FIND THE MISSING MEASUREMENTS OF PARALLELOGRAM

In the following parallelogram AB is base and the line drawn from D to AB is height.

Area of parallelogram = Base x height

Problem 1 :

For the parallelogram ABCD, what is DE to the nearest tenth?

Solution :

Area of parallelogram =  Base x height

Base (AB) = 9.4 in and height (DE) = ?

Base (AD) = 13 in and height (CF) = 9 in

9.4 x DE = 13 x 9

DE = (13 x 9)/9.4

DE = 12.4 in

Problem 2 :

Solution :

Area of the parallelogram = base x height

When base = 14, height = 5

When base = 10, height = ?

Let h be the required height.

14 x 5 = 10 x h

h = (14 x 5)/10

h = 7

So, the required height is 7.

Problem 3 :

Solution :

When base = 0.4, height = 0.3

When base = 0.5 and height =  h

0.4 x 0.3 = 0.5 x h

h = (0.4 x 0.3)/0.5

h = 0.24

So, the required height is 0.24.

Problem 4 :

Solution :

When base = 13, height = h

When base = 18, height = 12

13 x h = 18 x 12

h = (18 x 12)/13

h = 16.61

So, the required height is 16.61.

Problem 5 :

The area of a parallelogram is 24 in², and the height is 6 in. Find the length of the corresponding base.

Solution :

Area of parallelogram = 24 in²

height = 6 in

Let base = b

b x 6 = 24

b = 24/6

b = 4 in

So, the required base is 4 inches.

Problem 6 :

Find the area of the composite figure. Then find the area.

Solution :

The shape given above includes two basic shapes, parallelogram and right triangle.

Area of the shape = Area of parallelogram + area of triangle

Base of the parallelogram = 7 m and height = 6 m

Base of right triangle = 5 m and height = 6 m

 = 7 x 6 + (1/2) x 5 x 6

 = 42 + 15

 = 57 m²

Problem 7 :

The height of a parallelogram is 12 cm. If the area is 432 sq.cm. The base of the parallelogram is

a)  25   cm     b)  26 cm     c)  28 cm      d) 36 cm

Solution :

Height = 12 cm

Area of parallelogram = 432 square cm

base x height = 432

base x 12 = 432

base = 432/12

= 36 cm

So, the required base of the parallelogram is 36 cm, option d is correct.

Problem 8 :

Find the height of the parallelogram whose area is 210 cm² and base is 30 cm.

Solution :

base = 30 cm

Area of parallelogram = 210 cm²

base x height = 210

30 x height = 210

height = 210/30

= 7

So, the required height of the parallelogram is 7 cm.

Problem 9 :

The area of a parallelogram is equal to the area of a square whose perimeter is 120 m. If the height of the parallelogram is 20 m, find its corresponding base.

Solution :

Perimeter of square = 120 m

4(side length) = 120

Side length = 120/4

= 30 m

So, side length of square = 30 m

Area of square = side x side

= 30 x 30

= 900 square meter

Area of parallelogram = 900

base x height = 900

base x 20 = 900

base = 900/20

= 45 m

So, the required base of the parallelogram is 45 m.

Problem 10 :

Find the area of parallelogram with base 12 cm and altitude 11 cm.

Solution :

base = 12 cm and height = 11 cm

base x height = 12 x 11

= 132 square cm

Problem 11 :

Suresh won a parallelogram-shaped  trophy in a state level Chess tournament. He knows that the area of the trophy is 735 sq. cm and its base is 21 cm. What is the height of that trophy?

Solution :

Area of trophy = 735 square cm

base = 21 cm, height = ?

21 x height = 735

height = 735/21

= 35 cm

So, height of the trophy is 35 cm.

Problem 12 :

Janaki has a piece of fabric in the shape of a parallelogram. Its height is 12 m and its base is 18 m. She cuts the fabric into four equal parallelograms by cutting the parallel sides through its mid-points. Find the area of each new parallelogram.

Solution :

base = 12 m, height = 18 m

Area = 12 x 18

= 216 square meter

Area of one parallelogram shaped cloth = 216/4

= 54 square meter

Problem 13 :

A ground is in the shape of parallelogram. The height of the parallelogram is 14 meters and the corresponding base is 8 meters longer than its height. Find the cost of leveling the ground at the rate of ₹ 15 per sq. m.

Solution :

height = 14 m, base = 14 + 8

= 22 m

Area for levelling = 22 x 14

= 308 square meter

Cost of levelling = 308 x 15

= ₹4620