Question

The dimensions of a rectangle are such that its length is 3 in. more than its width. If the length were doubled and if the width were decreased by 1 in., the area would be increased by 66 in^2. What are the length and width of the rectangle?
The length of the rectangle is _______ in. and the width of the rectangle is _______ in.

Answer 1

Answer:

The length of the rectangle is 11 inch and its width is 8 inch.

Step-by-step explanation:

Given:

Rectangle $(R_1)$ and rectangle $(R_2)$.

Let the width of the rectangle be 'x' inch.

Its length = '(x+3)' inch

Area of the rectangle = $(A_1)$

⇒$A_1$ = $(x)(x+3)$

Now for rectangle $(R_2)$ its width is decreased by 1 inch and length is doubled,the area $(A_2)$ will increased by 66 sq-inch.

So,

$A_2=A_1+66$

According to the question.

⇒ Area of rectangle $(R_2)$  with width '(x-1)' and length '2(x+3)'='(2x+6)' .

⇒ $A_2=(x-1)(2x+6)$

Plugging the values of $A_2$ in terms of $A_1$.

⇒ $A_1+66= (x-1)(2x+6)$

Plugging  $A_1$ = $(x)(x+3)$ into the equation.

⇒ $(x)(x+3)+66=(x-1)(2x+6)$

⇒ $x^2+3x+66=2x^2+6x-2x-6$

⇒ $x^2-2x^2+3x-6x+2x+6+66=0$

⇒ $-x^2-x+72=0$

Now using quadratic formula we can find the value of 'x'.

Quadratic formula, (x)= $\frac{\pm b+\sqrt{b^2-4ac} }{2a}$

x= -9 and x= 8

Discarding the negative value.

Our width  = 'x' = 8 inches.

So the length of the rectangle = 11 inches and width of the rectangle = 8 inches.