Question

11. Refer to the rectangle at the right. Suppose the side lengths

Answer 1

Answer: a) The percent of change in perimeter is 100%.

b) The percent of change in area is 300%.

Step-by-step explanation:

Since we have given that

if the side lengths of rectangle get doubled.

Let the length be 'x'.

Let the breadth be 'y'.

So, Perimeter of rectangle would be

$2(x+y)$

Area of rectangle would be

$xy$

After doubling the lengths,

Length becomes '2x'.

Width becomes '2y'.

So, perimeter becomes

$2(2x+2y)\\\\=4(x+y)$

Area becomes

$2x\times 2y\\\\=4xy$

a. Find the percent of change in the perimeter.

$\dfrac{4(x+y)-2(x+y)}{2(x+y)}\times 100\\\\=\dfrac{2(x+y)}{2(x+y)}\times 100\\\\=100\%$

Hence, the percent of change in perimeter is 100%.

b. Find the percent of change in the area.

$\dfrac{4xy-xy}{xy}\times 100\\\\=\dfrac{3xy}{xy}\times 100\\\\=300\%$

Hence, the percent of change in area is 300%.