Given right triangle ABC, with right angle at C, if b = 8 and c = 17, using the Pythagorean Theorem, what is a?

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Answer 1 · 2018-05-17T00:22:18

A is 15.

So let write the Pythagorean Theorem equation out.

$a^{2} + b^{2} = c^{2}$ $a^2+b^2=c^2$

So our value of b is 8 and c is 17. So let plug those number into the equation.

$a^{2} + 8^{2} = 17^{2}$ $a^2+8^2=17^2$

Now you square the known value which you will get:

$a^{2} + 64 = 289$ $a^2+64=289$

Subtract 64 from both sides and you get:

$a^{2} = 225$ $a^2=225$

Square both side:

$\sqrt{a^{2}}$ $sqrt(a^2)$ = $\sqrt{225}$ $sqrt(225)$

When squaring $a^{2}$ $a^2$ inside the house, the exponent of 2 cancel out which is just a.

So the final answer is a = 15

So let double check to ensure that it adds up to 289. If it does not, then somewhere along the way I made a math calculation error.

$15^{2} + 8^{2} = c^{2}$ $15^2+8^2=c^2$
$225 + 64 = c^{2}$ $225+64=c^2$
$289 = c^{2}$ $289=c^2$
$\sqrt{289} =$ $sqrt(289)=$ sqrt(c^2)#

$17$ $17$ = $c$ $c$

So a = 15 work.