Question 940824: Please help me solve the following problem "The area of a certain rectangle is 288yds^2. the perimeter is 68 yd. If you double he length and width, what will be the area and the preimeter of the new rectangle?
Thank you in advance..!!
Found 2 solutions by lwsshak3, richard1234:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! "The area of a certain rectangle is 288yds^2. the perimeter is 68 yd. If you double he length and width, what will be the area and the preimeter of the new rectangle?
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Area varies as the square of the increase/decrease in length and width
If you double the length and width, the area increases by 4 times=4*288=1152 sq yds
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Perimeter varies directly as the increase/decrease in length and width
If you double the length and width, the perimeter increases by 2 times=2*68=136 yds
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! The area is multiplied by 4; I.e. new area = 288*4 = 1152 yd^2.
Note that we didn't need to solve for length, width. If the dimensions are l, w, then lw = 288. The new area is (2w)(2l) = 4lw = 4*288 yd^2.
Similarly, new perimeter is obtained by multiplying by 2, so new perimeter = 2*68 yd = 136 yd.
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