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Ksivusya [100]
3 years ago
15

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

Mathematics
1 answer:
NeX [460]3 years ago
7 0

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%.

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