MATH SOLVE

4 months ago

Q:
# 11. Refer to the rectangle at the right. Suppose the side lengthsare doubled.a. Find the percent of change in the perimeter.b. Find the percent of change in the area.What’s the answer?

Accepted Solution

A:

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 [tex]2(x+y)[/tex]Area of rectangle would be [tex]xy[/tex]After doubling the lengths, Length becomes '2x'.Width becomes '2y'.So, perimeter becomes[tex]2(2x+2y)\\\\=4(x+y)[/tex]Area becomes[tex]2x\times 2y\\\\=4xy[/tex]a. Find the percent of change in the perimeter.[tex]\dfrac{4(x+y)-2(x+y)}{2(x+y)}\times 100\\\\=\dfrac{2(x+y)}{2(x+y)}\times 100\\\\=100\%[/tex]Hence, the percent of change in perimeter is 100%.b. Find the percent of change in the area.
[tex]\dfrac{4xy-xy}{xy}\times 100\\\\=\dfrac{3xy}{xy}\times 100\\\\=300\%[/tex]Hence, the percent of change in area is 300%.