You can put this solution on YOUR website! find the dimension of a rectangle whose perimeter is 30 m and who area is 56 mē
2L + 2W = 30
simplify,divide by 2
L + W = 15
W = (15-L), use this form for substitution
:
L * W = 56
Replace W with (15-L)
L(15-L) = 56
-L^2 + 15L = 56
change the signs, multiply by -1
L^2 - 15L = -56
L^2 - 15L + 56 = 0
Factors to
(L-7)(L-8) = 0
Two solutions
L = 7, then W = 8
or
L = 8, then W = 7
:
:
Check this in the perimeter
2(8) + 2(7) = 30
You can put this solution on YOUR website! find the dimension of a rectangle whose perimeter is 30 m and who area is 56 mē
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Perimeter = 2L + 2W = 30
L + W = 15
L*W = 56
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Find a pair of factors of 56 whose sum is 15