Question

The length of a rectangle is 10m more than twice the width. The

area is 120 m². What are the dimensions of the rectangle to the
nearest tenth?

Answer 1

5.6, 21.3

Explanation:

Let x be the width.

We know that the length is 10+2x. The area of a rectangle is length times width, so substitute the values.

x*(10+2x)=120

10x+2x^2=120

Divide both sides by 2.

x^2+5x=60

Add 25/4 to both sides so we can make a square.

x^2+5x+ 25/4= 60+25/4=265 /4

Factor the left side.

(x+ 5 /2 )^2= 265 /4

x+2.5=(positive or negative) sqrt(265/4)

x=(positive or negative) sqrt(265/4)-2.5

But, we know that if the equation is negative, the width of the rectangle would be negative. Therefore, x=positive sqrt(265/4)-2.5

Then, we will find the sides.

sqrt66.25-2.5=5.63941, or about 5.6

10+(2(sqrt66.25-2.5))= 21.278821, or about 21.3