Figure it Out — Parallelograms and Rhombuses — Quadrilaterals

1. 14 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 102

   Find the remaining angles in the following quadrilaterals. (i) Parallelogram PEAR with ∠P = 40°. (ii) Parallelogram PQRS with ∠P = 110°. (iii) Rhombus UVWX in which diagonal XV makes an angle of 30° with side VU. (iv) Rhombus AOIE in which diagonal OE makes an angle of 20° with side EA.

   Hint. In a parallelogram opposite angles are equal and adjacent angles add to 180°. In a rhombus each diagonal also bisects the angles it passes through.

   Two facts carry all four parts: opposite angles of a parallelogram are equal, and adjacent angles add up to 180° — they add to 180° because a pair of adjacent angles sits as interior angles on the same side of a transversal cutting the two parallel opposite sides. A rhombus is a parallelogram, and additionally its diagonals bisect its angles.

   (i) Parallelogram PEAR, ∠P = 40° P and E are adjacent, so their angles are supplementary and ∠E = 180° − 40° = 140°. Opposite angles are equal, so ∠A = ∠P = 40° and ∠R = ∠E = 140°. Check: 40 + 140 + 40 + 140 = 360° ✓

   (ii) Parallelogram PQRS, ∠P = 110° Adjacent: ∠Q = 180° − 110° = 70°. Opposite: ∠R = ∠P = 110° and ∠S = ∠Q = 70°. Check: 110 + 70 + 110 + 70 = 360° ✓

   (iii) Rhombus UVWX, half-angle at V is 30° Since a rhombus is a parallelogram with all sides equal, each diagonal splits it into two isosceles triangles and therefore bisects the angles it passes through. So ∠XVU = ∠XVW = 30° and the whole angle is ∠UVW = 2 × 30° = 60°. Opposite angles are equal, so ∠WXU = ∠UVW = 60°, and the diagonal XV bisects that too: ∠UXV = ∠WXV = 30°. Adjacent angles add to 180°, so ∠U = 180° − 60° = 120°, and ∠W = ∠U = 120°. Check: 60 + 120 + 60 + 120 = 360° ✓

   (iv) Rhombus AOIE, half-angle at E is 20° The diagonal OE bisects ∠AEI, so ∠AEO = ∠OEI = 20° and ∠E = 40°. By the same bisection at the opposite vertex, ∠AOE = ∠EOI = 20°, so ∠O = 40°. Adjacent angles add to 180°, so ∠A = 180° − 40° = 140°, and ∠I = ∠A = 140°. Check: 40 + 140 + 40 + 140 = 360° ✓

   ✦ (i) ∠E = ∠R = 140°, ∠A = 40°. (ii) ∠Q = ∠S = 70°, ∠R = 110°. (iii) ∠XVU = ∠XVW = ∠UXV = ∠WXV = 30°, ∠UVW = ∠WXU = 60°, ∠U = ∠W = 120°. (iv) ∠OEI = ∠AOE = ∠EOI = 20°, ∠A = ∠I = 140°.

2. 23 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 102

   Using the diagonal properties, construct a parallelogram whose diagonals are of lengths 7 cm and 5 cm, and intersect at an angle of 140°.

   Hint. The one diagonal property of a parallelogram is that the diagonals bisect each other — they need not be equal.

   Step 1 — Recall which property to use. The diagonals of a parallelogram bisect each other, but unlike a rectangle's they need not be equal. So each diagonal must be cut in half at the point of intersection, and the two halves will differ in length between the diagonals.

   Step 2 — Construction.

   1. Draw AB = 7 cm — the first diagonal.
   2. Mark its midpoint O, so AO = OB = 3.5 cm.
   3. At O, draw a ray making an angle of 140° with OB.
   4. On that ray, cut off OC = OD = 2.5 cm on either side of O, so the second diagonal CD = 5 cm and is bisected at O.
   5. Join AC, CB, BD and DA.

   ADBC is the required parallelogram.

   Step 3 — Why it works. In triangles AOC and BOD, OA = OB, OC = OD, and the angles at O are vertically opposite and hence equal. So the triangles are congruent by SAS, giving AC = BD. The same argument on the other pair gives AD = CB. A quadrilateral whose opposite sides are equal is a parallelogram (proved in question 9 of the next set), so the construction is guaranteed rather than merely apparent.

   Check your figure: the diagonals should come out unequal (7 cm and 5 cm), so the angles must not be 90° — if your figure looks like a rectangle, the 140° has been drawn wrongly.

   ✦ Draw the 7 cm diagonal, bisect it at O, draw a ray at 140° through O and cut off 2.5 cm on each side for the 5 cm diagonal, then join the four endpoints.

3. 33 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 102

   Using the diagonal properties, construct a rhombus whose diagonals are of lengths 4 cm and 5 cm.

   Hint. No angle is given here, and none is needed — the diagonals of a rhombus always meet at one particular angle.

   Step 1 — Notice that the missing angle is not missing at all. No angle between the diagonals is stated, and none is needed, because the chapter proves that the diagonals of a rhombus always bisect each other at right angles. So the angle is fixed at 90° by the shape itself.

   Step 2 — Construction.

   1. Draw AB = 5 cm — the longer diagonal.
   2. Mark its midpoint O, so AO = OB = 2.5 cm.
   3. At O, construct a perpendicular to AB (compass and ruler, no protractor needed).
   4. On that perpendicular, cut off OC = OD = 2 cm on either side of O, so the second diagonal CD = 4 cm and is bisected at O.
   5. Join AC, CB, BD and DA.

   ADBC is the required rhombus.

   Step 3 — Why all four sides come out equal. The four triangles round O all have legs 2.5 cm and 2 cm with the included angle 90°, so they are congruent by SAS. Their hypotenuses are therefore equal, and those hypotenuses are the four sides of the quadrilateral.

   Each side measures √(2.5² + 2²) = √(6.25 + 4) = √10.25 ≈ 3.2 cm, which is a useful check on your drawing.

   ✦ Draw the 5 cm diagonal, bisect it at O, erect a perpendicular at O and cut off 2 cm on each side for the 4 cm diagonal, then join the four endpoints; each side should measure about 3.2 cm.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp104.pdf). The chapter builds every property by deduction (congruence and transversal arguments) rather than assertion, and its three 'Figure it Out' blocks sit on pages 94, 102 and 107-109. Every angle answer here was independently recomputed from the stated configuration and then checked against the book's own printed answer key. TWO DEVIATIONS ARE FLAGGED IN PLACE: (1) the key's answer to the Venn-diagram question 4(ii) contradicts its own answer to 4(i) and its own printed diagram — the mathematically correct answer under the book's stated definition of a kite is given, with the discrepancy explained; (2) question 5 (∠IOD in rectangles PAIR and RODS) has an answer in the key but its derivation depends on the exact printed figure, which cannot be recovered from the text, so the answer is cited rather than derived and this is stated openly in the solution.. Questions are referenced from the NCERT textbook for identification.