Construction of Polygons Class 9 ICSE — Maths Chapter 15
This chapter, Construction of Polygons Class 9 ICSE, teaches you how to draw quadrilaterals, parallelograms, trapeziums, rectangles, rhombuses, squares and regular hexagons with a ruler and compasses. It is Chapter 15 of the Class 9 Mathematics book. Every construction follows a fixed order of steps. You locate vertices one at a time. Each shape type has its own data patterns, explained section by section on this page. The topics follow the official CISCE syllabus for Class 9 Mathematics.
ICSE Class 9 Mathematics Chapter 15 Construction of Polygons PDF
What is inside Chapter 15
7 pages · 7 sections · 39 exercise questions
| Section | Title |
|---|---|
| 15.1 | CONSTRUCTION OF QUADRILATERALS |
| 15.2 | CONSTRUCTION OF PARALLELOGRAMS |
| 15.3 | CONSTRUCTION OF TRAPEZIUM |
| 15.4 | CONSTRUCTION OF RECTANGLES |
| 15.5 | CONSTRUCTION OF RHOMBUS |
| 15.6 | CONSTRUCTION OF SQUARE |
| 15.7 | TO CONSTRUCT A REGULAR HEXAGON |
- EXERCISE 15: 39 questions
15.1 CONSTRUCTION OF QUADRILATERALS
To construct a quadrilateral means to locate its four vertices on the page. The textbook gives two ground rules before any construction. First, decide where each of the four corners belongs. Second, always draw a rough freehand sketch of the shape. The sketch shows which sides, angles and diagonals you have, and where each vertex should land.
Four data patterns make a quadrilateral constructible. Each pattern names the triangle you draw first. The freehand sketch tells you which pattern you are in before you pick up the compasses.
| Case | Data given | First move |
|---|---|---|
| 1 | Four sides, one angle | Draw base, set angle, cut side |
| 2 | Three sides, two consecutive angles | Draw base, set both angles |
| 3 | Four sides, one diagonal | Triangle ABC, then triangle ADC |
| 4 | Three sides, two diagonals | Triangle ABC, then triangle ABD |
Case 1 gives four sides and one angle. Draw the base, set the angle, cut the side, then intersect two arcs for the last vertex. Case 2 gives three sides and two consecutive angles. Draw the base, set both angles, cut one side, then arc from the third vertex.
Case 3 gives four sides and one diagonal. Construct triangle ABC first, then triangle ADC on the other side of the diagonal. Case 4 gives three sides and two diagonals. Construct triangle ABC, then triangle ABD, then join C and D. The shape properties behind these constructions come from the earlier chapter on rectilinear figures in the Class 9 book.
Worked pattern for case 1. Let AB = 4.2 cm, BC = 3.8 cm, CD = 5.0 cm, DA = 4.5 cm and \angle B = 50^\circ. Draw BC = 3.8 cm. Through B, draw a ray at 50^\circ. Cut BA = 4.2 cm on that ray.
With A as centre and radius 4.5 cm, draw an arc. With C as centre and radius 5.0 cm, draw a second arc. The arcs meet at D. Join AD and CD. ABCD is the required quadrilateral.
15.2 CONSTRUCTION OF PARALLELOGRAMS
A parallelogram has two properties that drive every construction. Opposite sides are equal. The diagonals bisect each other. The textbook uses these properties to build the shape from two triangles, or from half-diagonals at the centre.
Five data patterns appear in this section. Each names the triangle or the centre point you draw first.
- Two consecutive sides and the included angle: triangle ABC, then triangle ADC.
- One side and both diagonals: triangle OBC from half-diagonals, then extend to A and D.
- Two consecutive sides and one diagonal: triangle BCD, then triangle BAD.
- Two diagonals and the included angle: mid-point O, line BOD at the angle, cut equal halves.
- Two adjacent sides and the height: perpendicular at C, parallel line at the height, arcs for A and D.
Cases 2 and 4 use the half-diagonal step. Since the diagonals bisect each other, OB = OD = \frac{1}{2} BD and OA = OC = \frac{1}{2} AC. Construct triangle OBC first, then extend BO to D and CO to A.
Worked pattern for case 2. Let side BC = 4.0 cm, diagonal AC = 6.0 cm and diagonal BD = 5.0 cm. Then OB = \frac{1}{2} \times 5.0 = 2.5 cm and OC = \frac{1}{2} \times 6.0 = 3.0 cm. Draw triangle OBC. Extend BO to D so OD = 2.5 cm, and CO to A so OA = 3.0 cm. Join AB, AD and CD. ABCD is the required parallelogram.
The textbook flags one common mistake. If BD = 4.5 cm, then \frac{1}{2} BD = 2.25 cm. A standard scale cannot measure 2.25 cm correctly. The fix is to start with the other diagonal. Draw BD = 4.5 cm first, find its mid-point, then place AC across it at the given angle. This way you halve AC instead, which gives a measurable length.
Case 5 works from the height. Draw BC, raise a perpendicular at C, cut the height, and draw a line parallel to BC through that point. Arcs from B and C, each of radius AB, fix A and D on that parallel. The other parallelogram methods in this book are gathered on the Class 9 Mathematics book hub.
15.3 CONSTRUCTION OF TRAPEZIUM
This chapter covers one trapezium case: four sides given, with one pair of sides parallel. Let AD be parallel to BC. The key move is to cut BE = AD from BC. The leftover piece EC = BC - AD becomes one side of triangle DEC.
Triangle DEC carries three known lengths. DE = AB, CD is given, and EC = BC - AD. Once triangle DEC stands, two arcs find vertex A. With B as centre and radius AB, draw one arc. With D as centre and radius AD, draw another. The arcs meet at A. Join AB and AD.
Worked example. Let AD = 3.2 cm, AB = 2.8 cm, BC = 5.6 cm and CD = 3.0 cm, with AD parallel to BC. Draw BC = 5.6 cm. Cut BE = AD = 3.2 cm from BC. Then EC = 5.6 - 3.2 = 2.4 cm.
Build triangle DEC with DE = 2.8 cm, CD = 3.0 cm and EC = 2.4 cm. With B as centre and radius 2.8 cm, draw an arc. With D as centre and radius 3.2 cm, draw another arc. The arcs meet at A. Join AB and AD. ABCD is the required trapezium.
15.4 CONSTRUCTION OF RECTANGLES
A rectangle has two properties that make its construction simple. Each angle is 90°. Opposite sides are equal. The textbook splits the rectangle into two right-angled triangles along a diagonal.
Case 1 gives the adjacent sides. Draw triangle ABC from the two sides and the right angle at B. Then draw triangle ADC on the other side of the diagonal AC. The two right angles match the rectangle’s property that every corner is 90°.
Case 2 gives one side and one diagonal. Draw a right-angled triangle ABC first. The given side and diagonal form a leg and the hypotenuse. Then draw the second right-angled triangle ADC. The diagonal AC is the hypotenuse of both triangles.
Worked example for case 2. Let BC = 4.0 cm and diagonal AC = 5.5 cm. Draw BC = 4.0 cm. At B, erect a perpendicular. With C as centre and radius 5.5 cm, cut the perpendicular at A. Triangle ABC is right-angled at B.
Now find D. With A as centre and radius 4.0 cm, draw an arc. With C as centre and radius AB, draw another arc. The arcs meet at D. Join AD and CD. ABCD is the required rectangle.
15.5 CONSTRUCTION OF RHOMBUS
A rhombus’s diagonals bisect each other at right angles. This one property gives the whole construction. You draw one diagonal, bisect it perpendicularly, then place half of the second diagonal on each side of the crossing point.
The half lengths follow the rule OD = OB = \frac{1}{2} BD. The perpendicular bisector of AC cuts AC at O. From that bisector you cut OD and OB equal to half of BD.
Worked example. Let the diagonals be AC = 6.4 cm and BD = 4.8 cm. Then OA = OC = \frac{1}{2} \times 6.4 = 3.2 cm and OD = OB = \frac{1}{2} \times 4.8 = 2.4 cm.
Draw AC = 6.4 cm. Draw its perpendicular bisector, meeting AC at O. From the bisector, cut OD = OB = 2.4 cm. Join AB, BC, CD and DA. ABCD is the required rhombus.
The textbook closes with two practice prompts. What steps change if the diagonals are 6.3 cm and 4.6 cm? What if they are 6.2 cm and 4.7 cm? The steps stay identical. Only the half lengths change. Halve each diagonal and follow the same order.
15.6 CONSTRUCTION OF SQUARE
A square’s diagonals are equal and bisect each other at right angles. That is exactly the rhombus property. So the square whose diagonal is given uses the same method as the rhombus construction.
Worked example. Construct a square with diagonal 5.6 cm. Half the diagonal is \frac{1}{2} \times 5.6 = 2.8 cm. Draw AC = 5.6 cm. Draw its perpendicular bisector, meeting AC at O. Cut OB = OD = 2.8 cm. Join AB, BC, CD and DA. ABCD is the required square.
Why does the transfer work? A square is a rhombus whose angles are all 90°. Both shapes share the diagonal bisection property. The only difference is that a square’s diagonals are also equal, which the construction never needs to check.
15.7 TO CONSTRUCT A REGULAR HEXAGON
Three methods build a regular hexagon of a given side. Each rests on a different property of the shape. All three end with the same six equal sides.
Method 1 uses the interior angle. Each interior angle of a regular hexagon is 120°, and opposite sides are parallel. Draw AB. Through A and B, draw rays at 120°. Cut AF and BC equal to the side. Continue the pattern at C and F. Join the final two vertices.
Method 2 uses the circumcircle. The side of a regular hexagon equals the radius of its circumcircle. Draw a circle of the given radius. Step arcs of the same radius around the circumference, dividing it into six equal parts. Join the six points.
Method 3 uses the central angle. Each side subtends \frac{360^{\circ}}{6} = 60^\circ at the centre. Construct an isosceles triangle AOB with base AB equal to the side and \angle AOB = 60^\circ. Draw the circumcircle. Step arcs of length AB around it to find C, D, E and F. Join them.
Worked example with Method 3. Let the side be 2.8 cm. Construct isosceles triangle AOB with AB = 2.8 cm and \angle AOB = 60^\circ. With O as centre and OA as radius, draw the circle.
With B as centre and radius 2.8 cm, cut the circle at C. Continue stepping around to find D, E and F. Join BC, CD, DE, EF and FA. ABCDEF is the required regular hexagon.
The textbook notes a shortcut for Method 3. You may draw the circle first, with radius equal to the side. Then construct \angle AOB = 60^\circ so its arms cut the circle at A and B. This saves drawing the triangle as a separate step.
What to practise from this chapter
Exercise 15 holds 39 questions. The questions are grouped by shape. Each group drills one construction method from this chapter.
- Quadrilaterals: Q1–7
- Parallelograms: Q8–14
- Rectangles: Q15–19
- Trapeziums: Q20–21
- Rhombuses: Q22–27 and Q31
- Squares: Q28–30
- Regular hexagon: Q32
- Ruler and compasses only: Q33–39
Questions 33 to 39 carry one restriction. They state ruler and compasses only. No protractor is allowed. The earlier questions let you use a protractor where the data includes an angle. Every chapter in the Class 9 Mathematics book hub is explained in this same section-by-section format.
Frequently asked questions
How do you construct a rhombus when only its two diagonals are given?
Draw one diagonal and bisect it perpendicularly to find its mid-point. Cut half of the second diagonal on each side of the mid-point. Join the four vertices to complete the rhombus.
Why must I draw a rough freehand sketch before constructing a quadrilateral?
The sketch shows which sides, angles and diagonals you have, and where each vertex goes. It prevents arcs from the wrong centres and helps you pick the right triangle to draw first.
What should I do when half of a diagonal length cannot be measured on my scale?
Start the construction with the other diagonal. For instance, if BD = 4.5 cm, half is 2.25 cm, which a standard scale cannot measure. Draw BD first, find its mid-point, then place AC across it at the given angle.
Why does the square construction use the same method as the rhombus?
A square’s diagonals are equal and bisect each other at right angles, the same property a rhombus’s diagonals have. So a square built from its diagonal follows the rhombus steps exactly.
Which of the three methods for a regular hexagon is quickest in an exam?
Method 2 is usually the quickest. It needs only one circle and repeated arcs of the same radius. Methods 1 and 3 require an angle or an isosceles triangle before the arcs begin.
Reference: ICSE Class 9 Mathematics textbook material, chapter 15: Construction of Polygons — CISCE-aligned.