Asked by kristina
the perimeter of a piece of cardboard is 34 inches. squares measuring 2 inches on a side are cut from each corner so that when the sides are folded up, the diagonal of the resulting box has the length 7 inches.what are the original dimensions of the card board?
Answers
Answered by
Reiny
let the width of the cardboard be x
then it's length is 17-x
2 inches cut off each dimensions leaves the base of the box to be
x-4 and 17-x-4
x-4 and 13-x
(x-4)^2 + (13-x)^2 = 49
x^2 - 8x + 16 + 169 - 26x + x^2 = 49
2x^2 - 34x + 136 = 0
x^2 - 17x + 68 = 0
x = (17 ± √17)/2
x = appr 10.56 inches or x = appr 6.438 inches
mmmhh, was expecting "nicer" numbers
check:
perimeter = 2(10.56+6.44) = 34 , check
new base is 6.56 by 2.438
hypotenuse??
5.56^2 + 2.438^2
= 6.998 , close enough to 7 based on round-off
then it's length is 17-x
2 inches cut off each dimensions leaves the base of the box to be
x-4 and 17-x-4
x-4 and 13-x
(x-4)^2 + (13-x)^2 = 49
x^2 - 8x + 16 + 169 - 26x + x^2 = 49
2x^2 - 34x + 136 = 0
x^2 - 17x + 68 = 0
x = (17 ± √17)/2
x = appr 10.56 inches or x = appr 6.438 inches
mmmhh, was expecting "nicer" numbers
check:
perimeter = 2(10.56+6.44) = 34 , check
new base is 6.56 by 2.438
hypotenuse??
5.56^2 + 2.438^2
= 6.998 , close enough to 7 based on round-off
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