FIND THE MISSING SIDE OF RECTANGLE WHEN PERIMETER IS GIVEN

To find perimeter of rectangle, we use the formula 

Perimeter = 2(length + width)

By applying the known values, we can solve for unknown.

Problem 1 :

A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 square feet, how many feet of fencing will be required ?

Solution :

Since it is rectangle, the opposite sides will be equal.

From the given information, we know that one of the side is 20 feet. Let x be the unknown side.

So, sides of the rectangle are x, x, 20, 20.

Area = 680

20x = 680

x = 680/20

x = 34

Length of fencing = Perimeter of the rectangular field (including three sides)

= x + x + 20

= 2x + 20

= 2(34) + 20

= 68 + 20

= 88

So, required length of fencing is 88 feet.

Problem 2 :

The ratio between the perimeter and the width of a rectangle is 5 : 1. If the area of the rectangle is 216 sq.cm. What is the length of the rectangle ?

Solution :

Perimeter = 5x and width = x

Let length = l and width = w

Perimeter : width = 5 : 1

l = 18

So, the required length is 18 cm.

Problem 3 :

A farmer wishes to start a 100 sq.m rectangular vegetable garden. Since he has only 30 m barbed wire he fences three sides of the garden letting his house compound wall act as the fourth side fencing. The dimension of the garden is.

Solution :

Let l be the length and w be the width of the rectangle.

Area of the vegetable garden = 100 sq.m

length x width = 100

l w = 100

w = 100/l ----(1)

length of fencing covering three sides = 30 m

l + l + w = 30

2l + w = 30

2l + (100/l) = 30

2l² + 100 = 30l

2l² - 30l + 100 = 0

l² - 15l + 50 = 0

(l - 10) (l - 5) = 0

l = 10 and l = 5

So, the required dimension of the rectangle is 5 m x 20 m.

Problem 4 :

The length of a rectangle is three times of its width. If the length of the diagonal is 8√10 cm, then the perimeter of the rectangle is :

a)  15 √10 cm    b)  16 √10 cm    c)  24√10 cm     d)  64 cm

Solution :

Let x be the width of the rectangle, then length = 3x

length of the diagonal = 8√10

x² + (3x)² = (8√10)²

x² + 9x² = 64(10)

10x² = 64(10)

x² = 64

x = 8

So, width of the rectangle is 8 cm, length = 3(8) ==> 24 cm.

Perimeter of the rectangle = 2(24 + 8)

= 2(32)

= 64 cm

So, option d is correct.

Problem 5 :

The length of a rectangle is halved, while its width is tripled. What is the percentage change in area ?

a) 25% increase     b)  50% increase     c) 50% decrease     d)  75% decrease

Solution :

Let l and w be the length and width of original rectangle.

When length is halved, the new length will be l/2.

When width is tripled, the new width will be 3w.

Area of rectangle = (l/2)(3w)

= 3lw/2

Percentage change in area = (1.5lw - lw) / lw x 100%

= (0.5lw/lw) x 100%

= 0.5 x 100%

= 50%

50% increase. Option c is correct.

Problem 6 :

The length of a blackboard is 8 cm more than its width. If the length is increased by 7 cm and width is decreased by 4 cm, the area remains the same. The length and width of the blackboard will be

a) 28, 20    b)  34, 26   c)  40, 32     d)  56, 48

Solution :

Let x be the width of the rectangle.

Length = x + 8

After increasing the length by 7 cm, the new length = x + 8 + 7

= x + 15

After decreasing the width by 4 cm, the new width = x - 4

x(x + 8) = (x + 15)(x - 4)

x² + 8x = x² - 4x + 15x - 60

8x = 11x - 60

8x - 11x = -60

3x = 60

x = 20

x + 8 ==> 20 + 8 ==> 28 cm

So, the length and width of the rectangle is 28 cm and 20 cm.