QUADRATIC-4

Length of a rectangle is a+5 while the width is 2a-2. If the area is 14cm, find its perimeter.

Solution

Area = length x width

14 = (a+5) x (2a-2)

(a+5) (2a-2) = 14

2a² – 2a + 10a – 10 = 14

2a² + 8a – 10 = 14

2a² + 8a – 10 - 14 = 0

2a² + 8a - 24 = 0 - - - - - - - (i)

We are going to solve equation (i) by factorization.

2a² + 8a - 24 = 0

2a² + 12a – 4a - 24 = 0 [splitting 8a into 12a-4a]

(2a² + 12a) – (4a – 24) = 0

2a(a + 6) – 4(a + 6) = 0

(2a - 4)(a + 6) = 0

2a – 4 = 0 OR a + 6 = 0

a = 2 OR a = -6

Length = a+5

            = 2+5

            = 7

Width = 2a-2

           = 2(2)- 2

           = 4 – 2

           = 2

Perimeter = 2(length + width)

Perimeter = 2(7 + 2)

                  = 2 x 14

                  = 28

Hence perimeter is 28cm