SOLVE UNKNOWN VALUES IN EACH PARALLELOGRAM

Find the value of x in each parallelogram.

Problem 1 :

Solution :

By observing the figure,

PQ ∥ RS

So, PQ = RS (Opposite sides of parallel are equal)

Here PQ = x/2 in and RS = 12 in

(x/2) = 12

x = 12 × 2

x = 24 in²

Problem 2 :

Solution :

By observing the figure,

PS ∥ QR

So, PS = QR (Opposite sides of parallel are equal)

Here PQ = 4x ft and RS = 8 ft

4x = 8

x = 8/4

x = 2 ft

Problem 3 :

Solution :

By observing the figure,

PQ ∥ RS

So, PQ = RS (Opposite sides of parallel are equal)

Here PQ = 43 yd and RS = (7 + 3x) yd

43 = (7 + 3x)

Subtracting 7 on both sides.

43 - 7 = 7 + 3x - 7

36 = 3x

36/3 = x

12 yd = x

Problem 4 :

Solution :

By observing the figure,

PS ∥ QR

So, PS = QR (Opposite sides of parallel are equal)

Here PS = 38 in and QR = (x + 43) in

38 = (x + 43)

Subtracting 43 on both sides.

38 - 43 = x + 43 – 43

-5 in. = x

Find the value of x and y in each parallelogram.

Problem 5 :

Solution :

By observing the figure,

PS ∥ QR and PQ ∥ RS

So, PS = QR and PQ = RS (Opposite sides of parallel are equal)

Here PS = (-7x)ft, QR = 21 ft, PQ = (2y)ft, and RS = 10 ft

-7x = 21 and 2y = 10

Therefore, x = -3 and y = 5.

Problem 6 :

Solution :

By observing the figure,

PQ ∥ RS and PS ∥ QR

So, PQ = RS and PS = QR (Opposite sides of parallel are equal)

Here PQ = 27 yd, RS = (63 – 6x) yd, PS = (4 + y) yd, and QR = 15 yd

27 = (63 – 6x) and (4 + y) = 15

Therefore, x = 6 and y = 11.

Problem 7 :

Solution :

By observing the figure,

PS ∥ QR and PQ ∥ RS

So, PS = QR and PQ = RS (Opposite sides of parallel are equal)

Here PS = 64 in, QR = (19 + 5x) in, PQ = (3y - 3) in, and RS = 72 in

64 = (19 + 5x) and (3y – 3) = 72

Therefore, x = 9 and y = 25.

Problem 8 :

Solution :

By observing the figure,

PS ∥ QR and PQ ∥ RS

So, PS = QR and PQ = RS (Opposite sides of parallel are equal)

Here PS = 18 ft, QR = (-x + 7) ft, PQ = 36 ft, and RS = (6y) ft

18 = (-x + 7) and 36 = 6y

Therefore, x = -11 and y = 6.

Problem 9 :

For the parallelogram given below, find the value of x, y and z

Solution :

In triangle BEC,

∠E + ∠B + ∠C = 180

90 + x + 50 = 180

140 + x = 180

x = 180 - 140

x = 40

z = 50

 ∠DFC = 90

 ∠DCF = ?

In triangle FDC,

 ∠FDC + ∠DCF + ∠CFD = 180

50 + ∠DCF + 90 = 180

∠DCF + 140 = 180

∠DCF = 180 - 140

∠DCF = 40

Sum of co-interior angle is 180 degree.

40 + y + 50 + x = 180

40 + y + 50 + 40 = 180

130 + y = 180

y = 180 - 130

y = 50

Problem 10 :

In a parallelogram PQRS, if ∠P = (3x − 5)° and ∠Q = (2x + 15)°, then find the value of x.

Solution :

∠P and ∠Q are co-interior angles, the sum of co-interior angles is 180.

3x - 5 + 2x + 15 = 180

5x + 10 = 180

5x = 180 - 10

5x = 170

x = 170 / 5

x = 34

Problem 11 :

In the given figure, ABCD is a parallelogram. Find x, y, z and p.

Solution :

110 + z = 180

z = 180 - 110

z = 70

y and z are co-interior angles.

y + z = 180

y + 70 = 180

y = 180 - 70

y = 110

p = 70 and y = x = 110