# Class 8 NCERT Solutions- Chapter 3 Understanding Quadrilaterals – Exercise 3.3

**Question 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°

**Question 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°

â‡’ x = 80Â°x = z = 80Â° {opposite angles of a parallelogram}

Therefore,

x = 80Â°, y = 100Â° and z = 80Â°

**(ii)**

50Â° + x = 180Â°

â‡’ x = 180Â° â€“ 50Â° = 130Â° {Adjacent angles of a parallelogram}

â‡’ x = y = 130Â° {opposite angles of a parallelogram}

â‡’ 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)**

y = 112Â° {opposite angles of a parallelogram}

x = 180Â° â€“ (y + 40Â°) {angle sum property of a triangle}

x = 28Â°

z = 28Â° {alternate angles}

**Question 3. Can a quadrilateral ABCD be a parallelogram if **

**(i) âˆ D + âˆ B = 180Â°?**

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

**(iii)âˆ A = 70Â° and âˆ C = 65Â°?**

**Solution:**

(i)Yes,The quadrilateral ABCD be a parallelogram if âˆ D + âˆ B = 180Â°,

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 the same length.Here, AD â‰ BC

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

**Question 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 which 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.

**Question 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 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°

As we know 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°

**Question 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.

The 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 = 90Â°

**Question 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 = 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°

**Question 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

â‡’ x =6

3y â€“ 1 = 26 and,

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

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

**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°

**Question 10. Explain how this figure is a trapezium. Which of its two sides **is** parallel? (Figure)**

**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 we have, âˆ M + âˆ L = 100Â° + 80Â° = 180Â°

Hence, MN || LK

As the quadrilateral KLMN has one pair of parallel line therefore it is a trapezium.

MN and LK are parallel lines.

**Question 11. Find mâˆ C in Figure if AB || DC?**

**Solution:**

mâˆ C + mâˆ B = 180Â° {angles on the same side of transversal}

â‡’ mâˆ C + 120Â° = 180Â°

â‡’ mâˆ C = 180Â° âˆ’ 120Â° = 60Â°

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

Hence, 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 of a quadrilateral is 360Â°.

Thus, as 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°

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