The width of a rectangle is one less than three-fourths the length and the perimeter is 96 inches.?

Question

Find the AREA of the rectangle.

Answer 1

W=width
L=length
w=3/4L -1
2W+2L=96 or W+L=48 dividing by 2,
substituting the first into the second
(3/4L-1)+L=48
3/4L+L-1=48
1 3/4 L=49 (7/4)L=49 multiply by 4
7L=216
L=28
W=3/4 L-1 or 20
checking the math
2(28)+2(20)=96
56+40=96 it checks out.
Area=28*20=560 sq inches

Answer 2

Let L be the length, W be the Width.

given that W = ¾L - 1
also given: perimeter = 96 inches.

formula for perimeter of rectangle = L + W + L + W = 2(L + W) = 96

since W = ¾L - 1,
perimeter = 2(L + ¾L - 1) = 96
2L + 3/2L - 2 = 96
7/2L = 96 + 2 = 98
L = 96/(7/2) = 28 inches

Substitute the value of L into W,
W = ¾(28) - 1 = 21 - 1 = 20 inches

therefore area of rectangle, A = L x W = 28 * 20 = 560 inches²

Answer 3

Area of a rectangle = L X W
Perimeter = 2(l + w) = 96
Let,
length = l
width = w

Therefore, width = 3/4*l - 1

Perimeter = 2(l + w)
2(l + 3/4*l - 1)
2(7/4*l - 1) = 96
7/4*l - 1 = 96/2
7/4*l = 48 + 1
l = 49 X 4/7
l = 7 X 4
l = 28.

w = 3/4 X l - 1
= 3/4 X 28 - 1
= 21 - 1
= 20.

Therefore, area = 20 X 28
= 560

Hope that helps.. :-) Cheers!!

Answer 4

width = 3/4 L - 1 where L is length.

2(3/4 L - 1) + 2 L = 96
3/2 L - 2 + 2L = 96
3/2 L + 2L = 96 + 2
7/2 L = 98
7 L = 196
L = 28
w = 20

area = 28 * 20 = 560 square inches

Answer 5

permit the dimensions of the rectangle be (3x+4). permit the width of the rectangle be x bear in mind my costly the fringe is the finished distance around the rectangle. ninety six=6x+8+2x ninety six=8x+8 88=8x x=11 the answer is length =33+4=37 ,the width is 11

Answer 6

Let length = L
Width, W = 3/4 L - 1
Perimeter = 2(L+W)
2(L + 0.75L - 1)=96
1.75L - 1 = 48
L=49/1.75=28
W=0.75*W - 1
=21-1=20