Question

A rectangle is 3 times as long as it is wide. If the length is increased by 6 and the width is increased by 8, its area is increased by 108. Find the original dimensions.

Answer 1

Answer:

Length = 6, Breadth = 2

Step-by-step explanation:

Given:

A rectangle is 3 times as long as it is wide.

If the length is increased by 6 and the width is increased by 8, its area is increased by 108.

Question asked:

Find the original dimensions.

Solution:

Let width of rectangle = $x$

As given that a rectangle is 3 times as long as it is wide.

Length of rectangle = $3x$

$Area\ of\ rectangle=length\times breadth$

                             $=x\times3x=3x^{2}$

Now, as given that length is increased by 6 and the width is increased by 8,

New length = $3x+6$

New breadth = $x+8$

New area = $(3x+6)(x+8)$

                $=3x(x+8)+6(x+8)\\\\=3x^{2} +24x+6x+48\\=3x^{2} +30x+48$

As new area increased  by 108, we can say:-

New area - old area = 108

$3x^{2} +30x+48-(3x^{2} )=108\\3x^{2} +30x+48-3x^{2} =108\\\\30x+48=108$

Subtracting both sides by 48

$30x+48-48=108-48\\30x=60$

Dividing both sides by 30

$x=2$

Width of rectangle = $x$ = 2

Length of rectangle = $3x$ = $3\times2=6$

Therefore, original length of rectangle was 6 and original width of rectangle was 2.