Question

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

The length of the sandbox is 10 and the width is 6.

a.Find the percent of change in the perimeter.

b. Find the percent of change in the area.

Answer 1

the percent change is 200% increase of the original perimeter. the percent increase of the area is 400% of the original

Answer 2

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 rectangle

The perimeter of the rectangle is the twice of the sum of its side.The perimeter P of the sandbox is,

$P =2\times (10+6)$

$P =2\times (16)$

$P=32$

When the length and the width of the sandbox are doubled the perimeter $P_d$ of the box is,

$P_d=2\times(20+12)$

$P_d=2\times(32)$

$P_d=64$

Percentage $\DeltaP$$\Delta P$ change in the perimeter,

$\Delta P=\dfrac{64-32}{32}\times 100$

$\Delta P=100$

b) The percent of change in the area.

Area of the rectangle

The area of the rectangle is the product of its side.The area A of the sandbox is,

$A = (10\times6)$

$A=60$

When the length and the width of the sandbox are doubled the area $A_d$ of the box is,

$A_d=20\times12$

$P_d=240$

Percentage $\DeltaP$$\Delta A$ change in the area,

$\Delta A=\dfrac{240-60}{60}\times 100$

$\Delta P=300$

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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