Question

The length of a rectangle is increased by 60%.by what percent would the width have to be decreased to maintain the same area

Answer 1

the width has to be decreased 37.5%...refer to the image attached for solution

Answer 2

Answer:

37.5% of the width has to be decreased to maintain the same area.

Step-by-step explanation:

Area of the rectangle = l x b

Given the length is increased by 60%

Hence the new length of rectangle becomes  $l+\frac{60+l}{100}=\frac{8 l}{5}$

Now the area of the new rectangle becomes $\bold{=\frac{8 l}{5} \times b 1}$

Let b1 be the new width of the rectangle

Since the area remains same,

$\begin{array}{l}{\left\l\times b=\frac{8 l}{5} \times b 1\right.} \\ {b=\frac{8}{5} \times b 1} \\ {b 1=\frac{5}{8} \times b}\end{array}$

% Decrease in width $\bold{=\frac{b-\frac{5}{8} b}{b} x 100}$

$\begin{aligned} &=\frac{3}{8} \times 100 \\ &=37.5 \% \end{aligned}$