NCERT Solutions for Class 8 Maths Chapter 3 - Understanding Quadrilaterals Exercise 3.3

The NCERT solutions for Class 8 Maths Chapter 3 – Understanding Quadrilaterals includes the solutions to all the questions present in the NCERT textbook. The subject experts at BYJU’S have solved each question of NCERT exercises with utmost care in order to help the students in solving any question from the NCERT textbook. NCERT Class 8 Exercise 3.3 is based on the different types of quadrilaterals. Students can download the NCERT Solutions of Class 8 mathematics to understand the concept more deeply.

NCERT Solutions for Class 8 Maths Chapter 3 – Understanding Quadrilaterals Exercise 3.3

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Access other exercise solutions of Class 8 Maths Chapter 3 – Understanding Quadrilaterals

Exercise 3.1 Solutions 7 Questions (1 Long Answer Question, 6 Short Answer Questions)

Exercise 3.2 Solutions 6 Questions (6 Short Answer Questions)

Exercise 3.4 Solutions 6 Questions (1 Long Answer Question, 5 Short Answer Questions)

Access Answers to Maths NCERT Class 8 Chapter 3 – Understanding Quadrilaterals Exercise 3.3 Page Number 50

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

(i) AD = …… (ii) ∠DCB = ……

(iii) OC = …… (iv) m ∠DAB + m ∠CDA = ……

Solution:

(i) AD = BC (Opposite sides of a parallelogram are equal)

(ii) ∠DCB = ∠DAB (Opposite angles of a parallelogram are equal) (iii) OC = OA (Diagonals of a parallelogram are equal)

(iv) m ∠DAB + m ∠CDA = 180°

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

Solution:

(i)

y = 100° (Opposite angles of a parallelogram)

x + 100° = 180° (Adjacent angles of a parallelogram)

⇒ x = 180° – 100° = 80°

x = z = 80° (Opposite angles of a parallelogram)

∴, x = 80°, y = 100° and z = 80°

(ii)

50° + x = 180° ⇒ x = 180° – 50° = 130° (Adjacent angles of a parallelogram) x = y = 130° (Opposite angles of a parallelogram)

x = z = 130° (Corresponding angle)

(iii)

x = 90° (Vertical opposite angles)

x + y + 30° = 180° (Angle sum property of a triangle)

⇒ 90° + y + 30° = 180°

⇒ y = 180° – 120° = 60°

also, y = z = 60° (alternate angles)

(iv)

z = 80° (Corresponding angle) z = y = 80° (Alternate angles) x + y = 180° (Adjacent angles)

⇒ x + 80° = 180° ⇒ x = 180° – 80° = 100°

(v)

x=28o

y = 112o z = 28o

3. Can a quadrilateral ABCD be a parallelogram if (i) ∠D + ∠B = 180°?

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

(iii)∠A = 70° and ∠C = 65°?

Solution:

(i) Yes, a quadrilateral ABCD can be a parallelogram if ∠D + ∠B = 180°. But it should also fulfill some conditions which are:

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

(b) Opposite angles must be equal.

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

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

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

Solution:

ABCD is a figure of quadrilateral that is not a parallelogram but has exactly two opposite

angles that is ∠B = ∠D of equal measure. It is not a parallelogram because ∠A ≠ ∠C.

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.

Solution:

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°

We know that opposite sides of a parallelogram are equal.

∠A = ∠C = 3x = 3 × 36° = 108°

∠B = ∠D = 2x = 2 × 36° = 72°

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

Solution:

Let ABCD be a parallelogram.

Sum of adjacent angles of a parallelogram = 180°

∠A + ∠B = 180°

⇒ 2∠A = 180°

⇒ ∠A = 90°

also, 90° + ∠B = 180°

⇒ ∠B = 180° – 90° = 90°

∠A = ∠C = 90°

∠B = ∠D = 90°

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

Solution:

y = 40° (Alternate interior angle)

∠P = 70° (Alternate interior angle)

∠P = ∠H = 70° (Opposite angles of a parallelogram)

z = ∠H – 40°= 70° – 40° = 30°

∠H + x = 180°

⇒ 70° + x = 180°

⇒ x = 180° – 70° = 110°

8. The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)

Solution:

(i) SG = NU and SN = GU (opposite sides of a parallelogram are equal) 3x = 18

x = 18/3

⇒ x =6

3y – 1 = 26 and,

⇒ 3y = 26 + 1

⇒ y = 27/3=9

x = 6 and y = 9

(ii) 20 = y + 7 and 16 = x + y (diagonals of a parallelogram bisect each other) y + 7 = 20

⇒ y = 20 – 7 = 13 and,

x + y = 16

⇒ x + 13 = 16

⇒ x = 16 – 13 = 3

x = 3 and y = 13

9. In the above figure, both RISK and CLUE are parallelograms. Find the value of x.

Solution:

∠K + ∠R = 180° (adjacent angles of a parallelogram are supplementary)

⇒ 120° + ∠R = 180°

⇒ ∠R = 180° – 120° = 60°

also, ∠R = ∠SIL (corresponding angles)

⇒ ∠SIL = 60°

also, ∠ECR = ∠L = 70° (corresponding angles) x + 60° + 70° = 180° (angle sum of a triangle)

⇒ x + 130° = 180°

⇒ x = 180° – 130° = 50°

10. Explain how this figure is a trapezium. Which of its two sides are parallel? (Fig 3.32)

Solution:

When a transversal line intersects two lines in such a way that the sum of the adjacent angles on the same side of transversal is 180°, then the lines are parallel to each other. Here, ∠M + ∠L = 100° + 80° = 180°

Thus, MN || LK

As the quadrilateral KLMN has one pair of parallel lines, it is a trapezium. MN and LK are parallel lines.

11. Find m∠C in Fig 3.33 if AB || DC ?

Solution:

m∠C + m∠B = 180° (Angles on the same side of transversal)

⇒ m∠C + 120° = 180°

⇒ m∠C = 180°- 120° = 60°

12. Find the measure of ∠P and ∠S if SP || RQ in Fig 3.34. (If you find m∠R, is there more than one method to find m∠P?)

Solution:

∠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 are more than one method to find m∠P.

PQRS is a quadrilateral. 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°

Exercise 3.3 of NCERT Solutions for Class 8 Maths Chapter 3- Understanding Quadrilaterals is based on the following topics:

1. Kinds of Quadrilaterals
   1. Trapezium
   2. Kite
   3. Parallelogram
   4. Elements of a parallelogram
   5. Angles of parallelogram
   6. Diagonals of a parallelogram

Also, explore – 

NCERT Solutions for Class 8 Maths

NCERT Solutions for Class 8