# Diagonal of a rectangle

I need help solving this problem:

The diagonal of a rectangle is $18$ cm longer than the shorter leg. If the area is $168 \ \text{cm}^2$, find the dimensions of the rectangle.

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Hints Let's name the shorter leg, say let it be $a$. Then use Pythagorean theorem to determine the other one (say you call it $b$). Then the area is $ab$ (in cm${}^2$).

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Diagonal is $18$ cm longer than shorter side.:) – Inceptio Mar 23 '13 at 21:04

Let $x$ = length of shorter leg

Let $y$ = length of longer leg

Let $d$ = length of diagonal

$$d = x + 18$$ $$xy = 168\tag{Area}$$ $$x^2 + y^2 = d^2 = (x+18)^2\tag{Pythagorean Theorem}$$

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Hint:

Take shorter leg as '$x$'

Longer leg = $\sqrt{(x+18)^2-x^2}$

$168= \sqrt{(x+18)^2-x^2}(x)$. Solve for $x$. I hope $\sqrt{(x+18)^2-x^2}$ gives something pretty.;)

Or try out factorizing $168$, that makes sense and problem will be solved easily.

$168=7\times2^3\times3$

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Please use $\cdot$ (which you get with \cdot ) for multiplication instead of periods. Particularly if you just have one, it looks like a decimal point. In some cases you might want $\times$, which you get with \times – Ross Millikan Mar 23 '13 at 22:10
@RossMillikan: Sure. I will use those from now. – Inceptio Mar 24 '13 at 3:55