A rectangle has a lenght that is 9 inches less than twice its width. The area of the rectangle is 180 square inches. What is the width of the rectangle?

In this question, we are asked to calculate the width of a rectangle having a length 9 inches less than twice it’s width and given the area of the rectangle.

First, we identify that the length is 9 inches less than twice the width

meaning if length is l and width is w; then l = (2w-9) inches

Mathematically the area of the rectangle is l * w ; meaning w * (2w-9)

This has a value equal to 180

w * (2w-9) = 180

opening the bracket;

2w^2 -9w = 180

2w^2 -9w - 180 = 0

solving this quadratic equation;

w = 12 or -7.5

since width cannot be negative, w = 12 inches

oeerivona oeerivona · Answer 2 · 2020-04-24T02:12:41+02:00

Answer:

The width of the triangle is 18.65 inches

Step-by-step explanation:

From the question, we have;

Length, L = Width, W - 9

Also the area is given by

L × W = 180

Therefore, we have

(W - 9) × W= 180

W² - 9·W = 180

W² - 9·W - 180 = 0

Factorizing or solving with quadratic formula

[tex]\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}[/tex]

a = 1

b = -9 and

c = -180

we get

(W + 9.65)(W-18.65) =0

Therefore W = - 9.65 or 18.65

Therefore, the Width, W = 18.65 and the length = W - 9 = 18.65 - 9 =9.65

The width of the triangle = 18.65 inches.