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°