PERIMETER AND AREA PROBLEMS ON CUSTOMARY UNITS OF LENGTH

Find the perimeter and the area of the rectangle. When possible, express your answer as a whole or mixed number.

Problem 1 :

Solution :

Area of the square = (side)²

= (1/2)²

= 1/4 cm²

So, area of the square is 1/4 cm²

Perimeter of the square = 4 × side

= 4 × 1/2

= 4/2

= 2 cm

So, perimeter of the square is 2 cm.

Problem 2 :

Solution :

By observing the figure,

Length = 54/5

Width = 2 m

Area of the rectangle = l × w

= (54/5) × 2

=  10.8 × 2

= 21.6 m²

So, area of the rectangle is 21.6 m²

Perimeter of the rectangle = 2 × (l + w)

= 2 × (54/5  + 2)

= 2 × (10.8 + 2)

= 2 × 12.8

= 25.6 m

So, perimeter of the rectangle is 25.6 m.

Problem 3 :

Solution :

By observing the figure,

Length = 4 in.

Width = 5 in.

Area of the rectangle = l × w

= 4 × 5

= 20 in².

So, area of the rectangle is 20 in².

Perimeter of the rectangle = 2 × (l + w)

= 2 × (4  + 5)

= 2 × 9

= 18 in.

So, perimeter of the rectangle is 18 in.

Problem 4 :

Solution :

By observing the figure,

Length = 9/5 cm

Width = 2/3 cm

Area of the rectangle = l × w

= 9/5 × 2/3

= 18/15

= 6/5 cm²

So, area of the rectangle is 6/5 cm².

Perimeter of the rectangle = 2 × (l + w)

= 2 × (9/5  + 2/3)

= 2 × (9/5 × 3/3 + 2/3 × 5/5)

= 2 × (27/15 + 10/15)

= 2 × (37/15)

= 74/15 cm

So, perimeter of the rectangle is 74/15 cm.

Find the side length of the rectangle.

Problem 5 :

Solution :

By observing the figure,

Length = ___ m

Width = 4.8 m

area of the rectangle = 29.76 sq m

Area of the rectangle = l × w

29.76 = l × 4.8

l = 29.76/4.8

l = 6.2 m

So, side length of the rectangle is 6.2 m.

Perimeter of the rectangle = 2 × (l + w)

= 2 × (6.2  + 4.8)

= 2 × 11

= 22 m

So, perimeter of the rectangle is 22 m.

Problem 6 :

Solution :

By observing the figure,

Length = 1/3 ft

Width = ___ ft

area of the rectangle = 2 sq ft

Area of the rectangle = l × w

2 = 1/3 × w

2= w/3

w = 6 ft

So, width of the rectangle is 6 ft.

Perimeter of the rectangle = 2 × (l + w)

= 2 × (1/3  + 6)

= 2 × (1/3 + 18/3)

= 2 × (19/3)

= 38/3

So, perimeter of the rectangle is 38/3 ft.

Problem 7 :

Solution :

By observing the figure,

Length = _____ cm

Width = 5.6 cm

area of the rectangle = 62.72 sq cm

Area of the rectangle = l × w

62.72 = l × 5.6

l = 62.72/5.6

l = 11.2

So, length of the rectangle is 11.2 cm².

Perimeter of the rectangle = 2 × (l + w)

= 2 × (11.2 + 5. 6)

= 2 × 16.8

= 33.6

So, perimeter of the rectangle is 33.6 cm.

Problem 8 :

Solution :

By observing the figure,

Length = _____ yd

Width = 1/2 yd

area of the rectangle = 7 sq yd

Area of the rectangle = l × w

7 = l × 1/2

l = 14 

So, length of the rectangle is 14 yd².

Perimeter of the rectangle = 2 × (l + w)

= 2 × (14  + 1/2)

= 2 × (14/1 ×  2/2 + 1/2)

= 2 × (28/2 + 1/2)

= 2 × (29/2)

= 58/2

So, perimeter of the rectangle is 29 yd.