Suppose that the width of a certain rectangle is 4 inches less than its length. The area is numerically 11 less than twice the perimeter. Find the length and width of the rectangle.
A) 7 in., 11 in.
B) 5 in., 9 in.
C) 4 in., 8 in.
D) 10 in., 14 in.

Question

Quizplus · Accepted Answer

We can use algebra to solve the problem. Let x be the length of the rectangle. Then the width is x-4. The area of the rectangle is x(x-4) = x^2 - 4x. The perimeter of the rectangle is 2(x + x-4) = 4x-8. Twice the perimeter is 8x-16. We are told that the area is numerically 11 less than twice the perimeter, so we can write the equation:
x^2 - 4x = 2(4x-8) - 11
Simplifying this equation gives:
x^2 - 4x = 8x - 27
Bringing all the terms to one side gives:
x^2 - 12x + 27 = 0
This quadratic equation factors as:
(x-3)(x-9) = 0
So x = 3 or x = 9. We reject x=3 since the width would be negative. Therefore, the length of the rectangle is 9 inches and the width is 5 inches (which is 4 inches less than the length). So the answer is (B) 5 in., 9 in.

Suppose That the Width of a Certain Rectangle Is 4