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 + 100o = 180o
Z = 180° – 100° = 80°
∠A = ∠C [Opposite angles of parallelogram are equal]
X = 80o
Hence,
X = 80o
Y= 100o
Z = 80o
(ii)
∠S + ∠P = 180o (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 = 100o + 80o
∠M and ∠L are adjacent angles and the sum of adjacent interior angles is 180o
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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