If a=5 and c=13, how do you find b?

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Answer 1 · 2015-10-17T21:23:47

Use Pythagoras and rearrange to find $b = 12$

Assuming we're dealing with a right angled triangle with legs of lengths $a$ , $b$ and hypotenuse of length $c$ , Pythagoras theorem tells us:

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

Subtracting $a^2$ from both sides, we get:

$b^2 = c^2-a^2$

Then taking the square root of both sides, we get:

$b = sqrt(c^2 - a^2)$

We are told that $a = 5$ and $c = 13$ , so

$b = sqrt(13^2-5^2) = sqrt(169-25) = sqrt(144) = 12$

Bonus

The $5, 12, 13$ triangle is the second one in a sequence of right angled triangles that starts with the $3, 4, 5$ triangle.

$a = 2k + 3$

$b = (a^2 - 1) / 2 = 2k^2+6k+4$

$c = (a^2 + 1) / 2 = 2k^2+6k+5$

This gives us right angled triangles with sides:

$k=0:$ $3, 4, 5$

$k=1:$ $5, 12, 13$

$k=2:$ $7, 24, 25$

$k=3:$ $9, 40, 41$

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