Question

An architect wants to draw a rectangle with a diagonal of 20 inches. The length of the rectangle is to be 8 inches more than twice the width. What dimensions should she make the rectangle?

Answer 1

Given:

Diagonal of the rectangle = 20 inches

The length of the rectangle is to be 8 inches more than twice the width.

To find:

The dimensions of the rectangle.

Solution:

Let width of the rectangle be x inches.

Then, length = 2x+8 inches

We know that, diagonal of a rectangle is

$Diagonal=\sqrt{length^2+width^2}$

$20=\sqrt{(2x+8)^2+x^2}$

Taking square both sides.

$400=4x^2+32x+64+x^2$

$0=5x^2+32x+64-400$

$0=5x^2+32x-336$

Splitting the middle term, we get

$5x^2+60x-28x-336=0$

$5x(x+12)-28(x+12)=0$

$(x+12)(5x-28)=0$

Using zero product property, we get

$x=-12, x=\dfrac{28}{5}=5.6$

Side length cannot be negative. So, only value of x is $5.6$.

Now,

Width = 5.6 inches

Length $=2(5.6)+8$

            $=11.2+8$

            $=19.2$ inches

Therefore, the length of rectangle is 19.2 inches and width of the rectangle is 5.6 inches.