Question

Suppose the length and the width of the sandbox are doubled. 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 % in

Answer 1

a. The percent of change in perimeter is 100% increase

b. The percent of change in Area in 300% increase.

Explanation:

The diagram was missing we have attached for your reference.

Given:

Length of the sandbox = 10 ft

Width of the sandbox = 6 ft

We need to find the change in percentage of perimeter and area when the dimensions are doubled.

Solution:

First we will find the Perimeter and area of the sand box with current dimension.

Now we know that;

Perimeter of rectangle is twice the sum of length and width.

Perimeter of rectangle =
`2(6+10)=2*16 =32\ ft`

Area of rectangle is length times width.

Perimeter of rectangle =
`6* 10 = 60\ ft^2`

Now When the dimension are doubled.

Length of the sand box = 20 ft

width of the sandbox = 12 ft

Perimeter of sandbox when dimensions are doubled =
`2(20+12)=2* 32 =64 \ ft`

Area of the sand box when dimension are doubled =
`20* 12= 240 \ ft^2`

Now we need to find the Percent of change in Perimeter and Area.

Percent of change can be calculated by new value minus original value divided by original value multiplied by 100.

Percent of change in Perimeter =
`(64-32)/(32) * 100 = 100\%`

Percent of change in area =
`(240-60)/(60)*100 = (180)/(60)* 100 = 300\%`

Hence When the dimension are doubled then the percent of change in perimeter is 100% increase and the percent of change in Area in 300% increase.