Question

The area of a certain rectangle is 288 yd2. The perimeter is 68 yd. What are the dimensions of the rectangle?

A) 24 yd by 10 yd


B) 18 yd by 16 yd


C) 22 yd by 12 yd


D) 36 yd by 8 yd

Please show work-THX

Answer 1

The rectangle is 16*18

Explanation:

The area of a rectangle is length * width and the perimeter is 2 times length * width

A = l*w

P = 2(l+w)

Replacing A and P

288 = l*w

68 = 2(l+w) => 34 = l+w => 34 - w = l

replacing l in the area

288 = (34 - w) w

w^2 - 34w + 288 = 0

(w - 16)(w - 18)

w = 16

w = 18

replacing in 34 - w = l

you get that when w = 16, l = 18 and when w = 18, l = 16