The length of a rectangle is 59 inches greater than twice the width. If the diagonal is 2 inches more than the length, find the
dimensions of the rectangle.
1 answer:
L = 59+2w
diagonal = 2+l
diagonal^2 = l^2 +w^2
(2+l)^2 = l^2 +w^2
4 +4l +l^2 = l^2 +w^2
4+4l =w^2
l = 59+2w
4+4(59+2w)=w^2
4+236+8w=w^2
w^2 -8w -240 =0
D = 64 +960 = 1024
w_1,2 = (8+/-sqrt1024)/2 = -12 and 20
l=59+2*20=59+40=99
l=99
w=20
hope helped
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