$a$ $a$ and $b$ $b$ are the legs of a right triangle. $c$ $c$ is the hypotenuse. If $c = 13, b = 2 a$ $c=13, b=2a$ , what is $a$ $a$ and $b$ $b$ ?

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$a$ $a$ and $b$ $b$ are the legs of a right triangle. $c$ $c$ is the hypotenuse. If $c = 13, b = 2 a$ $c=13, b=2a$ , what is $a$ $a$ and $b$ $b$ ?

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Answer 1 · 2017-02-26T07:33:53

$a = \frac{13 \sqrt{5}}{5}$ $a=(13sqrt5)/5$
$b = \frac{26 \sqrt{5}}{5}$ $b=(26sqrt5)/5$

Explanation:

By Pythagorean Theorem, we know that:

$c^{2} = a^{2} + b^{2}$ $c^2=a^2+b^2$ , where $c$ $c$ is the hypotenuse and $a and b$ $a and b$ are the legs.

Given that $c = 13, and b = 2 a$ $c=13, and b=2a$ ,

$\Rightarrow 13^{2} = a^{2} + {(2 a)}^{2}$ $=> 13^2=a^2+(2a)^2$
$\Rightarrow 13^{2} = a^{2} + 4 a^{2}$ $=> 13^2=a^2+4a^2$
$\Rightarrow 13^{2} = 5 a^{2}$ $=> 13^2=5a^2$ ,
$\Rightarrow a^{2} = \frac{13^{2}}{5}$ $=> a^2=13^2/5$
$\Rightarrow a = \frac{13}{\sqrt{5}} = \frac{13 \sqrt{5}}{5}$ $=> a=13/sqrt5=(13sqrt5)/5$

$\Rightarrow b = 2 a = 2 \times \frac{13 \sqrt{5}}{5} = \frac{26 \sqrt{5}}{5}$ $=> b=2a=2xx(13sqrt5)/5=(26sqrt5)/5$

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$a$ $a$ and $b$ $b$ are the legs of a right triangle. $c$ $c$ is the hypotenuse. If $c = 13, b = 2 a$ $c=13, b=2a$ , what is $a$ $a$ and $b$ $b$ ?

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aa and bb are the legs of a right triangle. cc is the hypotenuse. If c=13,b=2ac=13, b=2a, what is aa and bb?

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$a$ $a$ and $b$ $b$ are the legs of a right triangle. $c$ $c$ is the hypotenuse. If $c = 13, b = 2 a$ $c=13, b=2a$ , what is $a$ $a$ and $b$ $b$ ?