**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**

**(i)**

**Solution:**

∠B = ∠D = 100° (opposite angles of a parallelogram are equal)

So, y = 100°

∠A + ∠B = 180° (Adjacent angles of a parallelogram are supplementary)

Z + 100^{o} = 180^{o}

Z = 180° – 100° = 80°

∠A = ∠C [Opposite angles of parallelogram are equal]

X = 80^{o}

Hence,

X = 80^{o}

Y= 100^{o}

Z = 80^{o}

** (ii)**

∠S + ∠P = 180^{o }(Adjacent angles of a parallelogram)

50° + x = 180°

x = 180° – 50° = 130°

x = y = 130° (opposite angles of a parallelogram are equal)

x = z = 130° (corresponding angles are equal)

Hence,

X = 130°

Y = 130°

Z = 130°

**(iii)**

**Solution:**

x = 90° (vertical opposite angles are equal)

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

90° + y + 30° = 180°

y = 180° – 120° = 60°

also, y = z = 60° (alternate angles are equal)

Hence,

X = 90°

Y = 60°

Z = 60°

**(iv)**

**Solution:**

z = ∠B = 80° (corresponding angles are equal)

z = y = 80° (alternate angles are equal)

x + y = 180° (adjacent angles are equal)

x + 80° = 180°

x = 180° – 80° = 100°

Hence,

X = 100°

Y = 80°

Z = 80°

**(v)**

**Solution:**

∠A + ∠B = 180° [ sum of adjacent angle is 180°]

40° + z + 112° = 180°

So, z = 28°

z = x = 28° [alternate angles]

∠D = ∠B = 112° [ vertically opposite angle]

Hence,

x=28°

y = 112°

z = 28°

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

**(i) ∠D + ∠B = 180°?**

**Solution:**

Yes, a quadrilateral ABCD be a parallelogram if ∠D + ∠B = 180°. But it must fulfill some conditions like:

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

(b) Opposite angles must be equal.

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

**Solution:**

No, the given ABCD cannot be a parallelogram. Here, AD ≠ BC, In parallelogram opposite sides should be of same length.

**(iii)∠A = 70° and ∠C = 65°?**

**Solution:**

Opposite angles should be of same measures. Here ∠A ≠ ∠C. Hence, ABCD is not a parallelogram.

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

**Solution:**

We draw a rough figure of a quadrilateral ABCD 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. It is a kite.

**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:**

According to the given question,

Let ABCD is a parallelogram in which 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.

So,

∠A = ∠C = 3x = 3 × 36° = 108°

∠B = ∠D = 2x = 2 × 36° = 72°

Hence, the measures of the angles of the parallelograms are 108°, 72°, 108°, 72°.

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

**Solution:**

Let us consider 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 parallelograms are equal)

z = ∠H – 40°

70° – 40° = 30°

∠H + x = 180°

Now,

70° + x = 180°

x = 180° – 70° = 110°

Hence,

x = 110°

y = 40°

z = 30°

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

**(i)**

SG = NU (opposite sides of a parallelogram are equal)

and SN = GU (opposite sides of a parallelogram are equal)

3x = 18

x = 18/3

x =6

Now,

3y – 1 = 26

and,

3y = 26 + 1

y = 27/3=9

Hence,

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

Hence,

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 are equal)

x + 60° + 70° = 180° (by the property of angle sum of a triangle)

x + 130° = 180°

x = 180° – 130° = 50°

Hence, x = 50°

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

**Solution:**

∠M + ∠L = 100^{o} + 80^{o}

∠M and ∠L are adjacent angles and the sum of adjacent interior angles is 180^{o}

KL||NM

Hence, KLMN is a trapezium.

**11. Find m∠C in below figure if AB ll DC.**

Solution:

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

m∠C + 120° = 180°

m∠C = 180°- 120° = 60°

Hence, m∠C = 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. We know that sum of measures of all angles is 360°.

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

Now,

∠Q = 130°, ∠R = 90° and ∠S = 90°

∠P + 130° + 90° + 90° = 360°

∠P + 310° = 360°

Hence, ∠P = 360° – 310° = 50°

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