# The base of the rectangle is 3 less than twice the height. If the area is 135 square inches, find the dimensions of the rectangle.

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Let:

x = Height of rectangle

Then as per given conditions:

Base of rectangle = 2x - 3

Area of rectangle = Height*Base

= x*(2x - 3) = 2x^2 - 3x

It is given that:

Area = 135 square Inches

Therefor:

2x^2 - 3x = 135

==> 2x^2 - 3x - 135 = 0

==> 2x^2 - 18x + 15x - 135 = 0

==> 2x(x - 9) + 15(x - 9) = 0

==> (x - 9)(2x + 15)

Therefor:

x = 9 and -7.5

As height cannot be negative, we take the value of x as 9.

Then:

Base of rectangle = 2x - 3 = 2*9 - 3 = 15

Answer:

Height = 9 inches

Base = 15 inches

Let the height of the rectangle be h. Then the base of the rectangle = -3+2h.

The area of rectangle A = h(-3+2h) = 135 sq inches.

-3h+2h^2 = 135

2h^2-3h-135 = 0

2h^2 -18h+15h -135 = 0

2h(h-9) +15(h-9) = 0

(h-9)(2h+15).

Therefore h -9 = 0 and 2h+15 = 0.

We take h -9 = 0 to get h = 9 inch.

Therefore the dimension of the rectangle : h = 9 and base b = -3+2h = -3+2*9 = 15.

Check: Area = h*b = 9*15 = 135 sq inch

We'll note the height as x and the base as 2x - 3.

We'll calculate the area of the rectangle as the product of the base and height:

A = x(2x-3)

We know the value of the area and we'll substitute in the relation above:

135 = x(2x-3)

We'll remove the brackets:

135 = 2x^2 - 3x

We'll subtract 135 both sides and we'll use the symmetric property:

2x^2 - 3x - 135 = 0

We'll apply the quadratic formula:

x1 = [3+sqrt(9 + 1080)]/4

x1 = (3+33)/4

x1 = 9

x2 = (-30)/4

x2 = -7.5

Since a measure of a side cannot be negative, the value -*7.5 will be rejected.

**The height is 9 inches and the base is 2*9 - 3 = 15 inches.**