Question

The length of a rectangle is 6 yd more than twice the width x. The area is 416 yd2. Find the dimensions of the rectangle.

Answer 1

13 yds * 32 yds

Explanation:

A=area, l=length, w=width

A=w*l

l=6+2w

A=w*(6+2w)

A=6w+2w^2

416=6w+2w^2

416/2=(6w+2w^2)/2

208=3w+w^2

208-208=3w+w^2-208

w^2+3w-208=0

w^2+16w-13w-208=0

w(w+16)-13(w+16)=0

(w-13)(w+16)=0 ==> w+16=0 ==> w=-16, w can't be negative.

w-13=0

w=13

l=6+2w

l=6+2(13)

l=6+26

l=32

Dimensions: 13 * 32