Question 684555: the diagonal of a rectangle is 2 centimeters longer than its length and 9 centimeters greater than its width. find the dimensions of the rectangle.
Answer by Jolliano(16)   (Show Source):
You can put this solution on YOUR website! Let d be length of diagonal,h be height and w be width.
The length,width and diagonal form a right angled triangle with the diagonal as the hypotenuse.
From the question,
d = h + 2 .........(1) and
d = w + 9 ........(2).
Equating both,
h + 2 = w + 9
h = w + 9 - 2
h = w +7..............(3)
Using Pythagoras theorem,
h^2 + w^2 = d^2
Sub h with (3) and d with (2)
(w + 7)^2 + w^2 = (w + 9)^2
w^2 + 14w + 49 + w^2 = w^2 +18w + 81
Collecting like terms,
w^2 - 4w - 32= 0
Factorizing,
(w + 4)(w - 8) = 0
w = -4 or 8.
The Width cannot be negative so
Width = 8cm
From (3)
h = w + 7 = 8 + 7 = 15cm.
From (2)
d = w + 9 = 8 + 9 = 17cm.
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