, O is the center of the circle. The tangents at B and D intersect each other at point P. If AB is parallel to CD and ∠ABC = 55°, find :

(i) ∠BOD

(ii) ∠BPD

(i) In the given figure, since AB is parallel to CD, we have ∠BCD = ∠ABC = 55° due to the property of alternate angles being equal.

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

∴ ∠BOD = 2 × ∠BCD = 2 × 55° = 110°.

Hence, ∠BOD = 110°.

(ii) A tangent to a circle forms a 90° angle with the radius at the point of contact. Thus, ∠OBP = 90° and ∠ODP = 90°.

Considering quadrilateral ODPB, the sum of interior angles is 360°.

⇒ ∠BOD + ∠OBP + ∠ODP + ∠BPD = 360°

⇒ 110° + 90° + 90° + ∠BPD = 360°

⇒ ∠BPD = 360° – 290° = 70°.

Hence, ∠BPD = 70°.

Question 25

In the following figure, PQ = QR, ∠RQP = 68°, PC and CQ are tangents to the circle with center O.

Calculate the values of :

(i) ∠QOP

(ii) ∠QCP

(i) Here, since PQ = QR, the angles opposite these sides in triangle PQR are equal.

∴ ∠PRQ = ∠QPR.

In △PQR, the sum of angles is 180°:

⇒ ∠PRQ + ∠QPR + ∠RQP = 180°

⇒ ∠PRQ + ∠PRQ + 68° = 180°

⇒ 2∠PRQ = 180° – 68°

⇒ 2∠PRQ = 112°

⇒ ∠PRQ = \dfrac{112°}{2}

⇒ ∠PRQ = 56°.

Notice that the angle subtended by an arc at the centre of the circle is twice that subtended at any other point on the circle:

∴ ∠QOP = 2∠PRQ = 2 \times 56° = 112°.

Hence, ∠QOP = 112°.

(ii) Recall that a tangent to a circle is perpendicular to the radius at the point of contact.

∴ ∠OPC = 90° and ∠OQC = 90°.

In quadrilateral OQCP, the sum of angles is 360°:

⇒ ∠QOP + ∠OPC + ∠OQC + ∠QCP = 360°

⇒ 112° + 90° + 90° + ∠QCP = 360°

⇒ ∠QCP = 360° – 292° = 68°.

Hence, ∠QCP = 68°.

Question 26

In the figure, given below, AC is a transverse common tangent to two circles with centers P and Q and of radii 6 cm and 3 cm respectively. Given that AB = 8 cm, calculate PQ.

Since AC acts as a tangent to the circle centered at P at point A, we have:

∴ ∠PAB = 90°.

Similarly, AC is a tangent to the circle with center Q at point C, so:

∴ ∠QCB = 90°.

In triangles △PAB and △QCB, we notice:

⇒ ∠PAB = ∠QCB (Both are right angles)

⇒ ∠PBA = ∠QBC (Since they are vertically opposite angles)

Thus, △PAB is similar to △QCB.

In right-angled triangle △PAB, by applying the Pythagorean theorem, we get:

Calculating further:

\Rightarrow \text{PB} = \sqrt{\text{PA}^2 + \text{AB}^2}
\Rightarrow \text{PB} = \sqrt{6^2 + 8^2}
\Rightarrow \text{PB} = \sqrt{36 + 64}
\Rightarrow \text{PB} = \sqrt{100}
\Rightarrow \text{PB} = 10 \text{ cm}

We know that in similar triangles, the ratio of corresponding sides is equal:

\Rightarrow \frac{PA}{QC} = \frac{PB}{QB}
\Rightarrow \frac{6}{3} = \frac{10}{QB}
\Rightarrow QB = \frac{3 \times 10}{6}
\Rightarrow QB = \frac{30}{6}
\Rightarrow QB = 5 \text{ cm}.

From the figure, we can see:

QP = QB + PB = 5 + 10 = 15 cm.

Hence, QP = 15 cm.

Question 27

In the figure, given below, O is the center of the circumcircle of triangle XYZ. Tangents at X and Y intersect at point T. Given ∠XTY = 80° and ∠XOZ = 140°, calculate the value of ∠ZXY.

Since YT and XT are tangents to the circle, we have:

∴ ∠OYT = 90° and ∠OXT = 90°.

Now, consider the quadrilateral OYTX. The sum of the interior angles of a quadrilateral is 360°.

⇒ ∠XOY + ∠OYT + ∠OXT + ∠XTY = 360°

Substituting the known values:

⇒ ∠XOY + 90° + 90° + 80° = 360°

⇒ ∠XOY = 360° – 260° = 100°.

Looking at the figure, we have:

⇒ ∠XOZ + ∠YOZ + ∠XOY = 360°

Substituting the known angles:

⇒ 140° + ∠YOZ + 100° = 360°

⇒ ∠YOZ = 360° – 240° = 120°.

According to the circle theorem, the angle subtended at the center is twice the angle subtended at the circumference by the same arc.

∴ ∠YOZ = 2∠ZXY

Thus, we find:

⇒ ∠ZXY = \dfrac{1}{2}∠YOZ = \dfrac{1}{2} \times 120° = 60°.

Hence, ∠ZXY = 60°.

Question 28

In the given figure, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, find AE.

Since we are given that ∠CDE = 90°, its vertically opposite angle ∠ADB is also 90°. This makes triangle ABD a right-angled triangle. Now, we can apply the Pythagorean theorem:

Substitute the given values:

This leads to:

By rearranging, we find:

Taking the square root gives:

Now, using the property of intersecting chords, where the product of the segments of one chord equals the product of the segments of the other chord:

Substituting the known values:

Solving for DE:

From the figure, we add AD and DE to find AE:

Hence, AE = 15 cm.

Question 29

In the given circle with centre O, angle ABC = 100°, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.

Observe the diagram:

∴ ∠ADC + ∠ABC = 180° (because the sum of opposite angles in a cyclic quadrilateral is always 180°)

∴ ∠ADC + 100° = 180°

⇒ ∠ADC = 180° – 100°

⇒ ∠ADC = 80°.

Now, in △ADC,

∴ ∠ADC + ∠CAD + ∠ACD = 180° (using the angle sum property of triangles)

⇒ 80° + ∠CAD + 40° = 180°

⇒ ∠CAD = 180° – 120°

⇒ ∠CAD = 60°.

From the diagram, notice that:

∴ ∠DCT = ∠CAD = 60° (since angles in alternate segments are equal).

Therefore, ∠DCT = 60° and ∠ADC = 80°.

Question 30

In the figure given below, O is the center of the circle and SP is a tangent. If ∠SRT = 65°, find the values of x, y and z.

Since SP is a tangent to the circle, it implies that ∠TSR is 90°.

Consider △TSR:

∴ ∠TSR + ∠STR + ∠SRT = 180° (using the angle sum property of triangles)

⇒ 90° + x + 65° = 180°

⇒ x = 180° – 155° = 25°.

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 remaining part of the circle’s circumference.

∴ ∠SOQ = 2∠STQ

⇒ y = 2x = 2(25°) = 50°.

Now, look at △OSP:

∴ ∠OSP + ∠SOP + ∠SPO = 180° (again using the angle sum property of triangles)

⇒ 90° + y + z = 180°

⇒ 90° + 50° + z = 180°

⇒ z = 180° – 140° = 40°.

Thus, the values are x = 25°, y = 50°, and z = 40°.