The length of a rectangle is 9 centimeters more than the width. the area is 112 square centimeters. find the length and width of the rectangle

L = w + 9

lw = 112

Substitute l = w + 9 into the second equation

(w + 9) (w) = 112

w^2 + 9w - 112 = 0

(w+16) (w-7) = 0

w = - 16

or

w = 7

we have to take the positive value, so w = 7

l = w + 9

l = 7 + 9

l = 16

The length is 16, and the width is 7

Here,

let width (b) be x then,

length (l) = 9cm+x

area = 112 sq cm

now,

area of rectangle=l*b

or, 112 = (9+x) x

or, 112=9x+x^2

or, 0=x^2+9x-112

or, 0=x^2 + (16-7) x-112

or, 0=x^2+16x-7x-112

or, 0=x (x+16) - 7 (x+16)

or, 0 = (x-7) (x+16)

either,

0=x-7

or, 7=x

x=7cm

Or,

0=x+16

or, - 16=x

x = - 16[impossible, as distance is never negative] so,

x=7cm

therefore, length = 7cm + 9 cm = 16cm and width = 7cm.

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