The NCERT solutions for Class 8 maths Chapter 3- Understanding Quadrilaterals includes the solutions of all the questions present in the NCERT textbook. The subject experts at BYJUâ€™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 furthermore understand the concept.

### Download PDF of NCERT Solutions for class 8 Maths Chapter 3- Understanding Quadrilaterals Exercise 3.3

Â

Â

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

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

Exercise 3.2 Solutions 6 Questions (6 Short Answer Questions)

Exercise 3.4 Solutions 6 Questions (1 Long Answer Questions, 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 = â€¦â€¦ (ii) âˆ DCB = â€¦â€¦

(iii) OC = â€¦â€¦ (iv) m âˆ DAB + m âˆ CDA = â€¦â€¦

Solution:

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

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

(iv) m âˆ DAB + m âˆ CDA = 180Â°

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

Solution:

(i)

y = 100Â° (opposite angles of a parallelogram)

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

â‡’ x = 180Â° â€“ 100Â° = 80Â°

x = z = 80Â° (opposite angles of a parallelogram)

âˆ´, x = 80Â°, y = 100Â° and z = 80Â°

(ii)

50Â° + x = 180Â° â‡’ x = 180Â° â€“ 50Â° = 130Â° (Adjacent angles of a parallelogram) x = y = 130Â° (opposite angles of a parallelogram)

x = z = 130Â° (corresponding angle)

(iii)

x = 90Â° (vertical opposite angles)

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

â‡’ 90Â° + y + 30Â° = 180Â°

â‡’ y = 180Â° â€“ 120Â° = 60Â°

also, y = z = 60Â° (alternate angles)

(iv)

z = 80Â° (corresponding angle) z = y = 80Â° (alternate angles) x + y = 180Â° (adjacent angles)

â‡’ x + 80Â° = 180Â° â‡’ x = 180Â° â€“ 80Â° = 100Â°

(v)

x=28o

y = 112o z = 28o

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

Â

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

** (iii)âˆ A = 70Â° and âˆ C = 65Â°?**

Solution:

(i) Yes, a quadrilateral ABCD be a parallelogram if âˆ D + âˆ B = 180Â° but it should also

fulfilled some conditions which are:

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

(b) Opposite angles must be equal.

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

(iii) No, opposite angles should be of same measures. âˆ A â‰ âˆ C

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

Â

Solution:

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

angles that is âˆ B = âˆ D of equal measure. It is not a parallelogram because âˆ A â‰ âˆ 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.**

Â

Solution:

Let the measures of two adjacent angles âˆ A and âˆ B be 3x and 2x respectively in

parallelogram ABCD.

âˆ A + âˆ B = 180Â°

â‡’ 3x + 2x = 180Â°

â‡’ 5x = 180Â°

â‡’ x = 36Â°

We know that opposite sides of a parallelogram are equal.

âˆ A = âˆ C = 3x = 3 Ã— 36Â° = 108Â°

âˆ B = âˆ D = 2x = 2 Ã— 36Â° = 72Â°

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

Â

Solution:

Let ABCD be a parallelogram.

Sum of adjacent angles of a parallelogram = 180Â°

âˆ A + âˆ B = 180Â°

â‡’ 2âˆ A = 180Â°

â‡’ âˆ A = 90Â°

also, 90Â° + âˆ B = 180Â°

â‡’ âˆ B = 180Â° â€“ 90Â° = 90Â°

âˆ A = âˆ C = 90Â°

âˆ B = âˆ D = 90

Â°

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

Â

Solution:

y = 40Â° (alternate interior angle)

âˆ P = 70Â° (alternate interior angle)

âˆ P = âˆ H = 70Â° (opposite angles of a parallelogram)

z = âˆ H â€“ 40Â°= 70Â° â€“ 40Â° = 30Â°

âˆ H + x = 180Â°

â‡’ 70Â° + x = 180Â°

â‡’ x = 180Â° â€“ 70Â° = 110Â°

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

Â

Solution:

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

x = 18/3

â‡’ x =6

3y â€“ 1 = 26 an

d,

â‡’ 3y = 26 + 1

â‡’ 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

â‡’ y = 20 â€“ 7 = 13 and,

x + y = 16

â‡’ x + 13 = 16

â‡’ x = 16 â€“ 13 = 3

x = 3 and y = 13

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

Â

Solution:

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

â‡’ 120Â° + âˆ R = 180Â°

â‡’ âˆ R = 180Â° â€“ 120Â° = 60Â°

also, âˆ R = âˆ SIL (corresponding angles)

â‡’ âˆ SIL = 60Â°

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

â‡’ x + 130Â° = 180Â°

â‡’ x = 180Â° â€“ 130Â° = 50Â°

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

Â

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 180Â° then the lines are parallel to each other. Here, âˆ M + âˆ L = 100Â° + 80Â° = 180Â°

Thus, MN || LK

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

**11. Find mâˆ C in Fig 3.33 if AB || DC ?**

Â

Solution:

mâˆ C + mâˆ B = 180Â° (angles on the same side of transversal)

â‡’ mâˆ C + 120Â° = 180Â°

â‡’ mâˆ C = 180Â°- 120Â° = 60Â°

**12. Find the measure of âˆ P and âˆ S if SP || RQ ? in Fig 3.34. (If you find mâˆ R, is there more than one**

Â

**method to find mâˆ P?)**

Solution:

âˆ P + âˆ Q = 180Â° (angles on the same side of transversal)

â‡’ âˆ P + 130Â° = 180Â°

â‡’ âˆ P = 180Â° â€“ 130Â° = 50Â°

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

â‡’ 90Â° + âˆ S = 180Â°

â‡’ âˆ S = 180Â° â€“ 90Â° = 90Â°

Thus, âˆ P = 50Â° and âˆ S = 90Â°

Yes, there are more than one method to find mâˆ P.

PQRS is a quadrilateral. Sum of measures of all angles is 360Â°.

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

âˆ Q = 130Â°, âˆ R = 90Â° and âˆ S = 90Â°

âˆ P + 130Â° + 90Â° + 90Â° = 360Â°

â‡’ âˆ P + 310Â° = 360Â°

â‡’ âˆ P = 360Â° â€“ 310Â° = 50Â°

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

- Kinds of Quadrilaterals
- Trapezium
- Kite
- Parallelogram
- Elements of a parallelogram
- Angles of parallelogram
- Diagonals of a parallelogram