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Question
Suppose the length and the width of the sandbox are doubled.

the width is 6ft and the length is 10ft

a. Find the percent of change in the perimeter.

The percent of change in the perimeter is a
( %) increase.

b. Find the percent of change in the area.

The percent of change in the area is a
( %) increase.

Answer :

Answer:

Rectangle is the closed shaped polygon with 4 sides. Opposite sides of the rectangle are equal. when the length and the width of the sandbox are doubled the percent of change in the perimeter is 100 percent and the percent of change in the area is 300 percent.Given-The length of the sandbox is 10.The width of the sandbox is 6.a) The percent of change in the perimeter.Perimeter of the rectangleThe perimeter of the rectangle is the twice of the sum of its side.The perimeter P of the sandbox is,When the length and the width of the sandbox are doubled the perimeter of the box is,Percentage  change in the perimeter,b) The percent of change in the area.Area of the rectangleThe area of the rectangle is the product of its side.The area A of the sandbox is,When the length and the width of the sandbox are doubled the area  of the box is,Percentage  change in the area,Thus when the length and the width of the sandbox are doubled the percent of change in the perimeter is 100 percent and the percent of change in the area is 300 percent.

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