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

Question

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.

Savior · Accepted Answer

a. 100% increase b. 300% increase boom.

uhh · Answer

idk :P

Chloe · Answer

its so annoying they wont answer my question either

Ms. Sue · Answer

Please be patient. I can't help you, but math tutors should be on Jiskha within an hour or two.

Henry2 · Answer

a. P1 = 2L + 2W. = 2(L+W).
P2 = 4L + 4W.= 4(L+W).
P2/P1 = 4(L+W)/2(L+W) = 2.0 = 200%.
200% - 100% = 100% Increase.

b. A1 = L * W.
A2 = 2L * 2W = 4(L*W).
A2/A1 = 4(L*W)/(L*W) = 4.0 = 400%.
400% - 100% = 300% Increase.