one half area of a rectangular is 54m^2 . if the triple of the width exceeds twice the length by 3 meter find the perimeter.

Question

one half area of a rectangular is 54m² . if the triple of the width exceeds twice the length by 3 meter find the perimeter.

Answer 1

You should start by writing what you know.  That includes what is given in the problem and the formulas for the area and the perimeter of a rectangle.

 

Formulas:

P = 2W + 2L

A = WL

 

Problem information

0.5A = 54m²

3W = 2L + 3

 

The next important part of the problem: what do we want to know?  It's the perimeter.

Start manipulating:

0.5A = 54m²

A = 108m²

 

3W = 2L + 3

3W - 3 = 2L

L= (3W - 3)/2

 

108 = WL

108 = W(3W-3)/2

216 = 3W² -3W

3W² -3W - 216 = 0

Divide by 3

W² -W - 72 = 0

(W-9)(W+8) = 0

So the roots are -8 and 9, but only 9 is possible because the answer, which is a distance, must be positive.

So we know that W = 9 and from substitution into both other equations we know that L = 12.

 

Perimeter = 2W + 2L = 2(9m) + 2(12m) = 42m