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NCERT Solutions Ganita Prakash (Part 1) Chapter 4 .4 Quadrilaterals with Equal Sidelengths — Figure it Out

Book page 1024 Updated on2026-09-05

Q1.
Find the remaining angles in the following quadrilaterals.
Answer

(i) Parallelogram PERA — vertices P (bottom-left), E (bottom-right), A (top-right), R (top-left); PE ∥ RA and PR ∥ EA; ∠P = 40°.

∠A = ∠P = 40°  (opposite angles of a parallelogram)
∠E = 180 – 40 = 140°  (adjacent angles are supplementary)
∠R = ∠E = 140°
Check: 40 + 140 + 40 + 140 = 360° ✓

(ii) Parallelogram PQRS — P (bottom-left), Q (bottom-right), R (top-right), S (top-left); SR ∥ PQ and SP ∥ RQ; ∠P = 110°.

∠R = ∠P = 110°  (opposite angles)
∠Q = 180 – 110 = 70°  and  ∠S = 70°
Check: 110 + 70 + 110 + 70 = 360° ✓

(iii) Rhombus XWVU — all four sides carry a tick, the diagonal XV is drawn, and ∠XVW = 30°.

A diagonal of a rhombus bisects its angles, so
∠XVU = ∠XVW = 30°  ⇒ ∠UVW = 30 + 30 = 60°
∠X = ∠UVW = 60°  (opposite angles), split by XV into 30° + 30°
∠U = 180 – 60 = 120°  and  ∠W = ∠U = 120°
Check: 60 + 120 + 60 + 120 = 360° ✓

(iv) Rhombus OIEA — all four sides ticked, the diagonal OE is drawn, and ∠OEI = 20°.

∠OEA = ∠OEI = 20°  (the diagonal bisects ∠E)
∠AOE = ∠EOI = 20°  (the diagonal bisects ∠O too, and ∠O = ∠E)
∠E = ∠AEI = 20 + 20 = 40°  and  ∠O = 40°
∠A = ∠I = 180 – 40 = 140°
Check: 40 + 140 + 40 + 140 = 360° ✓
Why one angle is enough every time: in a parallelogram, opposite angles are equal and adjacent ones add to 180°, so a single angle fixes all four. In a rhombus you get more: a diagonal cuts each of the two angles it meets into two equal halves, so even a half-angle like the 30° in (iii) determines the whole figure.
Q2.
Using the diagonal properties, construct a parallelogram whose diagonals are of lengths 7 cm and 5 cm, and intersect at an angle of 140°.
Answer

The one diagonal property that defines a parallelogram is: the diagonals bisect each other. So halve each length and build outwards from the crossing point.

  1. Draw AB = 7 cm and mark its midpoint O, so OA = OB = 3.5 cm.
  2. At O draw a ray making 140° with OB.
  3. On that ray mark C with OC = 2.5 cm; on the opposite ray mark D with OD = 2.5 cm. Then CD = 5 cm and O is its midpoint.
  4. Join AC, CB, BD and DA.

ACBD is the required parallelogram.

Diagonals AB = 7 cm and CD = 5 cm, unequal ⇒ not a rectangle
Angle at O = 140° ≠ 90° ⇒ not a rhombus
Diagonals bisect each other ⇒ a genuine parallelogram
Why bisecting is enough: if OA = OB and OC = OD, then ∆AOC ≅ ∆BOD by SAS (vertically opposite angles at O). So AC = BD, and the equal alternate angles make AC ∥ BD. The same argument on the other pair gives AD ∥ CB — both pairs of opposite sides parallel, which is the definition.
Q3.
Using the diagonal properties, construct a rhombus whose diagonals are of lengths 4 cm and 5 cm.
Answer

A rhombus needs its diagonals to bisect each other at right angles — and no protractor is needed, because a perpendicular bisector can be drawn with a compass.

  1. Draw AB = 5 cm.
  2. Construct the perpendicular bisector of AB with a compass; let it cut AB at O, so OA = OB = 2.5 cm.
  3. On the perpendicular mark C and D with OC = OD = 2 cm (so CD = 4 cm).
  4. Join AD, DB, BC and CA.

ADBC is the required rhombus.

Each side = √(2.5² + 2²) = √(6.25 + 4) = √10.25 ≈ 3.2 cm
All four sides are equal because all four triangles round O are congruent right triangles with legs 2.5 cm and 2 cm.
Why perpendicular bisecting forces equal sides: the four triangles AOC, COB, BOD, DOA all have legs 2.5 cm and 2 cm with a right angle between them, so they are congruent by SAS. Their hypotenuses — the four sides of the quadrilateral — must therefore be equal.
Check it yourself: measure the four sides; each should be a little over 3.2 cm. Measure the angle at O; it should read exactly 90°.
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