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Understanding Quadrilaterals : Exercise 3.3 (Mathematics NCERT Class 8th)

Q.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 = ......
Sol. (i) AD = BC
In a parallelogram, opposite sides are equal.

(ii) ∠ DCB = ∠ DAB
In a parallelogram, opposite angles are equal.

(iii) OC = OA
In a parallelogram, diagonals bisect each other.

(iv) m ∠DAB + m ∠CDA = 180ᵒ.
In a parallelogram, adjacent angles are supplementary to each other.

Q.2 Consider the following parallelograms. Find the values of the unknowns x, y, z. Sol. (i) In a parallelogram, adjacent angles are supplementary to each other.
Therefore, ∠B + ∠C = 180ᵒ
100ᵒ + x = 180ᵒ
x = 180ᵒ - 100ᵒ
x = 80ᵒ
In a parallelogram, opposite angles are equal.
Therefore, z = x = 80ᵒ
In a parallelogram, opposite angles are equal.
Therefore, y = ∠B = 100ᵒ

(ii) In a parallelogram, adjacent angles are supplementary to each other.
x + 50ᵒ = 180ᵒ
x = 180ᵒ - 50ᵒ
x = 130ᵒ
In a parallelogram, opposite angles are equal.
Therefore, y = x = 130ᵒ
In a parallelogram, corresponding angles are equal.
Therefore, z = x = 130ᵒ

(iii) In a parallelogram, vertically opposite angles are equal.
Therefore, x = 90ᵒ
We know that, sum of interior angles of triangle is 180ᵒ
y + x + 30ᵒ = 180ᵒ
y + 90ᵒ + 30ᵒ = 180ᵒ
y + 120ᵒ = 180ᵒ
y = 180ᵒ - 120ᵒ
y = 60ᵒ
In a parallelogram, alternate angles are equal.
Therefore, z = y = 60ᵒ

(iv) In a parallelogram, corresponding angles are equal.
Therefore, z = 80ᵒ
In a parallelogram, adjacent angles are supplementary to each other.
x + 80ᵒ = 180ᵒ
x = 180ᵒ - 80ᵒ
x = 100ᵒ
In a parallelogram, opposite angles are equal.
Therefore, y = 80ᵒ

(v) In a parallelogram, opposite angles are equal.
Therefore, y = 112ᵒ
We know that, sum of interior angles of triangle is 180ᵒ
40ᵒ + y + x = 180ᵒ
40ᵒ + 112ᵒ + x = 180ᵒ
152ᵒ + x = 180ᵒ
x = 180ᵒ - 152ᵒ
x = 28ᵒ
In a parallelogram, alternate angles are equal.
Therefore, z = x = 28ᵒ

Q.3 Can a quadrilateral ABCD be a parallelogram if
(i) ∠D + ∠B = 180°?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) ∠A = 70° and ∠C = 65°?
Sol. (i) For ∠D + ∠B = 180°, quadrilateral ABCD might be a parallelogram if (a) sum of the measures of adjacent angles is 180ᵒ (b) Opposite angles are equal.

(ii) For AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm, quadrilateral ABCD cannot be a parallelogram since, opposite sides AD and BC are of different lengths.

(iii) For ∠A = 70° and ∠C = 65°, quadrilateral ABCD cannot be a parallelogram since, opposite angles A and C are not equal.

Q.4 Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Sol. For the quadrilateral ABCD which is a kite, there are two interior angles ∠A and ∠C of same measure. But, still quadrilateral ABCD is not a parallelogram as the measure of the remaining pair of opposite angles ∠D and ∠B are not equal. Q.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.
Sol. Let the two adjacent angles be 3a and 2a.
We know that in a parallelogram adjacent angles are supplementary to each other.
Therefore, 3a + 2a = 180ᵒ
5a = 180ᵒ
a = 36ᵒ
Therefore, one angle = 3a = 3 x 36ᵒ = 108ᵒ
Other angle = 2a = 2 x 36ᵒ = 72ᵒ

Q.6 Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Sol. Let each adjacent angle be a.
We know that in a parallelogram adjacent angles are supplementary to each other.
Therefore, a + a = 180ᵒ
2a = 180ᵒ
a = 90ᵒ
Thus, each adjacent angle is 90ᵒ
We know that, sum of interior angles of triangle is 180ᵒ
Therefore, a + a + a = 180ᵒ
3a = 180ᵒ
a = 60ᵒ
Hence, the measure of each angle of parallelogram is 90ᵒ, 90ᵒ and 60ᵒ.

Q.7 The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them. Sol. Here, ∠EHP and ∠y forms alternate interior angles.
Therefore, ∠y = 40ᵒ
In a parallelogram, corresponding angles are equal.
Therefore, 70ᵒ = z + 40ᵒ
70ᵒ - 40ᵒ = z
z = 30ᵒ
In a parallelogram, adjacent pair of angles are equal.
x + (z + 40ᵒ) = 180ᵒ
x + 70ᵒ = 180ᵒ
x = 110ᵒ

Q.8 The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm). Sol. (i) In parallelogram GUNS,
In a parallelogram, opposite sides are equal.
Therefore, GS = UN
3x = 18
x = 6 cm
Similarly, GU = SN
3y -1 = 26
3y = 26 + 1
3y = 27
y = 9 cm
Hence, the measures of x and y are 6 cm and 9 cm respectively.

(ii) In a parallelogram, diagonals bisect each other.
Therefore, y + 7 = 20
y = 13 cm
Similarly, x + y = 16
x + 13 = 16
x = 3 cm
Hence, the measures of x and y are 3 cm and 13 cm respectively.

Q.9 In the above figure both RISK and CLUE are parallelograms. Find the value of x.
Sol. In a parallelogram, adjacent angles are supplementary.
For parallelogram RISK, ∠RSK + ∠ISK = 180ᵒ
120ᵒ + ∠ISK = 180ᵒ
∠ISK = 60ᵒ
In a parallelogram, opposite angles are equal.
For parallelogram CLUE, ∠ULC = ∠CEU = 70ᵒ
We know that the sum of measures of all the interior angles of a triangle is 180ᵒ
x + 60ᵒ + 70ᵒ= 180ᵒ
x = 50ᵒ

Q.10 Explain how this figure is a trapezium. Which of its two sides are parallel? Sol.
We know that if a transversal line is intersecting two given lines such that the sum of the measures of the angles on the same side of transversal is 180ᵒ, then the given two lines will be parallel to each other.
For the given trapezium,
∠M + ∠L =100ᵒ + 80ᵒ = 180ᵒ (Since sum of interior opposite angles is 180ᵒ)
Therefore, NM and KL are parallel.
This proves that KLMN is a trapezium.

Q.11 Find m∠C in Fig if || . Sol. Given, || .
∠B + ∠C = 180ᵒ (Since they angles on same side of transversal)
120ᵒ + ∠C = 180ᵒ
∠C = 180ᵒ - 120ᵒ
∠C = 60ᵒ

Q.12 Find the measure of ∠P and ∠S if || in Fig. (If you find m∠R, is there more than one method to find m∠P?) Sol. From the figure, we can see that ∠P and ∠Q are angles on the same side of transversal
Therefore, ∠P + ∠Q = 180ᵒ
∠P = 50ᵒ
From the figure, we can see that ∠R and ∠S are angles on the same side of transversal
Therefore, ∠R + ∠S = 180ᵒ
90ᵒ + ∠S = 180ᵒ
∠S = 90ᵒ

Alternative method to find ∠P:
Using the angle sum property of quadrilateral, we can write,
∠P + ∠Q + ∠R + ∠s = 360ᵒ
∠P + 130ᵒ + 90ᵒ + 90ᵒ = 360ᵒ
∠P = 360ᵒ - 310ᵒ
∠P = 50ᵒ

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