Math Question...?

Question

A rectangle has a perimeter of 32 inches. Its width(W) is 4 inches shorter than its length(L). Find both dimensions; length and width. Find the area

Answer 1

A = L x W

32 = w(w - 4)

32 = w² - 4w

32 - 32 = w² - 4w - 32

0 = w² - 4w - 32

0 = w² - 8w + 4w - 32

0 = w(w - 8) + 4(w - 8)

0 = (w + 4)(w - 8)

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Roots

w + 4 = 0

w + 4 - 4 = 0 - 4

w = - 4. . . .Not used

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Roots

w - 8 = 0

w - 8 + 8 = 0 + 8

w = 8. . . .<= . .use this factor

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Area formula

A = L x W

32= 8(8 - 4)

32 = 8(4)

32 = 32

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Perimeter Formula

P = 2L + 2W

P = 2(8) + 2(4)

P = 16 + 8

P = 24

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The length is 8 inches

The width is 4 inches

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Answer 2

Length-10, Width-6, Area-60

Answer 3

ok. easy enough. it's width is 4 inches shorter than it's length, so. we will call the length "x" so the width will be x-4. if you add all sides and have them equal to 32 you can solve for x. so...
(x-4)+ (x-4)+x+x= 32
4x-8=32
4x=40
x=10

so the length is equal to 10
and the width is equal to 10-4 which is 6
so now to figure out the area just multiply length by width
10 * 6= 60 inches
and there you go.

length = 10
width = 6
area = 60
good luck!

Answer 4

perimeter of rectangle = 2(short side) + 2(long side) =32

2(W)+2(L)=32
2(L-4)+2(L)=32

2L-8+2L=32
4L=32+8
4L=40
L=10 inches

W=L-4
W=10-4
W=6 inches

Area = L x W
Area = 10 x 6
Area = 60 inches

Answer 5

P = 2L + 2W

W = L - 4

32 = 2L + 2(L - 4)
32 = 2L + 2L - 8
32 = 4L - 8
40 = 4L
10 = L in.

W = L - 4
W = 10 - 4
W = 6 in.

A = L x W
A = 10 x 6
A = 60 in^2

Answer 6

perimeter of rectangle = 2(short side) + 2(long side) =32.
2(L-4)+2(L)=32. solve for L (and then W).

Answer 7

W=10
L=6
A=60

32=W+W+L+L
W-4=L

SO....

32=W+W+(W-4)+(W-4)
32=2W+(W-4)+(W-4)
32 = 4W-8
32+8 = 4W
40=4W
10=W
.............

W-4=L
10-4=L
6=L
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A=L*W
A=6*10
A=60

Answer 8

32=2(x-4)+2x
32=2x-8+2x
32=4x-8
40=4x
10=x

lenth=10
width=10-4=6

area=LxW=10X6=60in^2