Question

if the length and breadth of a rectangle are doubled by what percentage is its area increased​​

Answer 1

In context to question asked,

We have to determine the percentage increase in area when the length and breadth of rectangle are doubled.

So, let the length and breadth of rectangle are l and b units.

So, area of rectangle will be $=(l\times b) = lb \:unit^2$

As length and breadth of rectangle are doubled,

So, new length and breadth of rectangle will be 2l, 2b respectively.

So, new area will be $2l \times 2b = 4lb\:cm^2$

So, required percentage increase will be,

$\frac{4lb-lb}{lb}\times 100\% = \frac{3lb}{lb} \times100\% = 300\%$

Hence, when the length and breadth of rectangle are doubles then the area will be increased by 300%.

Answer 2

Answer:

300%

Step-by-step explanation:

Let initial length be L

Let initial breadth be B

Then initial area will be LW

Final length after doubling = 2L

Final breadth after doubling =2B

Therefore final area = 2L×2B = 4LB

Percentage change in area = (change in area/initial area) × 100% = (3LB/LB) ×100% = 300%