# The base of a rectangle is three less than twice the height. Determine the dimensions of the rectangle if the area is 135 square inches.

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Let the dimensions be b and h

We know that the base (b) is 3 less than twice the height:

==> b = 2h - 3 ........(1)

But the area (A) = 135

==> b*h = 135........(2)

Now substitute b :

==> (2h-3)*h = 135

==> 2h^2 - 3h = 135

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

==> h1= [3 + sqrt(1089)]/4 = (3+ 33)/4= 36/4 = 9

==> h2= (3-33)/4 = -30/4 ( impossiboe becatse h can not be negative values:

**Then the height h= 9**

But b = 2h-3

==> b = 2*9 - 3 = 18-3 = 15

**== b= 15**

Let the height of the rectangle be H. Now the base is three less than twice the height or base is 2H-3.

The area of the rectangle is base* height = H*( 2H-3)

H*( 2H-3) = 135

=> 2H^2 - 3H =135

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

The roots of the equation are

H1 = [3 + sqrt(9+1080)] / 4

= (3 + 33)/4

= 36 / 4

=9

H2 = [3 - sqrt(9+1080)] / 4

= (3 - 33)/4

= -30 / 4

=-7.5

Now the dimension of a rectangle cannot be negative, so we eliminate -7.5

The Height then is 9 and the base is 135/ 9 =15

**The dimensions of the rectangle are 9 and 15 inches.**

Let height of rectangle be x.

Then the base 3less than 2wice the height: base b = -3+2x

So the area of the rectangle = base height = (-3+2x)x = 135

-3x+2x^2 = 135

2x^2-3x-135 = 0

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

Therefore 2x+15 = 0 or x-9 = 0

So x = 9 inch is the height and base = -3+2*9 =15 inch

We'll note the height of the rectangle as h and the base of teh rectangle as 2h - 3.

We'll calculate the area of the rectangle:

A = base*height

A = (2h-3)*h

But, from enunciation, we know that A = 135 inches^2

135 = (2h-3)*h

We'll remove the brackets:

135 = 2h^2 - 3h

We'll subtract 135 both sides:

2h^2 - 3h - 135 = 0

We'll apply the quadratic formula:

h1 = [3+sqrt(9+1080)]/4

h1 = (3+33)/4

h1 = 36/4

h1 = 9

h2 = (3-33)/4

h2 = -30/4

h2 = -15/2

**The height could not be negative, so the only valid value for h = 9.**

**The base is:**

b = 2h - 3

b = 2*9 - 3

b = 18-3

**b = 15**