The length of a rectangle is three times its width. If the perimeter of the rectangle is 40 in?

Question

How do I find it's area?

Answer 1

Let the width of rectangle be x
it implies, length = 3x
now perimeter = 2(L+B) , ie , 40" = 2 (3x+x)
40"= 8x
therefore, x = 40/8 = 5"
Width = 5"
Length = 3*5= 15"
Area = L*B = 5" *15" = 75 sq. inch

Answer 2

Let l = the length of the rectangle.
w = the width of the rectangle.
l = 3(w) (1)
Perimeter, P =2l +2w
2l +2w = 40 (2)
Substitute the value of l in (1) to equation(2):
2(3w) +2w = 40
6w +2w = 40
8w= 40
w = 5 inches
l =3(5)
l= 15 inches.
check:: substitute the value of l and w in (2):
2l +2w = 40
2(15) +2(5) = 40 Area = l(w)
30 + 10 = 40 Area =15(5)= 75 sq. inches

Answer 3

If the length is 3 times its width, then L=3w

The perimeter is the distance round the rectangle, or the sum of the four sides of the rectangle. Therefore, 3w + w + 3w + w = 40.
Or, 8w = 40 in, therefore w = 5 in.

Area = w*L = w*3w = 3*w^2 = 75 sq in.

Answer 4

L = 3W

2W + 2L = 40
2W + 2(3W) = 40 substitue L
2W + 6W = 40
8W = 40
W = 5 in.

Plug the number in to find the length
L = 3W
L = 3(5) = 15 in.

Area of rectangle = L x W
15(5) = 75 in.

Answer 5

first you have to get the L and w
L = 3w
and 2L + 2w = 40
substitute
2(3w) + 2w = 40
8w = 40, w = 5 and L = 15
now, L * w = area = 5 * 15 = 75 sq in

Answer 6

L = 3W
2L + 2W = 40
6W + 2W = 40
8W = 40
W = 5
L = 15
A = 5 x 15 in²
A = 75 in²