Math Question #6

Question

The length of a rectangle is 3 cm more than its width. The area of the rectangle is 108 square centimeters. Find the dimensions of the rectangle.

Answer 1

Area=l*w

108=l*w

l=w+3 substitute the 2nd equation into the first

108=w(w+3) multiply

108=w²+3w subtract 108 from each side

108-108=w²+3w-108

0=w²+3w-108 factor

0=(w+12)(w-9) solve for w

w+12=0 subtract 12 from both sides

w+12-12=0-12

w=-12 The width can’t be negative so this is an extraneous solution.

w-9=0 add zero to each side

w-9+9=0+9

w=9. The width is 9 cm. Substitute into the equation for length l=w+3. length=9+3. The length is 12 cm.

The answer is the width is 9 cm and the length is 12 cm.

Answer 2

length(L) = W + 3 (Given length is 3 cm more than its width)

Are of rectangle = L * W

= (W+3 )* W

108 = W^2 + 3W

W^2 + 3W -108 =0

finding factors: W^2 - 9W +12 W - 108 = 0

W(W-9) + 12(W-9) = 0

(W-9) (W+12) = 0

W = 9 , -12

W=9 cm

L = W+3(given) => 9+3 = 12

Hence, the dimensions of the rectangle are: Length(L)= 12 cm, Width(W) = 9 cm

Hope this helps, for any further doubts, i am available to clarify