if the width is w, then the length is w+7, so the diagonal is w+14:
w^2 + (w+7)^2 = (w+14)^2
Suppose you want to draw a rectangle where the width is 7 inches less than the length and the diagonal is 7 inches longer than the length. What are the dimensions of the rectangle?
2 answers
right triangle with hypotenuse = diagonal
x^2 + (x-7)^2 = (x+7)^2
x^2 + x^2 - 14 x + 49 = x^2 +14 x +
49
x^2 -28 x = 0 = x(x-28)
x = 0 or x = 28
so
28 by 21
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try that length =
diagonal^2 = 28^2 + 21^2
= 784 + 411 = 1225
so diagonal = sqrt(1225) = 35
yes, that works
x^2 + (x-7)^2 = (x+7)^2
x^2 + x^2 - 14 x + 49 = x^2 +14 x +
49
x^2 -28 x = 0 = x(x-28)
x = 0 or x = 28
so
28 by 21
-----------------------
try that length =
diagonal^2 = 28^2 + 21^2
= 784 + 411 = 1225
so diagonal = sqrt(1225) = 35
yes, that works
1 answer