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If the width of a rectangle is 2 less than twice its length and the area of the rectangle is 161 square cm what is the length of the diagonal?
Suppose the length is L cmThen the width is 2L - 2 cm
Therefore, the area is L*W = L*(2L - 2) = 2L^2 - 2L sq cm.
Then 2L^2 - 2L = 161
or 2L^2 - 2L - 161 = 0
therefore L = [1+sqrt(323)]/2 [the other root of the quadratic equation is negative and so cannot be a length].
and then W = sqrt(323) - 1
By Pythagoras,
L^2 = 1/4 + 1/4*323 + sqrt(323)/4 = 81 + sqrt(323)/2
W^2 = 323 + 1 - 2*sqrt(323) = 324 - 2*sqrt(323)
and then D^2 = L^2 + W^2 = 405 - 3/2*sqrt(323) = 378.0417 approx
and so D = 19.4433 cm, approx.
What else can I help you with?
The width of a rectangle is 8 less than twice its length If the area of the rectangle is 28 cm2 what is the length of the diagonal?
Suppose the length is L cmThen the width is 2L - 8 cmTherefore, the area is L*W = L*(2L - 8) = 2L^2 - 8L sq cm.Then 2L^2 - 8L = 28or 2L^2 - 8L - 28 = 0that is L^2 - 4L - 14 = 0therefore L = 2+3*sqrt(2) [the other root of the quadratic equation is negative and so cannot be a length].and then W = 6*sqrt(2) - 4By Pythagoras,L^2 = 4 + 9*2 + 12*sqrt(2) = 22 + 12*sqrt(2)W^2 = 36*2 + 16 - 48*sqrt(2) = 88 - 48*sqrt(2)and then D^2 = L^2 + W^2 = 110 - 36*sqrt(2) = 59.0883 approxand so D = 7.6869 cm, approx.
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