Question

One side of a rectangle is 9 inches longer than another side. If the longer side of this rectangle decreases by 5 inches, and the shorter side increased by 3 inches, the area of the new rectangle equals the area of the original rectangle. Find the dimensions of the original rectangle.

Answer 1

Answer:

6, 15

Step-by-step explanation:

Let the shorter side be x

Then, longer side will be x+9

Area of the rectangle= x(x+9) = x² + 9x

Now, New parameters of the rectangle are x+9-5 and x+3

New area= (x+4)(x+3) = x² + 7x + 12

Since both areas are same.

x²+7x+12 = x²+9x

2x=12

x=6, x+9=15

Answer 2

Answer:

Dimensions of the original rectangle is :

• length = 15 inches
• breadth = 6 inches

Step-by-step explanation:

Step 1 : Let's assume the dimensions of the rectangle :

➪ Breadth = $x \sf \: inches$

➪ Length = $(x + 9) \sf \: inches$

Step 2 : Length decreases by 5 inches :

Length now will be :

$\to (x + 9) - 5 \sf \: inches$

$\to x + 9 - 5 \sf \: inches$

$\to (x + 4) \sf \: inches.$

Step 3 : Breadth increases by 3 inches :

Breadth now will be :

$\to (x + 3) \sf \: inches.$

Step 4 : Calculate the dimensions :

Given :

The area remains same for the original rectangle and the new rectangle.

Area of the original rectangle : $l \times b$

Where,

• $l {\sf {(length) }} = x + 9$
• $b {\sf{(breadth)}} = x$

So, the area of the original rectangle is :

$\to (x + 9) \times x$

$\to x^2 + 9x$

And also,

Area of the new rectangle is :

$\to (x + 4)(x + 3)$

$\to x^2 + 4x + 3x + 12$

$\to x^2 + 7x + 12$

${\underline{\boldsymbol{According \: to \: the \: question : }}}$

$\implies {x}^{2} + 7x + 12 = {x}^{2} + 9x$

$\implies \: 7x + 12= 9x$

$\implies \: 12 = 9x - 7x$

$\implies \: 12 = 2x$

$\implies \: \frac{12}{2} = x$

$\implies \: 6 = x$

${ \sf{\therefore{ breadth= 6 \: inches}}}$

Therefore, the length will be :

→ 6 + 9

→ 15 inches