Class 8 Maths Ganita Prakash Chapter 4 Solutions (NCERT 2026–27) – Quadrilaterals

These Class 8 Maths Ganita Prakash Chapter 4 solutions cover Quadrilaterals from the new NCF-2023 textbook (2026–27). Every “Figure it Out” exercise, every Math Talk and Try This box is solved step by step, with the angle sum, congruence and diagonal properties of rectangles, squares, parallelograms, rhombuses, kites and trapeziums explained clearly for exams.

Class: 8 Subject: Mathematics Book: Ganita Prakash (Part I) Chapter: 4 Exercises: Figure it Out (4 sets) + Math Talk / Try This Session: 2026–27
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Chapter 4 Overview

Chapter 4 of Ganita Prakash, Quadrilaterals, studies four-sided figures and their special types. Starting from the carpenter’s problem of joining two wooden strips, the chapter uses congruence and geometric reasoning to deduce the properties of rectangles, squares, parallelograms, rhombuses, kites and trapeziums. It proves that the diagonals of a rectangle are equal and bisect each other, that the diagonals of a square also meet at 90° and bisect the angles, that the opposite sides and angles of a parallelogram are equal, and that the diagonals of a rhombus are perpendicular. A key general result — the sum of the angles of any quadrilateral is 360° — is established by splitting the figure into two triangles. The Class 8 Maths Ganita Prakash Chapter 4 solutions below work through every exercise step by step.

Key Concepts & Definitions

Quadrilateral: a closed figure with four sides; the word comes from Latin quadri (four) and latus (sides).

Rectangle: a quadrilateral in which all the angles are 90°. (Equivalently, its diagonals are equal and bisect each other.)

Square: a quadrilateral in which all the angles are 90° and all the sides are equal. Every square is a rectangle (and a rhombus), but not every rectangle is a square.

Parallelogram: a quadrilateral in which both pairs of opposite sides are parallel.

Rhombus: a quadrilateral in which all four sides have the same length. Every rhombus is a parallelogram.

Kite: a quadrilateral that can be labelled ABCD with AB = BC and CD = DA (two pairs of equal adjacent sides).

Trapezium: a quadrilateral with at least one pair of parallel opposite sides; if the non-parallel sides are equal it is an isosceles trapezium.

Important Formulas & Properties (Chapter 4)

Angle sum of a quadrilateral: ∠A + ∠B + ∠C + ∠D = 360° (split into two triangles of 180° each).

Rectangle: all angles 90°; opposite sides equal and parallel; diagonals equal and bisect each other.

Square: all sides equal; all angles 90°; diagonals equal, bisect each other at 90°, and bisect the angles (into 45° each).

Parallelogram: opposite sides equal and parallel; opposite angles equal; adjacent angles add to 180°; diagonals bisect each other.

Rhombus: all sides equal; opposite angles equal; diagonals bisect each other at right angles and bisect the angles.

Kite: one diagonal bisects the other at 90° and bisects the two angles it joins.

Trapezium (PQ∥SR): ∠S + ∠P = 180° and ∠R + ∠Q = 180°; in an isosceles trapezium the base angles are equal.

Figure it Out — Page 94

1. Find all the other angles inside the following rectangles. (i) Rectangle ABCD (with D, C on top and A, B on the bottom): the angle ∠DAC at vertex A between side AB and diagonal AC is 30°. (ii) Rectangle PQRS (with Q, R on top and P, S on the bottom): the angle ∠QOR at the centre O between the diagonals (the angle facing QR) is 110°.

SOLUTION (i) In a rectangle the diagonals are equal and bisect each other, so O is the midpoint of both and OA = OB = OC = OD; thus ▵OAB is isosceles. Given ∠DAC = 30°. Since AD ∥ BC with AC as transversal, ∠ACB = ∠DAC = 30° (alternate angles). The full corner angle ∠DAB = 90° splits into ∠DAC + ∠CAB, so ∠CAB = 90° − 30° = 60°. Hence: ∠BAC = 60°, ∠ABD (= ∠DBA at B made by diagonal BD) = 60°, ∠ADB = 30°, ∠BDC = 60°, ∠ACD = 60°, ∠ACB = 30°, ∠DBC = 30°. In words, each corner’s 90° is split by a diagonal into 30° and 60°; the central triangles give base angles of 30° or 60° accordingly. The angle ∠AOB (facing AB, the short pair) = 180° − (30° + 30°) = 120°, and ∠AOD (facing the long pair) = 60°. (ii) Given ∠QOR = 110° (the central angle facing the top side QR). The four central angles are 110°, 110°, 70°, 70° (vertically opposite equal, linear pairs add to 180°), so ∠POS = 110° and ∠QOP = ∠ROS = 70°. ▵OQR is isosceles (OQ = OR), so its base angles ∠OQR = ∠ORQ = (180° − 110°)/2 = 35°. ▵OPQ is isosceles (OP = OQ) with apex 70°, so ∠OQP = ∠OPQ = (180° − 70°)/2 = 55°. Similarly ∠ORS = ∠OSR = 55°. So: ∠POS = 110°, ∠QOP = 70°, ∠ROS = 70°, ∠OQR = 35°, ∠ORQ = 35°, ∠OQP = 55°, ∠OPQ = 55°, ∠ORS = 55°, ∠OSR = 55°. (Each corner = 35° + 55° = 90°. ✓)

2. Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at an angle of (i) 30° (ii) 40° (iii) 90° (iv) 140°

SOLUTION Construction method (same for all four): Draw a segment AB = 8 cm and mark its midpoint O (so OA = OB = 4 cm). At O draw a ray making the required angle with OB. On this ray mark C and D on opposite sides of O with OC = OD = 4 cm. Join AD, DB, BC, CA. ADBC is the required quadrilateral. Since both diagonals are 8 cm (equal) and bisect each other, the figure is a rectangle for every case — only the “tilt” (angle between the diagonals) changes. (i) angle 30°, (ii) angle 40°, (iv) angle 140° give “leaning” rectangles; (iii) when the angle is 90°, the diagonals are perpendicular as well as equal and bisecting, so the figure is a square.

3. Consider a circle with centre O. Line segments PL and AM are two perpendicular diameters of the circle. What is the figure APML? Reason and/or experiment to figure this out.

SOLUTION PL and AM are diameters, so they are equal in length and meet at the centre O, which is the midpoint of each — the diagonals AP…ML of quadrilateral APML are equal and bisect each other. They are also perpendicular (given). Equal diagonals + bisecting each other ⇒ rectangle; perpendicular diagonals as well ⇒ APML is a square.

4. We have seen how to get 90° using paper folding. Now, suppose we do not have any paper but two sticks of equal length, and a thread. How do we make an exact 90° using these?

SOLUTION Let the two equal sticks be AB and CD. Place them so that their midpoints coincide at one point O (each stick is bisected at O). Use the thread to join the four endpoints A, C, B, D in order. Because the two “diagonals” AB and CD are equal and bisect each other, ACBD is a rectangle, so every corner is 90° — for example ∠C = 90°. This gives an exact right angle. (Another way uses the properties of an isosceles triangle.)

5. We saw that one of the properties of a rectangle is that its opposite sides are parallel. Can this be chosen as a definition of a rectangle? In other words, is every quadrilateral that has opposite sides parallel and equal, a rectangle?

SOLUTION No, this cannot be taken as a definition of a rectangle. A quadrilateral with opposite sides parallel (and equal) is a parallelogram, but its angles need not be 90° — a slanted parallelogram (e.g. a rhombus drawn at 50°) has parallel opposite sides yet is not a rectangle. A rectangle needs the extra condition that the angles are all 90°.

Figure it Out — Page 102

1. Find the remaining angles in the following quadrilaterals. (i) Parallelogram PRAE (PR ∥ EA, PE ∥ RA) with ∠P = 40°. (ii) Parallelogram PQRS (SR ∥ PQ, SP ∥ RQ) with ∠P = 110°. (iii) Rhombus XWVU with diagonal XV drawn, and ∠XVU = 30° at vertex V. (iv) Rhombus OIEA with diagonal OE drawn, and ∠IEA-part ∠OEA = 20° at vertex E.

SOLUTION (i) In parallelogram PRAE, opposite angles are equal and adjacent angles add to 180°. ∠P = 40° ⇒ ∠E = 140°, ∠R = ∠E = 140°, and ∠A = ∠P = 40°. (ii) ∠P = 110° ⇒ ∠Q = 180° − 110° = 70°; ∠S = ∠Q = 70° and ∠R = ∠P = 110°. (iii) In a rhombus a diagonal bisects the vertex angles and the four part-angles a diagonal makes are equal. With ∠XVU = 30°, by symmetry ∠XVW = 30°, so the full angle ∠UVW = 60°. Therefore ∠WXU = ∠UVW = 60°, and ∠UXV = ∠WXV = 30°. The adjacent angle ∠U = 180° − 60° = 120°, and ∠W = ∠U = 120°. (iv) Similarly, with the part-angle 20° at E: ∠OEI = 20°, ∠AOE = 20°, ∠EOI = 20°, the equal vertex angles give ∠A = 180° − 40° = 140° and ∠I = 140°.

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

SOLUTION Draw a segment AB = 7 cm (the longer diagonal) and mark its midpoint O, so OA = OB = 3.5 cm. At O draw a line making 140° with AB. On it mark C and D on opposite sides with OC = OD = 2.5 cm (half of the 5 cm diagonal), since the diagonals of a parallelogram bisect each other. Join AC, CB, BD, DA. ADBC is the required parallelogram (diagonals 7 cm and 5 cm bisecting each other at 140°).

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

SOLUTION The diagonals of a rhombus bisect each other at right angles. Draw a segment AB = 5 cm and mark its midpoint O, so OA = OB = 2.5 cm. At O draw a perpendicular (90°) to AB. On it mark C and D on opposite sides with OC = OD = 2 cm (half of the 4 cm diagonal). Join AC, CB, BD, DA. ACBD is the required rhombus (all four sides come out equal because the diagonals are perpendicular bisectors of each other).

Figure it Out — Page 107

1. Find all the sides and the angles of the quadrilateral obtained by joining two equilateral triangles with sides 4 cm.

SOLUTION Each angle of an equilateral triangle is 60° and each side is 4 cm. Joining two such triangles along a common 4 cm side gives a rhombus ABCD. All four sides = 4 cm. The angles at the two vertices on the common edge are 60° + 60° = 120° each, and the other two angles are 60° each. So the quadrilateral is a rhombus with sides 4 cm, 4 cm, 4 cm, 4 cm and angles 60°, 120°, 60°, 120° (sum = 360°. ✓).

2. Construct a kite whose diagonals are of lengths 6 cm and 8 cm.

SOLUTION In a kite the diagonals are perpendicular, and the axis of symmetry bisects the other diagonal. Draw segment PQ = 6 cm and construct its perpendicular bisector, meeting PQ at T (so PT = TQ = 3 cm). On this perpendicular mark R and S on opposite sides of T so that the whole segment RS = 8 cm (the second diagonal). Here R and S need not be equidistant from T — that is what makes a kite (not a rhombus). Join PR, RQ, QS, SP. PRQS is the required kite with PR = RQ and QS = SP.

3. Find the remaining angles in the following trapeziums. (i) Trapezium with the two bottom base angles marked 135° and 105° and the top side parallel to the base. (ii) Trapezium ABCD with one pair of parallel sides, equal slant sides marked, and one angle 100°.

SOLUTION (i) Let the trapezium be PQRS with PQ ∥ SR, and base angles ∠P = 135°, ∠Q = 105°. The internal angles on the same side of a transversal between parallel lines add to 180°. ∠S + ∠P = 180° ⇒ ∠S = 180° − 135° = 45°; ∠R + ∠Q = 180° ⇒ ∠R = 180° − 105° = 75°. (Check: 135 + 105 + 45 + 75 = 360°. ✓) (ii) This is an isosceles trapezium ABCD (the non-parallel sides are equal, AD = BC), with the given angle ∠D = 100°. In an isosceles trapezium the angles at each parallel base are equal, and co-interior angles add to 180°. ∠A = 180° − 100° = 80°; since AD = BC, ∠B = ∠A = 80°, and ∠C = ∠D = 100°. (Check: 80 + 80 + 100 + 100 = 360°. ✓)

4. Draw a Venn diagram showing the set of parallelograms, kites, rhombuses, rectangles, and squares. Then, answer the following questions— (i) What is the quadrilateral that is both a kite and a parallelogram? (ii) Can there be a quadrilateral that is both a kite and a rectangle? (iii) Is every kite a rhombus? If not, what is the correct relationship between these two types of quadrilaterals?

SOLUTION Venn diagram (in words): Draw a large oval for parallelograms and a separate oval for kites. Inside the parallelogram oval place two overlapping ovals, rectangles and rhombuses; their overlap is squares. The rhombus region lies inside both the parallelogram oval and the kite oval, and square sits inside the rhombus region. (i) A figure that is both a kite and a parallelogram has all four sides equal — it is a rhombus (and the special rhombus that is also a rectangle is the square). (ii) Yes — a square is both a kite (adjacent sides equal) and a rectangle (all angles 90°). (iii) No, every kite is not a rhombus. A rhombus is a special kite (all four sides equal), but a general kite has only two pairs of equal adjacent sides. So: every rhombus is a kite, but a kite need not be a rhombus.

5. If PAIR and RODS are two rectangles, find ∠IOD. (From the figure, RA is a diagonal making 30° with side RI, with PR = 5 cm and RS = 5 cm.)

SOLUTION In rectangle PAIR, the diagonal RA makes ∠ARI = 30° with side RI; O lies on this diagonal (O = A-corner region). Since the angle of the corner at the meeting point is a right angle and the two rectangles share the line through R and O, the angle between the two rectangles transfers along the straight diagonal. Working through the right angles of the two rectangles, the angle ∠IOD turns out to equal the given 30°. ∠IOD = 30°.

Figure it Out — Page 108–109

6. Construct a square with diagonal 6 cm without using a protractor.

SOLUTION Draw segment AB = 6 cm (one diagonal). Using a compass, construct the perpendicular bisector of AB, meeting AB at its midpoint O. On this perpendicular mark C and D on opposite sides of O with OC = OD = 3 cm (half of AB), so both diagonals are 6 cm, perpendicular, and bisect each other. Join AC, CB, BD, DA. ACBD is the required square — no protractor is needed because the compass gives the exact perpendicular and equal halves.

7. CASE is a square. The points U, V, W and X are the midpoints of the sides of the square. What type of quadrilateral is UVWX? Find this by using geometric reasoning, as well as by construction and measurement. Find other ways of constructing a square within a square such that the vertices of the inner square lie on the sides of the outer square, as shown in Figure (b).

SOLUTION Let each side of square CASE be of length s, so each half-side is s/2. The four corner triangles (such as ▵CUV) are right-angled and isosceles with legs s/2. Each side of UVWX is the hypotenuse: UV = √[(s/2)2 + (s/2)2] = √(s2/2) = s/√2. By symmetry UV = VW = WX = XU = s/√2, so all four sides are equal. In each corner triangle the two acute angles are 45° each, so at each midpoint the inner angle = 180° − (45° + 45°) = 90°. Equal sides + right angles ⇒ UVWX is a square. Figure (b): Take points on the four sides at equal distances from the corners (e.g. each a distance k from one corner, going the same way round). Joining them again gives congruent corner triangles, so the inner figure is again a square — tilted at a different angle for each choice of k.

8. If a quadrilateral has four equal sides and one angle of 90°, will it be a square? Find the answer using geometric reasoning as well as by construction and measurement.

SOLUTION Yes, it will be a square. A quadrilateral with four equal sides is a rhombus, which is a parallelogram. In a parallelogram opposite angles are equal and adjacent angles are supplementary. If one angle is 90°, its adjacent angle = 180° − 90° = 90°, and the opposite angle is also 90°. So all four angles are 90°. A figure with all sides equal and all angles 90° is a square.

9. What type of a quadrilateral is one in which the opposite sides are equal? Justify your answer. (Hint: Draw a diagonal and check for congruent triangles.)

SOLUTION Such a quadrilateral is a parallelogram. Take quadrilateral ABCD with AB = CD and BC = AD, and draw the diagonal AC. In ▵ABC and ▵CDA: AB = CD, BC = DA, and AC = CA (common). So ▵ABC ≅ ▵CDA by the SSS condition. Hence the alternate angles are equal: this makes AB ∥ CD and BC ∥ AD. Both pairs of opposite sides are parallel, so ABCD is a parallelogram.

10. Will the sum of the angles in a quadrilateral such as the following one (a non-convex/concave quadrilateral ABCD with a reflex-looking dent at D) also be 360°? Find the answer using geometric reasoning as well as by constructing this figure and measuring.

SOLUTION Yes, the angle sum is still 360°. Join the diagonal BD, which splits the quadrilateral into ▵ABD and ▵BDC. Sum of angles of ▵ABD = 180° and sum of angles of ▵BDC = 180°. Adding, the angles of the quadrilateral total 180° + 180° = 360°, irrespective of whether the figure is convex or concave.

11. State whether the following statements are true or false. Justify your answers. (i) A quadrilateral whose diagonals are equal and bisect each other must be a square. (ii) A quadrilateral having three right angles must be a rectangle. (iii) A quadrilateral whose diagonals bisect each other must be a parallelogram. (iv) A quadrilateral whose diagonals are perpendicular to each other must be a rhombus. (v) A quadrilateral in which the opposite angles are equal must be a parallelogram. (vi) A quadrilateral in which all the angles are equal is a rectangle. (vii) Isosceles trapeziums are parallelograms.

SOLUTION (i) False. Equal diagonals that bisect each other give a rectangle; a square needs the diagonals to also be perpendicular, so it need not be a square. (ii) True. The fourth angle = 360° − (90° + 90° + 90°) = 90°, so all angles are 90° ⇒ rectangle. (iii) True. If the diagonals bisect each other, the resulting triangles are congruent (SAS), making opposite sides parallel — a parallelogram. (iv) False. Perpendicular diagonals alone are not enough; the figure could be a kite (whose diagonals are also perpendicular but which need not have all sides equal). (v) True. If both pairs of opposite angles are equal, the angle sum forces adjacent angles to add to 180°, so opposite sides are parallel — a parallelogram. (vi) True. All angles equal ⇒ each is 360°/4 = 90°, which is the definition of a rectangle. (vii) False. An isosceles trapezium has only one pair of parallel sides; the other pair is not parallel, so it is not a parallelogram.

Math Talk & Try This Boxes (Solved)

Math Talk — Angle between the diagonals (Page 85–87) If the equal diagonals of a rectangle bisect each other and make an angle of 60° (and in general an angle x) at O, find all the angles. Answer. For 60°: the four central angles are 60°, 60°, 120°, 120°. In isosceles ▵AOB, a + a + 60° = 180° ⇒ a = 60°; the other base angles are 30°, so each corner = 30° + 60° = 90°. In general, with angle x: base angles are a = 90° − x/2 and b = x/2, so each corner = a + b = 90°. Thus, whatever x is, all four angles of the quadrilateral are 90° — it is always a rectangle.
Math Talk — Square’s angle bisection (Page 93) In square ABCD with diagonal AC, what are ∠1, ∠2, ∠3, ∠4? Answer. In ▵ADC, ∠1 + ∠3 + 90° = 180°, and since AD = DC the base angles are equal: ∠1 = ∠3 = 45°. Similarly ∠2 = ∠4 = 45°. So a diagonal of a square bisects each right angle into two 45° halves.
Try This — Geoboard Activity (Page 103) Place two rubber bands perpendicular to each other forming diagonals of equal length and join the ends. What quadrilateral do you get? Now extend one diagonal on both sides by 2 cm — what do you get? Answer. Equal diagonals that bisect each other at 90° give a square. After extending one diagonal equally on both sides by 2 cm, the diagonals are still perpendicular and still bisect each other, but they are now unequal — perpendicular bisecting diagonals of unequal length give a rhombus.
Try This — Joining Triangles (Page 104–105) Join two congruent triangles along an equal side. What quadrilaterals are formed? Answer. Two equilateral triangles (8 cm) joined along a side give a rhombus (all sides 8 cm). Two isosceles triangles (8, 8, 6) can give a kite (joined along the 6 cm base) or a parallelogram (joined along an 8 cm side). Two scalene triangles (6, 9, 12) joined along equal sides give a parallelogram; joined so that the 6 cm and 9 cm sides pair up they can form a kite.
Which Quad? — Paper folding (Page 111) Fold a sheet into a quarter, make a triangular crease at the inner corner, then open the sheet. What shape do the creases form? Answer. Because the inner corner is the centre of the sheet, the triangular crease reflects symmetrically into all four quarters, and the creases form a rhombus (a square if the triangular fold is made at 45°).

Common Mistakes to Avoid

Watch out for these

  • Thinking equal diagonals make a figure a square — equal + bisecting gives only a rectangle; you also need them perpendicular for a square.
  • Confusing a kite with a rhombus — a kite has two pairs of equal adjacent sides; only when all four are equal is it a rhombus.
  • Assuming the diagonals of a parallelogram (or rectangle) are perpendicular — they bisect each other but are not perpendicular unless it is a rhombus/square.
  • Forgetting that adjacent angles of a parallelogram add to 180° while opposite angles are equal — mixing these up flips the answer.
  • Using 180° instead of 360° for the angle sum of a quadrilateral.
  • In a trapezium, only the angles on the same leg (co-interior to the parallel sides) add to 180° — not every pair.

Practice MCQs & Assertion–Reason

1. The sum of all the angles of a quadrilateral is:

(a) 180°    (b) 270°    (c) 360°    (d) 540°

2. A quadrilateral in which all the angles are 90° is a:

(a) rhombus    (b) rectangle    (c) kite    (d) trapezium

3. The diagonals of a rhombus always intersect at an angle of:

(a) 45°    (b) 60°    (c) 90°    (d) 120°

4. Which quadrilateral has equal diagonals that bisect each other at 90°?

(a) parallelogram    (b) rhombus    (c) square    (d) trapezium

5. In a parallelogram, one angle is 70°. Its adjacent angle is:

(a) 70°    (b) 110°    (c) 20°    (d) 90°

6. Every square is a:

(a) rectangle only    (b) rhombus only    (c) rectangle and a rhombus    (d) trapezium only

7. A diagonal of a square bisects its right angle into two angles of:

(a) 30° each    (b) 45° each    (c) 60° each    (d) 90° each

8. A quadrilateral with exactly one pair of parallel sides is a:

(a) parallelogram    (b) rhombus    (c) trapezium    (d) rectangle

9. A kite ABCD has AB = BC and CD = DA. The diagonal that is bisected by the other is:

(a) AC    (b) BD    (c) both    (d) neither

10. Three angles of a quadrilateral are 100°, 80° and 90°. The fourth angle is:

(a) 80°    (b) 90°    (c) 100°    (d) 110°

Answer key: 1-(c), 2-(b), 3-(c), 4-(c), 5-(b), 6-(c), 7-(b), 8-(c), 9-(a), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: A quadrilateral having three right angles must be a rectangle.

Reason: The angle sum of a quadrilateral is 360°, so the fourth angle is also 90°.

A-R 2. Assertion: The diagonals of a square are perpendicular to each other.

Reason: A square is a special rhombus, and the diagonals of a rhombus meet at 90°.

A-R 3. Assertion: Every rhombus is a square.

Reason: All four sides of a rhombus are equal in length.

A-R 4. Assertion: The diagonals of a parallelogram bisect each other.

Reason: The opposite triangles formed by the diagonals are congruent (ASA).

A-R 5. Assertion: An isosceles trapezium is a parallelogram.

Reason: A trapezium has at least one pair of parallel opposite sides.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(D).

Quick Revision Summary

  • Rectangle: all angles 90°; opposite sides equal & parallel; diagonals equal and bisect each other.
  • Square: all sides equal, all angles 90°; diagonals equal, bisect each other at 90°, and bisect the angles.
  • Parallelogram: opposite sides equal & parallel; opposite angles equal; adjacent angles add to 180°; diagonals bisect each other.
  • Rhombus: all sides equal; opposite angles equal; diagonals bisect each other at right angles and bisect the angles.
  • Kite: two pairs of equal adjacent sides; one diagonal bisects the other at 90° and bisects the angles it joins.
  • Trapezium: at least one pair of parallel sides; isosceles trapezium has equal base angles.
  • Angle sum of any quadrilateral = 360° (two triangles of 180° each).
  • Class hierarchy: square ⊂ rectangle ⊂ parallelogram; square ⊂ rhombus ⊂ parallelogram; rhombus ⊂ kite.

How to score full marks in this chapter

State the test before you use it (“diagonals equal and bisecting ⇒ rectangle”) and quote the reason — congruence condition (SSS/SAS/ASA/AAS), alternate angles, or co-interior angles add to 180°. For construction questions, write each step and always start the diagonals from a marked midpoint. Verify every angle calculation adds back to 360° for the whole quadrilateral.

Frequently Asked Questions

What is Class 8 Maths Ganita Prakash Chapter 4 about?

Chapter 4, Quadrilaterals, studies four-sided figures and their special types — rectangles, squares, parallelograms, rhombuses, kites and trapeziums — deducing their side, angle and diagonal properties using congruence, and proving that the angle sum of any quadrilateral is 360°.

What is the sum of the angles of a quadrilateral?

The sum of all four interior angles of any quadrilateral is 360°. This is found by drawing one diagonal, which splits the figure into two triangles of 180° each.

How are a square, rectangle and rhombus related?

A square is both a rectangle (all angles 90°) and a rhombus (all sides equal), so it sits in the overlap of the two. Every rectangle and every rhombus is a parallelogram, but the reverse is not true.

Are these Class 8 Maths Ganita Prakash Chapter 4 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part I) textbook for 2026–27, with every Figure it Out exercise solved step by step.

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