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Case Study Questions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

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Here we are providing Case Study questions for Class 8 Maths Chapter 3 Understanding Quadrilaterals.

Case Study Questions

Related posts, learning outcomes.

• Convex and Concave Polygons.
• Regular and Irregular Polygons.
• Sum of Measures of the Exterior Angles of a Polygon.
• Kinds of QuadrilateralTrapezium; Kite; Parallelogram.
• Some Special ParallelogramsRhombus; Rectangle; Square.

Important Keywords

• Convex Polygon: Polygons that have any line segment joining any two different points in the interior and have no portions of their diagonals in their exteriors are called convex polygons.
• Concave Polygon: Polygons that have one diagonal outside it are called concave polygons.
• Regular Polygon: A polygon whose all sides, all angles are equal that is which is both equiangular and equilateral are called regular polygon. Example: Square; Equilateral triangle
• Irregular Polygon: Polygon whose all sides are not equal are called Irregular polygon. Example: Rectangle.

Fundamental Facts

• Convex Polygon has each angle either acute or obtuse.
• Concave Polygon has one angle as reflex angle.

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Important Questions Class 8 Maths Chapter 3

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Important Questions Class 8 Mathematics Chapter 3 – Understanding Quadrilaterals

Mathematics deals with numbers of various forms, shapes, logic, quantity and arrangements. Mathematics also teaches us to solve problems based on numerical calculations and find solutions.

Quick Links

Chapter 3 of Class 8 Mathematics is called ‘Understanding Quadrilaterals’. A quadrilateral is a closed shape and also a type of polygon that has four sides, four vertices and four angles. It is formed by joining four non-collinear points. The sum of all the interior angles of a quadrilateral is always equal to 360 degrees. In a quadrilateral, the sides are straight lines and are two-dimensional. Square, rectangle, rhombus, parallelogram, etc., are examples of quadrilaterals. The formula for the angle sum of a polygon = (n – 2) × 180°.

Extramarks is the best study buddy for students and helps them with comprehensive online study solutions from Class 1 to Class 12. Our team of expert Mathematics teachers have prepared a variety of NCERT solutions to help students in their studies and exam preparation. Students can refer to our Important Questions Class 8 Mathematics Chapter 3 to practise exam-oriented questions. We have collated questions from various sources such as NCERT textbooks and exemplars, CBSE sample papers, CBSE past year question papers, etc. Students can prepare well for their exams and tests by solving a variety of chapter questions from our Important Questions Class 8 Mathematics Chapter 3.

To ace their exams, students can register on the Extramarks website to access Class 8 Mathematics Chapter 3 important questions, CBSE extra questions, Mathematics formulas, and much more.

Important Questions Class 8 Mathematics Chapter 3 – With Solution

Mentioned below are some sets of questions and their answers from our Chapter 3 Class 8 Mathematics important questions.

Question 1: A quadrilateral has three acute angles, each measuring 80°. What is the measure of the fourth angle of the quadrilateral?

Answer 1: – Let x be the measure of the fourth angle of a quadrilateral.

 The sum of all the angles of a quadrilateral + 360°

80° + 80° + 80° + x = 360° …………(since the measure of all the three acute angles = 80°)

 240° + x = 360°

x = 360° – 240°

Hence, the fourth angle made by the quadrilateral is 120°.

Question 2: Find the measure of all the exterior angles of a regular polygon with 

                        (i) 9 sides and (ii) 15 sides.

Answer 2 : (i) Total measure of all exterior angles = 360° 

                    Each exterior angle =sum of exterior angle = 360° = 40°

                                                           number of sides 9                                              

                    Each exterior angle = 40°

                   (ii) Total measure of all exterior angles = 360° 

                      Each exterior angle = sum of exterior angle =360° = 24° 

                                                            number of sides 15

                      Each exterior angle = 24° 

Answer 3: a) The sum of all the angles of the triangle = 180°

                       One side of a triangle

                       = 180°- (90° + 30°) = 60°

                        In a linear pair, the sum of two adjacent angles altogether measures up to 180° 

                       x + 90° = 180° 

                       x = 180° – 90° 

                          = 90° 

                       Similarly,

                       y + 60° = 180°

                       y = 180° – 60°

                           = 120°

                       similarly,

                      z + 30° = 180° 

                     z = 180° – 30° 

                        = 150°

                     Hence,x + y + z 

                                = 90° + 120° + 150° 

                                = 360°

                  Thus, the sum of the angles x, y, and z is altogether 360°

• b) Sum of all angles of quadrilateral = 360°

                 One side of quadrilateral = 360°- (60° + 80° + 120°) = 360° – 260° = 100°

                 x + 120° = 180°

                 x = 180° – 120°

                    = 60°

                y + 80° = 180° 

                y = 180° – 80° 

                  = 100°

                 z + 60° = 180°

                 z = 180° – 60° 

                   = 120°

                w + 100° = 180° 

                w = 180° – 100° = 80°

                x + y + z + w = 60° + 100° + 120° + 80° = 360°

Question 4: Adjacent sides of a rectangle are in the ratio 5: 12; if the perimeter of the given rectangle is 34 cm, find the length of the diagonal.

Answer 4: The ratio of the adjacent sides of the rectangle is 5: 12

                    Let 5x and 12x be adjacent sides.

                    The perimeter is the sum of all the given sides of a rectangle.

                     5x + 12x + 5x + 12x = 34 cm ……(since the opposite sides of the rectangle are the 

                                                                               same)

                     34x = 34

                      x = 34/34

                      x = 1 cm

                      Therefore, the adjacent sides of the rectangle are 5 cm and 12 cm, respectively.

                       That is,

                       Length =12 cm

                       Breadth = 5 cm

                       Length of the diagonal = √( l2 + b2)

                                                                 = √( 122 + 52)

                                                                 = √(144 + 25)

                                                                 = √169

                                                                 = 13 cm

                      Hence, the length of the diagonal of a rectangle is 13 cm.

Question 5: How many sides do regular polygons consist of if each interior angle is 165 ° ?

Answer 5: A regular polygon with an interior angle of 165°

                   We need to find the sides of the given regular polygon:-

                   The sum of all exterior angles of any given polygon is 360°.

                    Formula Used: Number of sides = 360∘ /Exterior angle

                            Exterior angle=180∘−Interior angle

                   Thus,

                   Each interior angle =165°

                   Hence, the measure of every exterior angle will be

                    =180°−165°

                    =15°

                   Therefore, the number of sides of the given polygon will be

                  =360°/15°

                   =24° 

Question 6: Find x in the following figure.

Answer 6: The two interior angles in the given figures are right angles = 90°

70° + m = 180°

 m = 180° – 70° 

(In a linear pair, the sum of two adjacent angles altogether measures up to 180°)

60° + n = 180° 

 n = 180° – 60° 

     = 120°

 The given figure has five sides, and it is a pentagon.

Thus, the sum of the angles of the pentagon = 540° 

 90° + 90° + 110° + 120° + y = 540°

410° + y = 540° 

y = 540° – 410° = 130°

x + y = 180°….. (Linear pair)

x + 130° = 180°

x = 180° – 130° 

Question 7: ABCD is a parallelogram with ∠A = 80°. The internal bisectors of ∠B and ∠C meet each other at O. Find the measure of the three angles of ΔBCO.

Answer 7: The measure of angle A = 80°.

                    In a parallelogram, the opposite angles are the same.

                    Hence,

                    ∠A = ∠C = 80°

                    And

                    ∠OCB = (1/2) × ∠C

                    = (1/2) × 80°

                    = 40°

              ∠B = 180° – ∠A (the sum of interior angles situated on the same side of the transversal is supplementary)

                     = 180° – 80°

                     = 100°

                     Also,

              ∠CBO = (1/2) × ∠B

              ∠CBO= (1/2) × 100°

              ∠CBO= 50°. 

             By the property of the sum of the angle BCO, we get,

            ∠BOC + ∠OBC + ∠CBO = 180°

            ∠BOC = 180° – (∠OBC + CBO)

                        = 180° – (40° + 50°)

                        = 180° – 90°

                        = 90°

      Hence, the measure of all the angles of triangle BCO is 40°, 50° and 90°.

Question 8: The measure of the two adjacent angles of the given parallelogram is the ratio of 3:2. Then, find the measure of each angle of the parallelogram.

Answer 8:  A parallelogram with adjacent angles in the ratio of 3:2

 To find:- The measure of each of the angles of the parallelogram.

                     Let the measure of angle A be 3x

                     Let the measure of angle B be 2x

Since the sum of the measures of adjacent angles is 180° for a parallelogram,

 3x+2x=180°

 ∠A=∠C =3x=108°

∠B=∠D =2x=72° (Opposite angles of a parallelogram are equal).

Hence, the angles of a parallelogram are 108°, 72°,108°and 72°

Question 9: Is it ever possible to have a regular polygon, each of whose interior angles is 100?

Answer 9: The sum of all the exterior angles of a regular polygon is 360°

As we also know, the sum of interior and exterior angles are 180°

Exterior angle + interior angle = 180-100=80°

When we divide the exterior angle, we will get the number of exterior angles

since it is a regular polygon means the number of exterior angles equals the number of sides.

Therefore n=360/ 80=4.5

And we know that 4.5 is not an integer, so having a regular polygon is impossible.

Whose exterior angle is 100°

Question 10: ABCD is a parallelogram in which ∠A=110 ° . Find the measure of the angles B, C and D, respectively.

Answer 10: The measure of angle A=110°

the sum of all adjacent angles of a parallelogram is 180°

∠A + ∠B = 180

110°+ ∠B = 180°

∠B = 180°- 110°

      = 70°.

Also ∠B + ∠C = 180° [Since ∠B and ∠C are adjacent angles]

70°+ ∠C = 180°

∠C = 180°- 70°

        = 110°.

Now ∠C + ∠D = 180° [Since ∠C and ∠D are adjacent angles]

110o+ ∠D = 180°

∠D = 180°- 110°

        = 70°

Question11: A diagonal and a side of a rhombus are of equal length. Find the measure of the angles of the rhombus.

Answer 11: Let ABCD be the rhombus.

                   All the sides of a rhombus are the same.

                   Thus, AB = BC = CD = DA.

                  The side and diagonal of a rhombus are equal.

                  AB = BD 

                  Therefore, AB = BC = CD = DA = BD 

                  Consider triangle ABD,

                  Each side of a triangle ABD is congruent.

                  Hence, ΔABD is an equilateral triangle.

                  Similarly,

                  ΔBCD is also an equilateral triangle.

                  Thus, ∠BAD = ∠ABD = ∠ADB = ∠DBC = ∠BCD = ∠CDB = 60°

                  ∠ABC = ∠ABD + ∠DBC = 60° + 60° = 120°

                  And

                  ∠ADC = ∠ADB + ∠CDB = 60° + 60° = 120°

     Hence, all angles of the given rhombus are 60°, 120°, 60° and 120°, respectively.

Question 12: The two adjacent angles of a parallelogram are the same. Find the measure of each and every angle of the parallelogram.

Answer 12: A parallelogram with two equal adjacent angles.

To find:- the measure of each of the angles of the parallelogram.

The sum of all the adjacent angles of a parallelogram is supplementary.

∠B = ∠A = 90°

In a parallelogram, the opposite sides are the same.

Hence, each angle of the parallelogram measures 90°.

Question 13: The measures of the two adjacent angles of a parallelogram are in the given ratio 3: 2. Find the measure of every angle of the parallelogram.

Answer 13: Let the measures of two adjacent angles ∠A and ∠B be 3x and 2x, respectively, in parallelogram ABCD.

∠A + ∠B = 180°

⇒ 3x + 2x = 180°

⇒ 5x = 180°

The opposite sides of a parallelogram are the same.

∠A = ∠C = 3x = 3 × 36° = 108°

∠B = ∠D = 2x = 2 × 36° = 72°

Question 14: State whether true or false.

(a) All the rectangles are squares.

(b) All the rhombuses are parallelograms.

(c) All the squares are rhombuses and also rectangles.

(d) All the squares are not parallelograms.

(e) All the kites are rhombuses.

(f) All the rhombuses are kites.

(g) All the parallelograms are trapeziums.

(h) All the squares are trapeziums.

Answer 14: (a) This statement is false.

Since all squares are rectangles, all rectangles are not squares.

                     (b) This statement is true.

                     (c) This statement is true.

                     (d) This statement is false.

                     Since all squares are parallelograms, the opposite sides are parallel, and opposite angles are 

                     congruent.

                    (e) This statement is false.

                     Since, for example, the length of the sides of a kite is not the same length.

                     (f) This statement is true.

                     (g) This statement is true.

                     (h) This statement is true.

Question 15: Two adjacent angles of a parallelogram are equal. What is the measure of each of these angles?

Answer 15: Let ∠A and ∠B be two adjacent angles.

But we know that the sum of adjacent angles of a parallelogram is 180o

But given that ∠A = ∠B

Now substituting, we get

∠A + ∠A = 180°

∠A=180/2 = 90°

Question 16:Triangle ABC is a right-angled triangle, and O is the midpoint of the side opposite to the right angle. State why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).

Answer 16: AD and DC are drawn in such a way that AD is parallel to BC 

and AB is parallel to DC

AD = BC and AB = DC

ABCD is a rectangle since the opposite sides are equal and parallel to each other, and the measure of all the interior angles is altogether 90°.

In a rectangle, all the diagonals bisect each other and are of equal length. 

Therefore, AO = OC = BO = OD

Hence, O is equidistant from A, B and C.

Question 17: Is the quadrilateral ABCD a parallelogram if

 (i) the measure of angle D + the measure of angle B = 180°?

 (ii) AB = DC = 8 cm , the length of AD = 4 cm and the length of BC = 4.4 cm?

(iii)The measure of angle A = 70° and the measure of angle C = 65°?

Answer 17: (i) Yes, the quadrilateral ABCD can be a parallelogram if ∠D + ∠B = 180° but it should also fulfil certain conditions, which are as follows:

(a) The sum of all the adjacent angles should be 180°.

(b) Opposite angles of a parallelogram must be equal.

(ii) No, opposite sides should be of the same length. Here, AD ≠ BC

(iii) No, opposite angles should be of the same measures. ∠A ≠ ∠C

Question 18: Find the measure of angles P and S if SP and RQ are parallel. 

Answer 18: ∠P + ∠Q = 180° (angles on the same side of transversal)

 ∠P + 130° = 180°

 ∠P = 180° – 130° = 50°

also, ∠R + ∠S = 180° (angles on the same side of transversal)

⇒ 90° + ∠S = 180°

⇒ ∠S = 180° – 90° = 90°

Thus, ∠P = 50° and ∠S = 90°

Yes, there is more than one method to find m∠P.

PQRS is a quadrilateral. The sum of measures of all angles is 360°.

Since we know the measurement of ∠Q, ∠R and ∠S.

∠Q = 130°, ∠R = 90° and ∠S = 90°

∠P + 130° + 90° + 90° = 360°

⇒ ∠P + 310° = 360°

⇒ ∠P = 360° – 310° = 50°

Question 19: The opposite angles of a parallelogram are (3x + 5)° and (61 – x)°. Find the measure of four angles.

Answer 19: (3x + 5)° and (61 – x)° are the opposite angles of a parallelogram.

The opposite angles of a parallelogram are the same. 

Therefore, (3x + 5)° = (61 – x)°

                    3x + x = 61° – 5°

                    4x = 56°

                    x = 56°/4

                    x = 14°

The first angle of the parallelogram =3x + 5

                                                               = 3(14) + 5 

                                                               = 42 + 5 = 47°

The second angle of the parallelogram=61 – x 

                                                                     = 61 – 14 = 47°

The measure of angles adjacent to the given angles = 180° – 47° = 133°

Hence, the measure of the four angles of the parallelogram is 47°, 133°, 47°, and 133°.

Question 20: What is the maximum exterior angle possible for a regular polygon?

Answer 20:   To find:- The maximum exterior angle possible for a regular polygon.

A polygon with minimum sides is an equilateral triangle.

So, the number of sides =3

The sum of all exterior angles of a polygon is 360°

Exterior angle =360°/Number of sides 

Therefore, the maximum exterior angle possible will be

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Chapter 3 Class 8 Understanding Quadrilaterals

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Get NCERT Solutions of Chapter 3 Class 8 Understanding Quadrilaterals free at teachoo. Answers to all exercise questions and examples have been solved, with concepts of the chapter explained.

In this chapter, we will learn

• What are curves , open curves, closed curves, simple curves
• What are polygons , Different Types of Polygons
• Diagonal of a Polygon
• Convex and Concave Polygons
• Regular and Irregular Polygons
• Angle Sum Property of Polygons
• Sum of Exterior Angles of a Polygon
• Exterior Angles of a Regular Polygon
• What is a Quadrilateral
• Parallelogram
• Parallelogram propertie s - Opposite Angles are equal, Opposite sides are equal, Adjacent Angles are supplementary, Diagonals Bisect Each other
• Rhombus, Rectangle, Square are all parallelograms with additional properties

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Class 8 Maths Chapter 3 Important Question Answers - Understanding Quadrilaterals

Q1: What is the maximum exterior angle possible for a regular polygon? Sol:   To find:-  The maximum exterior angle possible for a regular polygon. A polygon with minimum sides is an equilateral triangle. So, the number of sides = 3 The sum of all exterior angles of a polygon is 360° Exterior angle = 360°/Number of sides  Therefore, the maximum exterior angle possible will be = 360°/3 = 120°

Q2: Find the measure of angles P and S if SP and RQ are parallel.  Sol:   ∠P + ∠Q = 180° (angles on the same side of transversal) ∠P + 130° = 180° ∠P = 180° – 130° = 50° also, ∠R + ∠S = 180° (angles on the same side of transversal) ⇒ 90° + ∠S = 180° ⇒ ∠S = 180° – 90° = 90° Thus, ∠P = 50° and ∠S = 90° Yes, there is more than one method to find m∠P. PQRS is a quadrilateral. The sum of measures of all angles is 360°. Since we know the measurement of ∠Q, ∠R and ∠S. ∠Q = 130°, ∠R = 90° and ∠S = 90° ∠P + 130° + 90° + 90° = 360° ⇒ ∠P + 310° = 360° ⇒ ∠P = 360° – 310°  = 50°

Sol:  AD and DC are drawn in such a way that AD is parallel to BC   and AB is parallel to DC AD = BC and AB = DC ABCD is a rectangle since the opposite sides are equal and parallel to each other, and the measure of all the interior angles is altogether 90°. In a rectangle, all the diagonals bisect each other and are of equal length.  Therefore, AO = OC = BO = OD Hence, O is equidistant from A, B and C. Q4: State whether true or false. (a) All the rectangles are squares. (b) All the rhombuses are parallelograms. (c) All the squares are rhombuses and also rectangles. (d) All the squares are not parallelograms. (e) All the kites are rhombuses. (f) All the rhombuses are kites. (g) All the parallelograms are trapeziums. (h) All the squares are trapeziums. Sol: (a)  This statement is false. A rectangle has opposite sides equal and all angles equal to 90 degrees, but for it to be a square, all four sides must be equal. Therefore, not all rectangles are squares.

(b) This statement is true.

A rhombus is a type of parallelogram where all four sides are of equal length. Since it has both pairs of opposite sides parallel, it is always a parallelogram. 

(c)  This statement is true.

A square has all properties of a rhombus (all sides equal) and a rectangle (all angles 90 degrees). Therefore, all squares are both rhombuses and rectangles.

(d)  This statement is false.

A square is a specific type of parallelogram where all sides are equal and all angles are right angles. Therefore, all squares are parallelograms.

(e)  This statement is false.

A kite has two pairs of adjacent sides equal, but not all four sides need to be equal. For it to be a rhombus, all four sides must be equal. Therefore, not all kites are rhombuses.

(f) This statement is true.

A rhombus has all four sides equal, which satisfies the condition of a kite having two pairs of adjacent sides equal. Therefore, all rhombuses are kites. 

(g)  This statement is false.

A parallelogram has both pairs of opposite sides parallel, and a trapezium is a quadrilateral with exactly one pair of parallel sides. Since parallelograms have two pairs of parallel sides, they do not meet this criterion and are not considered trapeziums. 

(h)  This statement is true.

A square has both pairs of opposite sides parallel, which means it satisfies the condition of a trapezium having at least one pair of parallel sides. Therefore, all squares are trapeziums. 

Q5: The two adjacent angles of a parallelogram are the same. Find the measure of each and every angle of the parallelogram. Sol:  A parallelogram with two equal adjacent angles. To find:-  the measure of each of the angles of the parallelogram. The sum of all the adjacent angles of a parallelogram is supplementary. ∠A + ∠B = 180° 2∠A = 180° ∠A = 90° ∠B = ∠A = 90° In a parallelogram, the opposite sides are the same. Therefore, ∠C = ∠A = 90° ∠D = ∠B = 90° Hence, each angle of the parallelogram measures 90°. Q6: ABCD is a parallelogram in which ∠A = 110 ° . Find the measure of the angles B, C and D, respectively. Sol:  The measure of angle A = 110° the sum of all adjacent angles of a parallelogram is 180° ∠A + ∠B = 180 110°+ ∠B = 180° ∠B = 180°- 110° = 70°. Also ∠B + ∠C = 180° [Since ∠B and ∠C are adjacent angles] 70° + ∠C = 180° ∠C = 180°- 70° = 110°. Now ∠C + ∠D = 180° [Since ∠C and ∠D are adjacent angles] 110° + ∠D = 180° ∠D = 180°- 110° = 70° Q7: The measure of the two adjacent angles of the given parallelogram is the ratio of 3:2. Then, find the measure of each angle of the parallelogram. Sol:   A parallelogram with adjacent angles in the ratio of 3:2  To find:-  The measure of each of the angles of the parallelogram. Let the measure of angle A be 3x Let the measure of angle B be 2x Since the sum of the measures of adjacent angles is 180° for a parallelogram, ∠A+∠B=180° 3x+2x=180° 5x=180° x=36° ∠A=∠C =3x=108° ∠B=∠D =2x=72° (Opposite angles of a parallelogram are equal). Hence, the angles of a parallelogram are 108°, 72°,108°and 72°

Q7: Find x in the following figure. Sol:  The two interior angles in the given figures are right angles = 90° 70° + m = 180° m = 180° – 70°  = 110° (In a linear pair, the sum of two adjacent angles altogether measures up to 180°) 60° + n = 180°  n = 180° – 60°  = 120° (In a linear pair, the sum of two adjacent angles altogether measures up to 180° The given figure has five sides, and it is a pentagon. Thus, the sum of the angles of the pentagon = 540°  90° + 90° + 110° + 120° + y = 540° 410° + y = 540° y = 540° – 410° = 130° x + y = 180°….. (Linear pair) x + 130° = 180° x = 180° – 130°  = 50°

Q8: Adjacent sides of a rectangle are in the ratio 5: 12; if the perimeter of the given rectangle is 34 cm, find the length of the diagonal. Sol:  The ratio of the adjacent sides of the rectangle is 5: 12 Let 5x and 12x be adjacent sides. The perimeter is the sum of all the given sides of a rectangle. 5x + 12x + 5x + 12x = 34 cm ……(since the opposite sides of the rectangle are the  same) 34x = 34 x = 34/34 x = 1 cm Therefore, the adjacent sides of the rectangle are 5 cm and 12 cm, respectively. That is, Length =12 cm Breadth = 5 cm Length of the diagonal = √( l2 + b2) = √( 122 + 52) = √(144 + 25) = √169 = 13 cm Hence, the length of the diagonal of a rectangle is 13 cm.

Q9: Find the measure of all the exterior angles of a regular polygon with  (i) 9 sides and  (ii) 15 sides. Sol:  (i)  Total measure of all exterior angles = 360°  Each exterior angle =sum of exterior angle = 360° = 40° number of sides 9                                               Each exterior angle = 40° (ii)  Total measure of all exterior angles = 360°  Each exterior angle = sum of exterior angle =360° = 24°  number of sides 15 Each exterior angle = 24° 

Q10: A quadrilateral has three acute angles, each measuring 80°. What is the measure of the fourth angle of the quadrilateral? Sol: – Let x be the measure of the fourth angle of a quadrilateral. The sum of all the angles of a quadrilateral + 360° 80° + 80° + 80° + x = 360° …………(since the measure of all the three acute angles = 80°)  240° + x = 360° x = 360° – 240° x = 120° Hence, the fourth angle made by the quadrilateral is 120°. Q11: How many sides do regular polygons consist of if each interior angle is 165 ° ? Sol:  A regular polygon with an interior angle of 165° We need to find the sides of the given regular polygon:- The sum of all exterior angles of any given polygon is 360°. Formula Used: Number of sides = 360 ∘ /Exterior angle Exterior angle = 180 ∘ −Interior angle Thus, Each interior angle = 165° Hence, the measure of every exterior angle will be = 180° − 165° = 15° Therefore, the number of sides of the given polygon will be = 360°/15° = 24°  Q12: ABCD is a parallelogram with ∠A = 80°. The internal bisectors of ∠B and ∠C meet each other at O. Find the measure of the three angles of ΔBCO. Sol:  The measure of angle A = 80°. In a parallelogram, the opposite angles are the same. Hence, ∠A = ∠C = 80° And ∠OCB = (1/2) × ∠C = (1/2) × 80° = 40° ∠B = 180° – ∠A (the sum of interior angles situated on the same side of the transversal is supplementary) = 180° – 80° = 100° Also, ∠CBO = (1/2) × ∠B ∠CBO= (1/2) × 100° ∠CBO= 50°.  By the property of the sum of the angle BCO, we get, ∠BOC + ∠OBC + ∠CBO = 180° ∠BOC = 180° – (∠OBC + CBO) = 180° – (40° + 50°) = 180° – 90° = 90° Hence, the measure of all the angles of triangle BCO is 40°, 50° and 90°. Q13: Is it ever possible to have a regular polygon, each of whose interior angles is 100? Sol:  The sum of all the exterior angles of a regular polygon is 360° As we also know, the sum of interior and exterior angles are 180° Exterior angle + interior angle = 180 - 100 = 80° When we divide the exterior angle, we will get the number of exterior angles  since it is a regular polygon means the number of exterior angles equals the number of sides. Therefore n = 360/ 80 = 4.5 And we know that 4.5 is not an integer, so having a regular polygon is impossible. Whose exterior angle is 100° Q14: A diagonal and a side of a rhombus are of equal length. Find the measure of the angles of the rhombus. Sol:  Let ABCD be the rhombus. All the sides of a rhombus are the same. Thus, AB = BC = CD = DA. The side and diagonal of a rhombus are equal. AB = BD  Therefore, AB = BC = CD = DA = BD  Consider triangle ABD, Each side of a triangle ABD is congruent. Hence, ΔABD is an equilateral triangle. Similarly, ΔBCD is also an equilateral triangle. Thus, ∠BAD = ∠ABD = ∠ADB = ∠DBC = ∠BCD = ∠CDB = 60° ∠ABC = ∠ABD + ∠DBC = 60° + 60° = 120° And ∠ADC = ∠ADB + ∠CDB = 60° + 60° = 120° Hence, all angles of the given rhombus are 60°, 120°, 60° and 120°, respectively. Q15: The measures of the two adjacent angles of a parallelogram are in the given ratio 3: 2. Find the measure of every angle of the parallelogram. Sol:  Let the measures of two adjacent angles ∠A and ∠B be 3x and 2x, respectively, in parallelogram ABCD. ∠A + ∠B = 180° ⇒ 3x + 2x = 180° ⇒ 5x = 180° ⇒ x = 36° The opposite sides of a parallelogram are the same. ∠A = ∠C = 3x = 3 × 36° = 108° ∠B = ∠D = 2x = 2 × 36° = 72°

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Extra Questions – Class 8 Maths Chapter 3 Understanding Quadrilaterals

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In Class 8 Mathematics, Chapter 3 focuses on Understanding Quadrilaterals , which are shapes with four sides. To help students practice and understand this chapter better, extra questions have been created. These extra questions are like bonus exercises that give students more practice and help them explore quadrilaterals in more detail.

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The Class 8 Maths Chapter 3 Extra Questions cover various aspects of quadrilaterals, such as their properties, types, and how they are used. By solving these extra questions, students can improve their knowledge and skills in working with quadrilaterals. The questions also come with solutions, making it easier for students to check their answers and learn from their mistakes.

These extra questions provide a fun and engaging way for students to learn more about quadrilaterals. They can practice identifying different types of quadrilaterals, understanding their features, and solving problems related to them. By working through these extra questions , students can boost their confidence in math and be better prepared for tests and exams.

Class 8 Maths Chapter 3 Extra Questions with Solutions – Understanding Quadrilaterals

For Class 8 students learning about quadrilaterals, extra questions with solutions are a helpful tool. These extra questions from chapter 3 class 8 maths cover different aspects of quadrilaterals and come with answers to check your work. By practicing with these questions, students can improve their understanding of quadrilaterals and how to solve related problems.

The solutions provided not only give the correct answers but also explain how to solve each question step by step. This resource helps students identify areas where they need more practice and enhances their overall grasp of quadrilaterals in a clear and straightforward manner.

Get Extra Questions for Class 8 Maths on Infinity Learn for free.

Very Short Answer Type

3x + 5 = 5x – 1

⇒ 3x – 5x = -1 – 5

x + y + z = 360°

(x + 10)° + (3x + 5)° + (2x + 15)° = 180°

⇒ x + 10 + 3x + 5 + 2x + 15 = 180

⇒ 6x + 30 = 180

⇒ 6x = 180 – 30

Question 4. The angles of a quadrilateral are in the ratio of 2 : 3 : 5 : 8. Find the measure of each angle.

Solution: The sum of a quadrilateral’s internal angles equals 360°.

Let the quadrilateral’s angles be 2x°, 3x°, 5x°, and 8x°.

2x + 3x + 5x + 8x = 360°

⇒ 18x = 360°

Hence the angles are

2 × 20 = 40°,

3 × 20 = 60°,

5 × 20 = 100°

and 8 × 20 = 160°.

Question 5. Find the measure of an interior angle of a regular polygon of 9 sides.

Solution: Measure of an interior angle of a regular polygon

Question 6. Length and breadth of a rectangular wire are 9 cm and 7 cm respectively. If the wire is bent into a square, find the length of its side.

Side of the square = \(\frac { 32 }{ 4 }\) = 8 cm.

Hence, the length of the side of square = 8 cm.

Then m∠S = 110° (Opposite angles are equal)

Since ∠P and ∠Q are supplementary.

Then m∠P + m∠Q = 180°

⇒ m∠P + 110° = 180°

⇒ m∠P = 180° – 110° = 70°

⇒ m∠P = m∠R = 70° (Opposite angles)

Hence m∠P = 70, m∠R = 70°

and m∠S = 110°

Since the diagonals of a rhombus bisect each other

z = 5 and y = 12

Hence, x = 13 cm, y = 12 cm and z = 5 cm.

⇒ 125° + ∠D = 180°

⇒ ∠D = 180° – 125°

⇒ 125° = y + 56°

⇒ y = 125° – 56°

∠z + ∠y = 180° (Adjacent angles)

⇒ ∠z + 69° = 180°

⇒ ∠z = 180° – 69° = 111°

Hence the angles x = 55°, y = 69° and z = 111°

Now, sum of exterior angles of a polygon is 360°, therefore,

x + 60° + 90° + 90° + 40° = 360°

⇒ x + 280° = 360°

Short Answer Type

3y + 2y – 5 = 180°

⇒ 5y – 5 = 180°

⇒ 5y = 180 + 5°

⇒ 5y = 185°

3y = 3x + 3

⇒ 3 × 37 = 3x + 3

⇒ 111 = 3x + 3

⇒ 111 – 3 = 3x

Hence, x = 36° and y – 37°.

∠ABC = ∠ADC (Opposite angles of a rhombus)

∠ADC = 126°

∠ODC = \(\frac { 1 }{ 2 }\) × ∠ADC (Diagonal of rhombus bisects the respective angles)

⇒ ∠ODC = \(\frac { 1 }{ 2 }\) × 126° = 63°

⇒ ∠DOC = 90° (Diagonals of a rhombus bisect each other at 90°)

∠OCD + ∠ODC + ∠DOC = 180° (Angle sum property)

⇒ ∠OCD + 63° + 90° = 180°

⇒ ∠OCD + 153° = 180°

⇒ ∠OCD = 180° – 153° = 27°

Hence ∠OCD or ∠ACD = 27°

x + 8 = 16 – x

⇒ x + x = 16 – 8

Similarly, OB = OD

5y + 4 = 2y + 13

Hence, x = 4 and y = 3

Question 15. Write true and false against each of the given statements.

(a) Diagonals of a rhombus are equal.

(b) Diagonals of rectangles are equal.

(c) Kite is a parallelogram.

(d) Sum of the interior angles of a triangle is 180°.

(e) A trapezium is a parallelogram.

(f) Sum of all the exterior angles of a polygon is 360°.

(g) Diagonals of a rectangle are perpendicular to each other.

(h) Triangle is possible with angles 60°, 80° and 100°.

(i) In a parallelogram, the opposite sides are equal.

Question 16. The sides AB and CD of a quadrilateral ABCD are extended to points P and Q respectively. Is ∠ADQ + ∠CBP = ∠A + ∠C? Give reason.

Join AC, then

∠CBP + ∠ADQ = ∠BCA + ∠BAC + ∠ACD + ∠DAC

= (∠BCA + ∠ACD) + (∠BAC + ∠DAC)

Higher Order Thinking Skills (HOTS)

Let AD = x cm

diagonal BD = 3x cm

In right-angled triangle DAB,

AD 2 + AB 2 = BD 2 (Using Pythagoras Theorem)

x 2 + AB 2 = (3x) 2

⇒ x 2 + AB 2 = 9x 2

⇒ AB 2 = 9x 2 – x 2

⇒ AB 2 = 8x 2

⇒ AB = √8x = 2√2x

Required ratio of AB : AD = 2√2x : x = 2√2 : 1

Question 18. If AM and CN are perpendiculars on the diagonal BD of a parallelogram ABCD, Is ∆AMD = ∆CNB? Give reason. (NCERT Exemplar)

AD = BC (opposite sides of parallelogram)

∠AMB = ∠CNB = 90°

∠ADM = ∠NBC (AD || BC and BD is transversal.)

So, ∆AMD = ∆CNB (AAS)

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NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals

NCERT solutions for class 8 maths chapter 3 understanding quadrilaterals define a polygon as a simple closed curve that is made up of straight lines. Thus, a quadrilateral can be defined as a polygon that has four sides, four angles, and four vertices. This chapter starts by introducing children to some very important concepts that they need to learn before moving on to studying quadrilaterals . These topics include the classification of polygons on the basis of sides, examining diagonals , concave, convex, regular, and irregular polygons as well as the angle sum property. The scope of NCERT solutions class 8 maths chapter 3 is very vast as there are several properties and types of quadrilaterals available. However, the explanation given in these solutions helps to simplify the learning process ensuring that students can build a strong geometrical foundation. 

Class 8 maths NCERT solutions chapter 3 elaborates on special quadrilaterals such as squares , rectangles , parallelograms , kites , and rhombuses . They show kids how to solve problems based on these figures and intelligently utilize the associated properties to remove the complexities from such questions. In the NCERT solutions Chapter 3 Understanding Quadrilaterals we will take an in-depth look at the basic elements and theories of these four-sided polygons and also you can find some of these in the exercises given below.

• NCERT Solutions Class 8 Maths Chapter 3 Ex 3.1
• NCERT Solutions Class 8 Maths Chapter 3 Ex 3.2
• NCERT Solutions Class 8 Maths Chapter 3 Ex 3.3
• NCERT Solutions Class 8 Maths Chapter 3 Ex 3.4

NCERT Solutions for Class 8 Maths Chapter 3 PDF

Using the NCERT solutions class 8 maths children can solidify several concepts of quadrilaterals. They understand the conditions under which a special quadrilateral such as a parallelogram becomes a square, how to find the measure of an interior or exterior angle , and so on. The links to all these brief and precise solutions are given below and kids can use them to improve their mathematical acumen.

☛ Download Class 8 Maths NCERT Solutions Chapter 3 Understanding Quadrilaterals

NCERT Class 8 Maths Chapter 3   Download PDF

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

Quadrilaterals form a vital shape contributing to geometrical studies. Thus, children need to develop a robust conceptual foundation as they will require it in higher classes for solving more complicated problems and constructing this figure. They can do this by revising the solutions given above regularly. The following sections deal with an exercise-wise detailed analysis of NCERT Solutions Class 8 Maths Chapter 3 understanding quadrilaterals.

• Class 8 Maths Chapter 3 Ex 3.1 - 7 Questions
• Class 8 Maths Chapter 3 Ex 3.2 - 6 Questions
• Class 8 Maths Chapter 3 Ex 3.3 - 12 Questions
• Class 8 Maths Chapter 3 Ex 3.4 - 6 Questions

☛ Download Class 8 Maths Chapter 3 NCERT Book

Topics Covered: Identifying the polygon, finding the measure of angles, and verifying the exterior angles of a polygon are topics under class 8 maths NCERT solutions chapter 3. Apart from this, there are many sections dealing with the various elements of trapeziums , parallelograms, rectangles, squares, etc.

Total Questions: There are a total of 31 fantastic sums in Class 8 maths chapter 3 Understanding Quadrilaterals. 7 are simple theory-based problems, 16 are in-between and 8 are higher-order thinking sums.

List of Formulas in NCERT Solutions Class 8 Maths Chapter 3

The questions in the NCERT solutions class 8 maths chapter 3 are not only based on some formulas but also see the use of various vital properties. The sum of interior and exterior angles , along with theorems give the keys to attempting these sums. The angle sum property states that the sum of all the interior angles of a polygon is a multiple of the number of triangles that make up that polygon. Such pointers covered in NCERT solutions for class 8 maths chapter 3 make up the crux of this lesson and are given below.

• Angle Sum Property of a Quadrilateral: a + b + c + d = 360°. (a, b, c, d are the interior angles).
• The opposite sides and opposite angles of a parallelogram are equal in length.
• The adjacent angles in a parallelogram are supplementary.
• The diagonals of a parallelogram bisect each other.
• The diagonals of a rhombus are perpendicular bisectors of one another.

Important Questions for Class 8 Maths NCERT Solutions Chapter 3

NCERT Solutions for Class 8 Maths Video Chapter 3

FAQs on NCERT Solutions Class 8 Maths Chapter 3

Do i need to practice all questions provided in ncert solutions class 8 maths understanding quadrilaterals.

All the sums in the NCERT Solutions Class 8 Maths Understanding Quadrilaterals cover different subtopics of the lesson. These sums also pave a foundation for the geometrical topics in grades that are to follow. Thus, it is crucial for kids to practice all questions so as to get a clear idea of all the components in a quadrilateral.

What are the Important Topics Covered in Class 8 Maths NCERT Solutions Chapter 3?

Each exercise is based on a different topic such as angles of a polygon, rhombus, square, and rectangles; thus, each section that falls under the NCERT Solutions Class 8 Maths Chapter 3 must be given equal importance. Kids need to strategize their studies to focus more on learning properties and then applying them to questions.

How Many Questions are there in NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals?

There are a total of 31 questions in the NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals that are distributed among 4 exercises. There are different types of questions such as true and false sums, identifying the type of shape based on certain properties, and finding the measure of a particular angle using formulas.

What are the Important Formulas in Class 8 Maths NCERT Solutions Chapter 3?

Formulas such as the angle sum property of a quadrilateral, exterior angle property of a polygon, and other associated theories form the foundation of the NCERT Solutions Class 8 Maths Chapter 3. Students must spend a good amount of time practicing questions so as to get a good understanding of their application.

How CBSE Students can utilize NCERT Solutions Class 8 Maths Chapter 3 effectively?

To effectively utilize NCERT Solutions Class 8 Maths Chapter 3 it is advised that students go through the theory and solved examples associated with each exercise. They should then try to attempt the problem on their own. Finally, to get the best out of these solutions kids should cross-check their answers and go through the steps so that they can organize their answers in a well-structured manner.

Why Should I Practice NCERT Solutions Class 8 Maths Understanding Quadrilaterals Chapter 3?

The only way to ensure that a student has perfected his knowledge of a chapter is by practicing the questions periodically. The NCERT Solutions Class 8 Maths Understanding Quadrilaterals Chapter 3 has been given by experts with certain tips included to simplify the problems. By regular revision, kids will be confident with the topic and can get an amazing score in their examination.

• CBSE-Understanding Quadrilaterals
• Sample Questions

Understanding Quadrilaterals-Sample Questions

• STUDY MATERIAL FOR CBSE CLASS 8 MATH
• Chapter 1 - Algebraic Expressions and Identities
• Chapter 2 - Comparing Quantities
• Chapter 3 - Cubes and Cube Roots
• Chapter 4 - Data handling
• Chapter 5 - Direct and Inverse Proportions
• Chapter 6 - Exponents and Powers
• Chapter 7 - Factorization
• Chapter 8 - Introduction to Graphs
• Chapter 9 - Mensuration
• Chapter 10 - Playing with Numbers
• Chapter 11 - Practical Geometry
• Chapter 12 - Squares and Square Roots
• Chapter 13 - Visualizing Solid Shapes
• Chapter 14 - Linear Equations in One Variable
• Chapter 15 - Rational Numbers
• Chapter 16 - Understanding Quadrilaterals

• Understanding Quadrilaterals Class 8 Assertion Reason Questions Maths Chapter 3

Last Updated on August 26, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 8 maths. Assertion Reason questions are the new question format that is introduced in CBSE board. The resources for assertion reason questions are very less. So, to help students we have created chapterwise assertion reason questions for class 8 maths. In this article, you will find assertion reason questions for CBSE Class 8 Maths Chapter 3 Understanding Quadrilaterals. It is a part of Assertion Reason Questions for CBSE Class 8 Maths Series.

Table of Contents

Assertion Reason Questions on Understanding Quadrilaterals

Questions :

Q. 1. Assertion (A): In a quadrilateral, there are 4 sides. Reason(R): A quadrilateral is a four-sided polygon, having four vertices and four edges. (a) Both A and R are true and R is the correct explanation of $A$ (b) Both $A$ and $R$ are true but $R$ is not the correct explanation of A (c) $A$ is true but $R$ is false (d) $A$ is false but $R$ is true

Difficulty Level: Medium

Ans. Option (a) is correct Explanation: Quadrilateral has 4 sides, 4 vertices and 4 edges.

Also read: Understanding Quadrilaterals Case Study Questions for Class 8

Q. 2. Assertion (A): All the parallelograms are rectangles. Reason(R): All the rhombuses are parallelograms. (a) Both $A$ and $R$ are true and $R$ is the correct explanation of $A$ (b) Both $A$ and $R$ are true but $R$ is not the correct explanation of A (c) $A$ is true but $R$ is false (d) $A$ is false but $R$ is true

Ans. Option (d) is correct Explanation: All parallelograms are not rectangles, but all rhombuses are parallelograms.

Linear Equations in One Variable Class 8 Assertion Reason Questions Maths Chapter 2

Rational numbers class 8 assertion reason questions maths chapter 1, you may also like.

• Case Study Questions for CBSE Class 8 Maths

Download eBooks for CBSE Class 8 Maths Understanding Quadrilaterals

• Understanding Quadrilaterals Topicwise Worksheet for CBSE Class 8 Maths

Topics from which assertion reason questions may be asked

• Convex and Concave Polygons.
• Regular and Irregular Polygons.
• Sum of Measures of the Exterior Angles of a Polygon.
• Kinds of QuadrilateralTrapezium; Kite; Parallelogram.
• Some Special ParallelogramsRhombus; Rectangle; Square.

Assertion reason questions from the above given topic may be asked.

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Frequently Asked Questions (FAQs) on Understanding Quadrilaterals Assertion Reason Questions Class 8

Q1: what are assertion reason questions.

A1: Assertion-reason questions consist of two statements: an assertion (A) and a reason (R). The task is to determine the correctness of both statements and the relationship between them. The options usually include: (i) Both A and R are true, and R is the correct explanation of A. (ii) Both A and R are true, but R is not the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true. or A is false, and R is also false.

Q2: Why are assertion reason questions important in Maths?

A2: Students need to evaluate the logical relationship between the assertion and the reason. This practice strengthens their logical reasoning skills, which are essential in mathematics and other areas of study.

Q3: How can practicing assertion reason questions help students?

A3: Practicing assertion-reason questions can help students in several ways: Improved Conceptual Understanding:  It helps students to better understand the concepts by linking assertions with their reasons. Enhanced Analytical Skills:  It enhances analytical skills as students need to critically analyze the statements and their relationships. Better Exam Preparation:  These questions are asked in exams and practicing them can improve your performance.

Q4: What strategies should students use to answer assertion reason questions effectively?

A4: Students can use the following strategies: Understand Each Statement Separately:  Determine if each statement is true or false independently. Analyze the Relationship:  If both statements are true, check if the reason correctly explains the assertion.

Q5: What are common mistakes to avoid when answering Assertion Reason questions?

A5: Common mistakes include: Not reading the statements carefully and missing key details. Assuming the Reason explains the Assertion without checking the logical connection. Confusing the order or relationship between the statements. Overthinking and adding information not provided in the question.

Q6: What resources can help me practice Assertion Reason Questions for Class 8 Maths?

A6: Use study guides specifically designed for Assertion-Reason questions. Online educational platforms and reference books for Class 8 Maths also offer practice questions and explanations. xamcontent.com also provides assertion reason questions for cbse class 8 maths.

Q7: How do you classify quadrilaterals based on their properties?

A7: Quadrilaterals can be classified based on their properties such as sides, angles, and diagonals. For example: (1) Parallelograms have opposite sides that are equal and parallel. (2) Rhombuses have all four sides equal in length. (3) Rectangles have all angles equal to 90 degrees.

Q8: Can a quadrilateral have equal sides and angles but still not be a square?

A8: Yes, a rhombus can have all sides equal and opposite angles equal, but its angles need not be right angles, unlike in a square.

Q9: What is the difference between a square and a rhombus?

A9: A square is a type of rhombus with all four sides equal and all angles equal to 90 degrees. However, a rhombus may have all sides equal but not necessarily all angles equal to 90 degrees.

Q10: Are there any online resources or tools available for practicing linear equations in one variable assertion reason questions?

A10: A9: We provide assertion reason questions for CBSE Class 8 Maths on our  website . Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit  Physics Gurukul  website. they are having a large collection of case study questions for all classes.

Related Posts

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

Class 8 Maths Chapter 3 Try These Class 8 Maths Exercise 3.1 Solutions Class 8 Maths Exercise 3.2 Solutions Class 8 Maths Exercise 3.3 Solutions Class 8 Maths Exercise 3.4 Solutions

The NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals and Class 8 Maths Chapter 3 Try These Solutions in English and Hindi Medium modified and updated for session 2024-25. As per the revised syllabus, the of exercises in chapter 3 of class 8th Maths are four only, which are given here with solutions.

8th Maths Chapter 3 Solutions in English Medium

• Class 8 Maths Chapter 3 Try These Solutions
• Class 8 Maths Exercise 3.1 in English
• Class 8 Maths Exercise 3.2 in English
• Class 8 Maths Exercise 3.3 in English
• Class 8 Maths Exercise 3.4 in English

8th Maths Chapter 3 Solutions in Hindi Medium

• Class 8 Maths Exercise 3.1 in Hindi
• Class 8 Maths Exercise 3.2 in Hindi
• Class 8 Maths Exercise 3.3 in Hindi
• Class 8 Maths Exercise 3.4 in Hindi
• Class 8 Maths Chapter 3 NCERT Book
• Class 8th Maths Solutions Page
• Class 8 all Subjects Solutions

Class VIII Mathematics chapter 3 is based on latest NCERT Books for current year, useful to all students. Download Prashnavali 3.1, Prashnavali 3.2, Prashnavali 3.3 and Prashnavali 3.4 in Hindi Medium and Exercise 3.1, Exercise 3.2, Exercise 3.3 and Exercise 3.4 in English Medium free to download in PDF format. These NCERT Solutions are applicable for all board using NCERT Books for their academic session. All the solutions are updated on the basis of latest CBSE Curriculum 2024-25. This chapter is based on closed figures like triangles, quadrilaterals and other polygons.

Free NCERT Solutions App for 8th

More about Class 8 Maths Chapter 3

What is a regular polygon state the name of a regular polygon of: (a) 3 sides (b) 4 sides (c) 6 sides.

A regular polygon: A polygon having all sides of equal length and the interior angles of equal size is known as regular polygon. (i) 3 sides Polygon having three sides is called a triangle.

(ii) 4 sides Polygon having four sides is called a quadrilateral.

(iii) 6 sides Polygon having six sides is called a hexagon.

How many sides does a regular polygon have, if the measure of an exterior angle is 24 degree?

Let number of sides be n. Sum of exterior angles of a regular polygon = 360 Number of sides = n = 360/24 = 15 Hence, the regular polygon has 15 sides.

Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1, Exercise 3.2, Exercise 3.3 and Exercise 3.4 in English Medium as well as Hindi Medium are given below to download in PDF form. Download Class 8 Maths App in English for offline use and Kaksha 8 Ganit App in Hindi Medium for the academic session 2024-25.

In Chapter 3 Understanding Quadrilaterals, we will learn about plane surface and plane figures, different types of polygons like Triangles, Quadrilaterals, Pentagon, Hexagon, Heptagon, etc. Number of diagonals in each polygon, a brief description about CONVEX and CONCAVE polygons. Regular and irregular polygons having 3, 4, 5 and 6. Angle sum property of triangle and the questions based on the same fact to find the missing term. Questions based on sum of measure of exterior angles and questions related to Trapezium, Parallelogram, Kite to find the missing side or angle.

Download NCERT Books and Offline apps based on new CBSE Syllabus. Ask your doubts and share your knowledge with your friends and other users.

There are exercises based on the properties of different types of quadrilaterals in this lesson. Just as all the sides and all the angles of the square are equal, the opposite sides of the rectangle are equal, the opposite angle and the diagonal are also equal. Quadrilateral and rectangle have the difference of angles and diagonals. If the diagonals of a rhombus are equal, then it is square. Every angle of a square is a right angle. All the questions in chapter 3 of 8th Maths are mainly based on the properties of polygons.

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NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals

• NCERT Solutions
• Chapter 3 Understanding Quadrilaterals

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals - FREE PDF Download

The NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals covers all the chapter's questions (All Exercises). These NCERT Solutions for Class 8 Maths have been carefully compiled and created in accordance with the most recent CBSE Syllabus 2024-25 updates. Students can use these NCERT Solutions for Class 8 to reinforce their foundations. Subject experts at Vedantu have created the continuity and differentiability class 8 NCERT solutions to ensure they match the current curriculum and help students while solving or practising problems.

Glance of NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals  | Vedantu

In this article, we will learn about different quadrilaterals like squares, rectangles, parallelograms, rhombuses, and trapeziums, along with their properties.

This chapter dives into the world of quadrilaterals, which are four-sided closed figures.

This chapter explains effective methods to solve problems concerning quadrilaterals.

Each type of quadrilateral is discussed in terms of its defining properties including side lengths, angle measurements, diagonals, and symmetry.

The chapter also highlights special properties of certain quadrilaterals, like the properties of diagonals in rectangles and squares, and the diagonals of rhombuses.

This article contains chapter notes, formulas, exercise links, and important questions for chapter 3 - Understanding Quadrilaterals.

There are four exercises (26 fully solved questions) in Class 8th Maths Chapter 3 Understanding Quadrilaterals.

Access Exercise Wise NCERT Solutions for Chapter 3 Maths Class 8

Exercises Under NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

Exercise 3.1 introduces polygons, covering their basic definition and classification based on the number of sides, such as triangles, quadrilaterals, pentagons, etc. It also distinguishes between convex and concave polygons, helping students understand the differences between these types of polygons.

Exercise 3.2 delves into the properties of quadrilaterals, exploring various types like trapeziums, kites, and parallelograms. This exercise helps students learn to identify different quadrilaterals and understand their specific properties.

Exercise 3.3 examines the properties of parallelograms, such as opposite sides being equal and parallel, opposite angles being equal, and diagonals bisecting each other. It includes problems for identifying parallelograms based on these properties and proving certain properties using theorems.

Exercise 3.4 looks at special parallelograms like rhombuses, rectangles, and squares. It highlights their unique properties, such as all sides being equal in a rhombus and all angles being 90 degrees in a rectangle. This exercise helps students understand and differentiate between these specific types of parallelograms.

List of Formulas

There are two major kinds of formulas related to quadrilaterals - Area and Perimeter. The following tables depict the formulas related to the areas and perimeters of different kinds of quadrilaterals.

Area of Quadrilaterals

Perimeter of Quadrilaterals

Perimeter of any quadrilateral is equal to the sum of all its sides, that is, AB + BC + CD + AD.

Access NCERT Solutions for Class 8 Maths Chapter 3 – Understanding Quadrilaterals

Exercise 3.1.

1. Given here are some figures.

Classify each of them on the basis of following.

Simple Curve

Ans: Given: the figures $(1)$to $(8)$

We need to classify the given figures as simple curves.

We know that a curve that does not cross itself is referred to as a simple curve.

Therefore, simple curves are $1,2,5,6,7$.

Simple Closed Curve

We need to classify the given figures as simple closed curves.

We know that a simple closed curve is one that begins and ends at the same point without crossing itself.

Therefore, simple closed curves are $1,2,5,6,7$.

We need to classify the given figures as polygon.

We know that any closed curve consisting of a set of sides joined in such a way that no two segments

cross is known as a polygon.

Therefore, the polygons are $1,2$.

Convex Polygon

We need to classify the given figures as convex polygon.

We know that a closed shape with no vertices pointing inward is called a convex polygon.

Therefore, the convex polygon is $2$.

Concave Polygon

We need to classify the given figures as concave polygon.

We know that a polygon with at least one interior angle greater than 180 degrees is called a concave

Therefore, the concave polygon is $1$.

2. What is a regular polygon?

State the name of a regular polygon of

(i) 3 sides (ii) 4 sides (iii) 6 sides

Solution: A regular polygon is a flat shape with all sides of equal length and all interior 5angles equal in measure.  In simpler terms, all the sides are the same size and all the corners look the same.

Here are the names of regular polygons based on the number of sides:

(i) 3 sides -  Equilateral Triangle (all three angles are also 60 degrees each)

(ii) 4 sides - Square

(iii) 6 sides - Hexagon

Exercise-3.2

1. Find ${\text{x}}$in the following figures.

We need to find the value of ${\text{x}}{\text{.}}$

We know that the sum of all exterior angles of a polygon is ${360^ \circ }.$

$  {\text{x}} + {125^ \circ } + {125^ \circ } = {360^ \circ } $

$   \Rightarrow {\text{x}} + {250^ \circ } = {360^ \circ } $

$   \Rightarrow {\text{x}} = {360^ \circ } - {250^ \circ } $

$   \Rightarrow {\text{x}} = {110^ \circ } $ 

$  {\text{x}} + {90^ \circ } + {60^ \circ } + {90^ \circ } + {70^ \circ } = {360^ \circ } $

$  \Rightarrow {\text{x}} + {310^ \circ } = {360^ \circ } $

$   \Rightarrow {\text{x}} = {360^ \circ } - {310^ \circ } $

 $  \Rightarrow {\text{x}} = {50^ \circ } $ 

2. Find the measure of each exterior angle of a regular polygon of 

Given: a regular polygon with $9$ sides

We need to find the measure of each exterior angle of the given polygon.

We know that all the exterior angles of a regular polygon are equal.

The sum of all exterior angle of a polygon is ${360^ \circ }$.

Formula Used: ${\text{Exterior}}\;{\text{angle}} = \dfrac{{{{360}^ \circ }}}{{{\text{Number}}\;{\text{of}}\;{\text{sides}}}}$

Sum of all angles of given regular polygon $ = {360^ \circ }$

Number of sides $ = 9$

Therefore, measure of each exterior angle will be

$   = \dfrac{{{{360}^ \circ }}}{9} $

 $  = {40^ \circ } $ 

Given: a regular polygon with $15$ sides

Number of sides $ = 15$

$   = \dfrac{{{{360}^ \circ }}}{{15}} $

$ = {24^ \circ } $ 

3. How many sides does a regular polygon have if the measure of an exterior angle is ${24^ \circ }$?

Ans: Given: A regular polygon with each exterior angle ${24^ \circ }$

We need to find the number of sides of given polygon.

We know that sum of all exterior angle of a polygon is ${360^ \circ }$.

Formula Used: ${\text{Number}}\;{\text{of}}\;{\text{sides}} = \dfrac{{{{360}^ \circ }}}{{{\text{Exterior}}\;{\text{angle}}}}$

Each angle measure $ = {24^ \circ }$

Therefore, number of sides of given polygon will be

$   = \dfrac{{{{360}^ \circ }}}{{{{24}^ \circ }}} $

 $  = 15 $ 

4. How many sides does a regular polygon have if each of its interior angles is ${165^ \circ }$?

Ans: Given: A regular polygon with each interior angle ${165^ \circ }$

We need to find the sides of the given regular polygon.

${\text{Exterior}}\;{\text{angle}} = {180^ \circ } - {\text{Interior}}\;{\text{angle}}$

Each interior angle $ = {165^ \circ }$

So, measure of each exterior angle will be

$   = {180^ \circ } - {165^ \circ } $

$   = {15^ \circ } $ 

Therefore, number of sides of polygon will be

$   = \dfrac{{{{360}^ \circ }}}{{{{15}^ \circ }}} $

$   = 24 $ 

Is it possible to have a regular polygon with measure of each exterior angle as ${22^ \circ }$?

Given: A regular polygon with each exterior angle ${22^ \circ }$

We need to find if it is possible to have a regular polygon with given angle measure.

We know that sum of all exterior angle of a polygon is ${360^ \circ }$. The polygon will be possible if ${360^ \circ }$ is a perfect multiple of exterior angle.

$\dfrac{{{{360}^ \circ }}}{{{{22}^ \circ }}}$ does not give a perfect quotient. 

Thus, ${360^ \circ }$ is not a perfect multiple of exterior angle. So, the polygon will not be possible.

Can it be an interior angle of a regular polygon? Why?

Ans: Given: Interior angle of a regular polygon $ = {22^ \circ }$

We need to state if it can be the interior angle of a regular polygon.

And, ${\text{Exterior}}\;{\text{angle}} = {180^ \circ } - {\text{Interior}}\;{\text{angle}}$

Thus, Exterior angle will be

$   = {180^ \circ } - {22^ \circ } $

 $  = {158^ \circ } $ 

$\dfrac{{{{158}^ \circ }}}{{{{22}^ \circ }}}$ does not give a perfect quotient. 

Thus, ${158^ \circ }$ is not a perfect multiple of exterior angle. So, the polygon will not be possible.

What is the minimum interior angle possible for a regular polygon?

Ans:   Given: A regular polygon

We need to find the minimum interior angle possible for a regular polygon.

A polygon with minimum number of sides is an equilateral triangle.

So, number of sides $ = 3$

${\text{Exterior}}\;{\text{angle}} = \dfrac{{{{360}^ \circ }}}{{{\text{Number}}\;{\text{of}}\;{\text{sides}}}}$

Thus, Maximum Exterior angle will be

$   = \dfrac{{{{360}^ \circ }}}{3} $

$   = {120^ \circ } $ 

We know, ${\text{Interior}}\;{\text{angle}} = {180^ \circ } - {\text{Exterior}}\;{\text{angle}}$

Therefore, minimum interior angle will be

$   = {180^ \circ } - {120^ \circ } $

$   = {60^ \circ } $ 

What is the maximum exterior angel possible for a regular polygon?

Ans: Given: A regular polygon

We need to find the maximum exterior angle possible for a regular polygon.

Therefore, Maximum Exterior angle possible will be

$ = {120^ \circ } $

 Exercise 3.3

1. Given a parallelogram ABCD. Complete each statement along with the definition or property used.

$\;{\text{AD}}$ = $...$

Given: A parallelogram ${\text{ABCD}}$ 

We need to complete each statement along with the definition or property used.

We know that opposite sides of a parallelogram are equal.

Hence, ${\text{AD}}$ = ${\text{BC}}$ 

$\;\angle {\text{DCB }} = $ $...$

Given: A parallelogram ${\text{ABCD}}$.

${\text{ABCD}}$ is a parallelogram, and we know that opposite angles of a parallelogram are equal.

Hence, $\angle {\text{DCB   =  }}\angle {\text{DAB}}$

${\text{OC}} = ...$ 

${\text{ABCD}}$ is a parallelogram, and we know that diagonals of parallelogram bisect each other.

Hence, ${\text{OC  =  OA}}$

$m\angle DAB\; + \;m\angle CDA\; = \;...$

Given : A parallelogram ${\text{ABCD}}$.

${\text{ABCD}}$ is a parallelogram, and we know that adjacent angles of a parallelogram are supplementary to each other.

Hence, $m\angle DAB\; + \;m\angle CDA\; = \;180^\circ $

 2. Consider the following parallelograms. Find the values of the unknowns x, y, z.

Given: A parallelogram ${\text{ABCD}}$

We need to find the unknowns ${\text{x,y,z}}$

The adjacent angles of a parallelogram are supplementary.

Therefore, ${\text{x} + 100^\circ  = 180^\circ }$

${\text{x} = 80^\circ }$ 

Also, the opposite angles of a parallelogram are equal.

Hence, ${\text{z}} = {\text{x}} = 80^\circ $ and ${\text{y}} = 100^\circ $

Given: A parallelogram.

We need to find the values of ${\text{x,y,z}}$

The adjacent pairs of a parallelogram are supplementary.

Hence, $50^\circ  + {\text{y}} = 180^\circ $

${\text{y}} = 130^\circ $

Also, ${\text{x}} = {\text{y}} = 130^\circ $(opposite angles of a parallelogram are equal)

And, ${\text{z}} = {\text{x}} = 130^\circ $ (corresponding angles)

(iii)  

Given: A parallelogram 

${\text{x}} = 90^\circ $(Vertically opposite angles)

Also, by angle sum property of triangles

${\text{x}} + {\text{y}} + 30^\circ  = 180^\circ $

${\text{y}} = 60^\circ $

Also,${\text{z}} = {\text{y}} = 60^\circ $(alternate interior angles)

Given: A parallelogram

Corresponding angles between two parallel lines are equal.

Hence, ${\text{z}} = 80^\circ $ Also,${\text{y}} = 80^\circ $ (opposite angles of parallelogram are equal)

In a parallelogram, adjacent angles are supplementary

Hence,${\text{x}} + {\text{y}} = 180^\circ $

$  {\text{x}} = 180^\circ  - 80^\circ  $

$  {\text{x}} = 100^\circ  $ 

As the opposite angles of a parallelogram are equal, therefore,${\text{y}} = 112^\circ $ 

Also, by using angle sum property of triangles

$  {\text{x}} + {\text{y}} + 40^\circ  = 180^\circ  $

$  {\text{x}} + 152^\circ  = 180^\circ  $

$  {\text{x}} = 28^\circ  $ 

And ${\text{z}} = {\text{x}} = 28^\circ $(alternate interior angles)

3. Can a quadrilateral ${\text{ABCD}}$be a parallelogram if 

(i) $\angle {\text{D}}\;{\text{ + }}\angle {\text{B}} = 180^\circ ?$

Given: A quadrilateral ${\text{ABCD}}$

We need to find whether the given quadrilateral is a parallelogram.

For the given condition, quadrilateral ${\text{ABCD}}$ may or may not be a parallelogram.

For a quadrilateral to be parallelogram, the sum of measures of adjacent angles should be $180^\circ $ and the opposite angles should be of same measures.

(ii) ${\text{AB}} = {\text{DC}} = 8\;{\text{cm}},\;{\text{AD}} = 4\;{\text{cm}}\;$and ${\text{BC}} = 4.4\;{\text{cm}}$

As, the opposite sides ${\text{AD}}$and ${\text{BC}}$are of different lengths, hence the given quadrilateral is not a parallelogram.

(iii) $\angle {\text{A}} = 70^\circ $and $\angle {\text{C}} = 65^\circ $

As, the opposite angles have different measures, hence, the given quadrilateral is a parallelogram.

4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.

Given: A quadrilateral.

We need to draw a rough figure of a quadrilateral that is not a paralleloghram but has exactly two opposite angles of equal measure.

A kite is a figure which has two of its interior angles, $\angle {\text{B}}$and $\angle {\text{D}}$of same measures. But the quadrilateral ${\text{ABCD}}$is not a parallelogram as the measures of the remaining pair of opposite angles are not equal.

5. The measures of two adjacent angles of a parallelogram are in the ratio 3:2. Find the measure of each of the angles of the parallelogram.

Ans: Given: A parallelogram with adjacent angles in the ratio $3:2$

We need to find the measure of each of the angles of the parallelogram.

Let the angles be $\angle {\text{A}} = 3{\text{x}}$and $\angle {\text{B}} = 2{\text{x}}$

As the sum of measures of adjacent angles is $180^\circ $ for a parallelogram.

$  \angle {\text{A}} + \angle {\text{B}} = 180^\circ  $

 $ 3{\text{x}} + 2{\text{x}} = 180^\circ  $

 $ 5{\text{x}} = 180^\circ  $

 $ {\text{x}} = 36^\circ  $ 

$~\angle A=$ $\angle {\text{C}}$ $= 3{\text{x}} = 108^\circ$and $~\angle B=$ $\angle {\text{D}}$ $= 2{\text{x}} = 72^\circ$(Opposite angles of a parallelogram are equal).

Hence, the angles of a parallelogram are $108^\circ ,72^\circ ,108^\circ $and $72^\circ $.

6. Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.

Given: A parallelogram with two equal adjacent angles.

The sum of adjacent angles of a parallelogram are supplementary.

$  \angle {\text{A}} + \;\angle {\text{B}} = 180^\circ  $

$  2\angle {\text{A}}\;{\text{ =  180}}^\circ  $

$  \angle {\text{A}}\;{\text{ = }}\;{\text{90}}^\circ  $

$  \angle {\text{B}}\;{\text{ = }}\angle {\text{A}}\;{\text{ = }}\;{\text{90}}^\circ  $

Also, opposite angles of a parallelogram are equal

$  \angle {\text{C}} = \angle {\text{A}} = 90^\circ  $

$  \angle {\text{D}} = \angle {\text{B}} = 90^\circ  $ 

Hence, each angle of the parallelogram measures $90^\circ $.

7. The adjacent figure ${\text{HOPE}}$is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.

Given: A parallelogram ${\text{HOPE}}$.

We need to find the measures of angles ${\text{x,y,z}}$and also state the properties used to find these angles.

$\angle {\text{y}} = 40^\circ $(Alternate interior angles)

And $\angle {\text{z}} + 40^\circ  = 70^\circ $(corresponding angles are equal)

$\angle {\text{z}} = 30^\circ $

Also, ${\text{x}} + {\text{z}} + 40^\circ  = 180^\circ $(adjacent pair of angles)

${\text{x}} = 110^\circ $

8. The following figures ${\text{GUNS}}$and ${\text{RUNS}}$are parallelograms. Find ${\text{x}}$and${\text{y}}$. (Lengths are in cm).

Given: Parallelogram ${\text{GUNS}}$.

We need to find the measures of ${\text{x}}$and ${\text{y}}$.

${\text{GU = SN}}$(Opposite sides of a parallelogram are equal).

$  3{\text{y }} - {\text{ }}1{\text{ }} = {\text{ }}26{\text{ }} $

$  3{\text{y }} = {\text{ }}27{\text{ }} $

$  {\text{y }} = {\text{ }}9{\text{ }} $ 

Also,${\text{SG = NU}}$

Therefore, 

$  3{\text{x}} = 18 $

$  {\text{x}} = 3 $ 

Given: Parallelogram ${\text{RUNS}}$

We need to find the value of ${\text{x}}$and ${\text{y}}{\text{.}}$

The diagonals of a parallelogram bisect each other, therefore, 

$  {\text{y }} + {\text{ }}7{\text{ }} = {\text{ }}20{\text{ }} $

$  {\text{y }} = {\text{ }}13 $

 $ {\text{x }} + {\text{ y }} = {\text{ }}16 $

$  {\text{x }} + {\text{ }}13{\text{ }} = {\text{ }}16 $

 $ {\text{x }} = {\text{ }}3{\text{ }} $ 

9. In the above figure both ${\text{RISK}}$and ${\text{CLUE}}$are parallelograms. Find the value of ${\text{x}}{\text{.}}$

Given: Parallelograms ${\text{RISK}}$and ${\text{CLUE}}$

As we know that the adjacent angles of a parallelogram are supplementary, therefore, 

In parallelogram ${\text{RISK}}$

$  \angle {\text{RKS  + }}\angle {\text{ISK}} = 180^\circ  $

 $ 120^\circ  + \angle {\text{ISK}} = 180^\circ  $ 

As the opposite angles of a parallelogram are equal, therefore,

In parallelogram ${\text{CLUE}}$,

$\angle {\text{ULC}} = \angle {\text{CEU}} = 70^\circ $

Also, the sum of all the interior angles of a triangle is $180^\circ $

$  {\text{x }} + {\text{ }}60^\circ {\text{ }} + {\text{ }}70^\circ {\text{ }} = {\text{ }}180^\circ  $

$  {\text{x }} = {\text{ }}50^\circ  $ 

10. Explain how this figure is a trapezium. Which of its two sides are parallel?

We need to explain how the given figure is a trapezium and find its two sides that are parallel.

If a transversal line intersects two specified lines in such a way that the sum of the angles on the same side of the transversal equals $180^\circ $, the two lines will be parallel to each other.

Here, $\angle {\text{NML}} = \angle {\text{MLK}} = 180^\circ $

Hence, ${\text{NM}}||{\text{LK}}$

Hence, the given figure is a trapezium.

11. Find ${\text{m}}\angle {\text{C}}$in the following figure if ${\text{AB}}\parallel {\text{CD}}$${\text{AB}}\parallel {\text{CD}}$.

Given: ${\text{AB}}\parallel {\text{CD}}$ and quadrilateral

We need to find the measure of $\angle {\text{C}}$

$\angle {\text{B}} + \angle {\text{C}} = 180^\circ $(Angles on the same side of transversal).

$  120^\circ  + \angle {\text{C}} = 180^\circ  $

$  \angle {\text{C}} = 60^\circ  $ 

12. Find the measure of $\angle {\text{P}}$and$\angle {\text{S}}$, if ${\text{SP}}\parallel {\text{RQ}}$in the following figure. (If you find${\text{m}}\angle {\text{R}}$, is there more than one method to find${\text{m}}\angle {\text{P}}$?)

Given: ${\text{SP}}\parallel {\text{RQ}}$and 

We need to find the measure of $\angle {\text{P}}$and $\angle {\text{S}}$.

The sum of angles on the same side of transversal is $180^\circ .$

$\angle {\text{P}} + \angle {\text{Q}} = 180^\circ $

$  \angle {\text{P}} + 130^\circ  = 180^\circ  $

$  \angle {\text{P}} = 50^\circ  

 $\angle {\text{R }} + {\text{ }}\angle {\text{S }} = {\text{ }}180^\circ {\text{ }} $

$  {\text{ }}90^\circ {\text{ }} + {\text{ }}\angle {\text{S }} = {\text{ }}180^\circ  $

  ${\text{ }}\angle {\text{S }} = {\text{ }}90^\circ {\text{ }} $ 

Yes, we can find the measure of ${\text{m}}\angle {\text{P}}$ by using one more method.

In the question,${\text{m}}\angle {\text{R}}$and ${\text{m}}\angle {\text{Q}}$are given. After finding ${\text{m}}\angle {\text{S}}$ we can find ${\text{m}}\angle {\text{P}}$ by using angle sum property.

Exercise 3.4

1. State whether True or False.

(a) All rectangles are squares.

(b) All rhombuses are parallelograms.

(c) All squares are rhombuses and also rectangles.

(d) All squares are not parallelograms.

(e) All kites are rhombuses.

(f) All rhombuses are kites.

(g) All parallelograms are trapeziums.

(h) All squares are trapeziums.

Every square is indeed a type of rectangle, not every rectangle can be called a square.

It's correct to say that all squares can be classified as parallelograms due to their shared characteristic of having opposite sides that are parallel and opposite angles that are equal.

Because, a kite shape is that its adjacent sides are not necessarily equal in length, unlike those of a square.

2. Identify all the quadrilaterals that have.

(a) four sides of equal length

(b) four right angles

(a) Rhombus and square have all four sides of equal length.

(b) Square and rectangles have four right angles. 

3. Explain how a square is

(i) a quadrilateral

(ii) a parallelogram

(iii) a rhombus

(iv) a rectangle

(i) Square is a quadrilateral because it has four sides.

(ii) A square is a parallelogram because its opposite sides are parallel and opposite angles are equal.

(iii) Square is a rhombus because all four sides are of equal length and diagonals bisect at right angles.

(iv)Square is a rectangle because each interior angle, of the square, is 90° 

4. Name the quadrilaterals whose diagonals.

(i) bisect each other 

(ii) are perpendicular bisectors of each other 

(iii) are equal

(i) Parallelogram, Rhombus, Square and Rectangle

(ii) Rhombus and Square

(iii)Rectangle and Square 

5. Explain why a rectangle is a convex quadrilateral.

A rectangle is a convex quadrilateral because both of its diagonals lie inside the rectangle.

6. ABC is a right-angled triangle and O is the mid-point of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).

AD and DC are drawn so that AD || BC and AB || DC

AD = BC and AB = DC

ABCD is a rectangle as opposite sides are equal and parallel to each other and all the interior angles are of 90°.

In a rectangle, diagonals are of equal length and also bisect each other.

Hence, AO = OC = BO = OD

Thus, O is equidistant from A, B and C.

Overview of Deleted Syllabus for CBSE Class 8 Maths Understanding Quadrilaterals

Class 8 Maths Chapter 3: Exercises Breakdown

In conclusion, NCERT Solutions for Class 8 Maths Chapter 3 - Understanding Quadrilaterals provides a comprehensive and detailed understanding of the properties and characteristics of various types of quadrilaterals. By studying this chapter and using the NCERT solutions, students can enhance their knowledge of quadrilaterals and develop their problem-solving abilities. The chapter starts by introducing quadrilaterals and their diverse types, including parallelograms, rectangles, squares, rhombuses, and trapeziums. It goes on to explain each type, detailing their characteristic properties like side lengths, angles, diagonals, and symmetry. Students that practice these kinds of questions will gain confidence and perform well on tests.

Other Study Material for CBSE Class 8 Maths Chapter 3

Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths . Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals

1. What is the Area of a Field in the Shape of a Rectangle with Dimensions of 20 Meters and 40 Meters?

We know that the field is rectangular. Hence, we can apply the area of a rectangle to find the field area.

Length of the field = 40 Metre

Width of the field = 20 Metre

Area of the rectangular field = Length × Width = 40 × 20 = 800 Sq. Meters.

We know if the length of the rectangle is L and breadth is B then,

Area of a rectangle = Length × Breadth or L × B

Perimeter = 2 × (L + B)

So, the properties and formulas of quadrilaterals that are used in this question:

Area of the Rectangle = Length × Width

So, we used only a specific property to find the answer.

2. Find the Rest of the Angles of a Parallelogram if one Angle is 80°?

For a parallelogram ABCD, as we know the properties:

Opposite angles are equal.

Opposite sides are equal and parallel.

Diagonals bisect each other.

The summation of any two adjacent angles = 180 degrees.

So, the angles opposite to the provided 80° angle will likewise be 80°.

Like we know, know that the Sum of angles of any quadrilateral = 360°.

So, if ∠A = ∠C = 80° then,

Sum of ∠A, ∠B, ∠C, ∠D = 360°

Also, ∠B = ∠D

Sum of 80°, ∠B, 80°, ∠D = 360°

Or, ∠B +∠ D = 200°

Hence, ∠B = ∠D = 100°

Now, we found all the angles of the quadrilateral, which are:

3. Why are the NCERT Solutions for Class 8 Maths Chapter 3 important?

The questions included in NCERT Solutions for Chapter 3 of Class 8 Maths are important not only for the exams but also for the overall understanding of quadrilaterals. These questions have been answered by expert teachers in the subject as per the NCERT (CBSE) guidelines. As the students answer the exercises, they will grasp the topic more comfortably and in a better manner.

4. What are the main topics covered in NCERT Solutions for Class 8 Maths Chapter 3?

All the topics of the syllabus of Class 8 Maths Chapter 3 have been dealt with in detail in the NCERT Solutions by Vedantu. The chapter is Understanding Quadrilaterals and has four exercises. All the important topics in Quadrilaterals have also been carefully covered. Students can also refer to the important questions section to get a good idea about the kind of questions usually asked in the exam.

5. Do I need to practice all the questions provided in the NCERT Solutions Class 8 Maths “Understanding Quadrilaterals”?

It helps to solve as many questions as possible because Mathematics is all about practice. If you solve all the practice questions and exercises given in NCERT Solutions for Class 8 Maths, you will be able to score very well in your exams comfortably. This will also help you understand the concepts clearly and allow you to apply them logically in the questions.

6. What are the most important concepts that I need to remember in Class 8 Maths Chapter 3?

For Class 8 Maths Chapter 3, you must remember the definition, characteristics and properties of all the quadrilaterals prescribed in the syllabus, namely, parallelogram, rhombus, rectangle, square, kite, and trapezium. Also know the properties of their angles and diagonals. Regular practise will help students learn the chapter easily.

7. Is Class 8 Maths Chapter 3 Easy?

Class 8 chapter 3 of Maths is a really interesting but critical topic. It's important not only for the Class 8 exams but also for understanding future concepts in higher classes. So, to stay focused and get a good grip of all concepts, it is advisable to download the NCERT Solutions for Class 8 Maths from the Vedantu website or from the Vedantu app at free of cost. This will help the students to clear out any doubts and allow them to excel in the exams. 

8. In Maths Class 8 Chapter 3, how are quadrilaterals used in everyday life?

Quadrilaterals are everywhere! Here are some examples:

Shapes in your house: Doors, windows, tabletops, picture frames, book covers, even slices of bread are quadrilaterals (mostly rectangles).

Construction and design: Architects use rectangles and squares for walls, floors, and windows. Roads and bridges often involve trapezoids and other quadrilaterals for support.

Everyday objects: Stop signs, traffic signals, and many sports fields (like baseball diamonds) are quadrilaterals.

9. How many quadrilaterals are there in Class 8 Chapter 3 Maths?

There are many types of quadrilaterals mentioned in Class 8 Understanding Quadrilaterals, but some of the most common include:

Rectangle (all four angles are 90 degrees, opposite sides are equal and parallel)

Square (a special rectangle with all sides equal)

Parallelogram (opposite sides are parallel)

Rhombus (all four sides are equal)

Trapezoid (one pair of parallel sides)

10. What are real examples of quadrilaterals in Class 8 Chapter 3 Maths?

Some real examples of quadrilaterals are:

Rectangle: Doorway, window pane, sheet of paper, tabletop, chocolate bar, playing card (most common)

Square: Dice, coaster, napkin, wall tiles (when all sides are equal)

Parallelogram: Textbook cover, kite (when opposite sides are parallel), solar panel

Rhombus: Traffic warning sign (diamond shape with all sides equal)

Trapezoid: Slice of pizza, roof truss (one pair of parallel sides)

Irregular Quadrilateral: Flag (many flags like the US flag are not perfectly symmetrical quadrilaterals)

11. How do you identify a quadrilateral in Maths Class 8 Chapter 3?

A quadrilateral has the following properties:

Four straight sides

Four angles (interior angles add up to 360 degrees)

Four vertices (corners where two sides meet)

NCERT Solutions for Class 8 Maths

Ncert solutions for class 8.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals are provided below. Our solutions covered each questions of the chapter and explains every concept with a clarified explanation. To score good marks in Class 8 Mathematics examination, it is advised to solve questions provided at the end of each chapter in the NCERT book.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals are prepared based on Class 8 NCERT syllabus, taking the types of questions asked in the NCERT textbook into consideration. Further, all the CBSE Class 8 Solutions Maths Chapter 3 are in accordance with the latest CBSE guidelines and marking schemes.

Class 8 Maths Chapter 3 Exercise 3.1 Solutions

Class 8 Maths Chapter 3 Exercise 3.2 Solutions

Class 8 Maths Chapter 3 Exercise 3.3 Solutions

Class 8 Maths Chapter 3 Exercise 3.4 Solutions

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NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

October 4, 2019 by Sastry CBSE

• Class 8 Maths Understanding Quadrilaterals Exercise 3.1
• Class 8 Maths Understanding Quadrilaterals Exercise 3.2
• Class 8 Maths Understanding Quadrilaterals Exercise 3.3
• Class 8 Maths Understanding Quadrilaterals Exercise 3.4
• Understanding Quadrilaterals Class 8 Extra Questions

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1

More CBSE Class 8 Study Material

• NCERT Solutions for Class 8 Maths
• NCERT Solutions for Class 8 Science
• NCERT Solutions for Class 8 Social Science
• NCERT Solutions for Class 8 English
• NCERT Solutions for Class 8 English Honeydew
• NCERT Solutions for Class 8 English It So Happened
• NCERT Solutions for Class 8 Hindi
• NCERT Solutions for Class 8 Sanskrit

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• Class 8 Maths MCQs
• Chapter 3 Understanding Quadrilaterals

Class 8 Maths Chapter 3 Understanding Quadrilaterals MCQs

Class 8 Maths Chapter 3 Understanding Quadrilaterals MCQs   (Questions and Answers) are provided here, online. These objective questions are designed for students, as per the CBSE syllabus (2022-2023) and NCERT guidelines. Solving the chapter-wise questions will help students understand each concept and help to score good marks in exams. Also, learn important questions for class 8 Maths here at BYJU’S.

Practice more and test your skills on Class 8 Maths Chapter 3 Understanding Quadrilaterals MCQs with the given PDF here.

MCQs on Class 8 Understanding Quadrilaterals

Multiple Choice Questions (MCQs) are available for Class 8 Understanding Quadrilaterals chapter. Each problem consists of four multiple options, out of which one is the correct answer. Students have to solve the problem and select the correct answer.

1. Which of the following is not a quadrilateral?

B. Rectangle

C. Triangle

D. Parallelogram

Explanation: A quadrilateral is a four-sided polygon but triangle is a three-sided polygon.

2. Which of the following quadrilaterals has two pairs of adjacent sides equal and its diagonals intersect at 90 degrees?

D. Rectangle

3. Which one of the following is a regular quadrilateral?

B. Trapezium

Explanation: A square has all its sides equal and angles equal to 90 degrees.

4. If AB and CD are two parallel sides of a parallelogram, then:

A. AB>CD

B. AB<CD

D. None of the above

5. The perimeter of a parallelogram whose parallel sides have lengths equal to 12 cm and 7 cm is:

Explanation: Perimeter of parallelogram = 2 (Sum of Parallel sides)

P = 2 (12 + 7)

6. If ∠A and ∠C are two opposite angles of a parallelogram, then:

A. ∠A > ∠C

C. ∠A < ∠C

Explanation: Opposite angles of a parallelogram are always equal.

7. If ∠A and ∠B are two adjacent angles of a parallelogram. If ∠A = 70 ° , then ∠B = ?

Explanation: The adjacent angles of parallelogram are supplementary.

∠A + ∠B = 180°

70° + ∠B = 180°

∠B = 180 – 70° = 110°

8. ABCD is a rectangle and AC & BD are its diagonals. If AC = 10 cm, then BD is:

Explanation: The diagonals of a rectangle are always equal.

9. Each of the angles of a square is:

A. Acute angle

B. Right angle

C. Obtuse angle

D. 180 degrees

Explanation: All the angles of square is at right angle.

10. The quadrilateral whose diagonals are perpendicular to each other is:

A. Parallelogram

C. Trapezium

11. Which of the following is not a regular polygon?

A. Square B. Equilateral triangle C. Rectangle D. Regular hexagon

Answer: C. Rectangle Explanation: A regular polygon is both equiangular and equilateral. But all four sides of a rectangle are not equal, thus it is not a regular polygon.

12. If the two angles of a triangle are 80° and 50°, respectively. Find the measure of the third angle. A. 50° B. 60° C. 70° D. 80°

Answer: A. 50°

Explanation: By the angle sum property of triangle, we know that; Sum of all the angles of a triangle = 180° Let the unknown angle be x 80° + 50° + x = 180° x = 180° – 130° x = 50°

13. In a parallelogram ABCD, angle A and angle B are in the ratio 1:2. Find the angle A. A. 30° B. 45° C. 60° D. 90°

Answer: C.60°

Explanation: As we know, the sum of adjacent angles of a parallelogram is equal to 180° and opposite angles are equal to each other. Thus, in parallelogram ABCD angle A and angle B are adjacent to each other Let angle A = x and angle B = 2x. So, x + 2x = 180° 3x = 180° x = 60°

14. The angles of a quadrilateral are in ratio 1:2:3:4. Which angle has the largest measure? A. 120° B. 144° C. 98° D. 36°

Answer: B.144°

Explanation: Suppose, ABCD is a quadrilateral. Let angle A is x Then, x + 2x + 3x + 4x = 360° [Angle sum property of quadrilateral] 10x = 360° x = 36° Hence, the greatest angle is 4x = 4 x 36 = 144°

15. The length and breadth of a rectangle is 4 cm and 2 cm respectively. Find the perimeter of the rectangle. A. 12 cm B. 6 cm C. 8 cm D. 16 cm

Answer: A. 12 cm Explanation: Given, length of rectangle is 4 cm Breadth of rectangle = 2cm By the formula of perimeter of rectangle, we know that; Perimeter = 2 (Length + Breadth) P = 2(4+2) P = 2 x 6 P = 12 cm

16. The diagonals of a rectangle are 2x + 1 and 3x – 1, respectively. Find the value of x. A. 1 B. 2 C. 3 D. 4

Answer: B.2

Explanation: The diagonals of a rectangle are equal in length. 2x + 1 = 3x -1 1 + 1 = 3x – 2x 2 = x Thus, the value of x is 2.

17. The diagonals of a kite: A. Bisects each other B. Are perpendicular to each other C. Does not bisect each other D. None of the above

Answer: B. Are perpendicular to each other

Explanation: The diagonals of a kite are perpendicular to each other. They intersect at 90 degrees but does not bisect.

18. A rhombus has a side length equal to 5 cm. Find its perimeter. A. 25 B. 10 C. 20 D. 30

Answer: C. 20

Explanation: A rhombus is a parallelogram that has all its four sides equal. Thus, the perimeter of rhombus, P = 4 x side-length P = 4 x 5 P = 20 cm

19. ABCD is a parallelogram. If angle A is equal to 45°, then find the measure of its adjacent angle. A. 135° B. 120° C. 115° D. 180°

Answer: A.135°

Explanation: The adjacent angles of a parallelogram sums up to 180°. Thus, 45° + x = 180° x = 180° – 45° x = 135°

20. The kite has exactly two distinct consecutive pairs of sides of equal length. A. True B. False

Answer: A. True

Explanation: A kite is a quadrilateral that has exactly two distinct consecutive pairs of sides of equal length.

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NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

Ncert solutions for class 8 maths chapter 3 understanding quadrilaterals| pdf download.

• Exercise 3.1 Chapter 3 Class 8 Maths NCERT Solutions
• Exercise 3.2 Chapter 3 Class 8 Maths NCERT Solutions
• Exercise 3.3 Chapter 3 Class 8 Maths NCERT Solutions
• Exercise 3.4 Chapter 3 Class 8 Maths NCERT Solutions

NCERT Solutions for Class 8 Maths Chapters:

How many exercises in Chapter 3 Understanding Quadrilaterals

What is equilateral triangle, in a quadrilateral abcd, the angles a, b, c and d are in the ratio 1 : 2 : 3 : 4. find the measure of each angle of the quadrilateral., the interior angle of a regular is 108°. find the number of sides of the polygon., contact form.