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## 8th grade (Illustrative Mathematics)

### Unit 8: Lesson 8

Lesson 8: Finding unknown side lengths# Pythagorean theorem example

CCSS.Math:

Sal uses the Pythagorean theorem to find the height of a right triangle with a base of 9 and a hypotenuse of 14. Created by Sal Khan and Monterey Institute for Technology and Education.

## Video transcript

Say we have a right triangle. Let me draw my right triangle
just like that. This is a right triangle. This is the 90 degree
angle right here. And we're told that
this side's length right here is 14. This side's length right
over here is 9. And we're told that
this side is a. And we need to find
the length of a. So as I mentioned already,
this is a right triangle. And we know that if we have a
right triangle, if we know two of the sides, we can always
figure out a third side using the Pythagorean theorem. And what the Pythagorean theorem
tells us is that the sum of the squares of the
shorter sides is going to be equal to the square of the
longer side, or the square of the hypotenuse. And if you're not sure about
that, you're probably thinking, hey Sal, how do I know
that a is shorter than this side over here? How do I know it's
not 15 or 16? And the way to tell is that the
longest side in a right triangle, and this only applies
to a right triangle, is the side opposite the
90 degree angle. And in this case, 14 is opposite
the 90 degrees. This 90 degree angle kind of
opens into this longest side. The side that we call
the hypotenuse. So now that we know that that's
the longest side, let me color code it. So this is the longest side. This is one of the
shorter sides. And this is the other of
the shorter sides. The Pythagorean theorem tells us
that the sum of the squares of the shorter sides, so a
squared plus 9 squared is going to be equal
to 14 squared. And it's really important that
you realize that it's not 9 squared plus 14 squared is
going to be equal to a squared. a squared is one
of the shorter sides. The sum of the squares of these
two sides are going to be equal to 14 squared, the
hypotenuse squared. And from here, we just
have to solve for a. So we get a squared plus 81
is equal to 14 squared. In case we don't know
what that is, let's just multiply it out. 14 times 14. 4 times 4 is 16. 4 times 1 is 4 plus 1 is 5. Take a 0 there. 1 times 4 is 4. 1 times 1 is 1. 6 plus 0 is 6. 5 plus 4 is 9, bring
down the 1. It's 196. So a squared plus 81 is equal
to 14 squared, which is 196. Then we could subtract 81 from
both sides of this equation. On the left-hand side, we're
going to be left with just the a squared. These two guys cancel out, the
whole point of subtracting 81. So we're left with a squared
is equal to 196 minus 81. What is that? If you just subtract
1, it's 195. If you subtract 80,
it would be 115 if I'm doing that right. And then to solve for a, we just
take the square root of both sides, the principal square
root, the positive square root of both sides
of this equation. So let's do that. Because we're dealing with
distances, you can't have a negative square root, or a
negative distance here. And we get a is equal to
the square root of 115. Let's see if we can break
down 115 any further. So let's see. It's clearly divisible by 5. If you factor it out, it's
5, and then 5 goes in the 115 23 times. So both of these are
prime numbers. So we're done. So you actually can't
factor this anymore. So a is just going to be equal
to the square root of 115. Now if you want to get a sense
of roughly how large the square root of 115 is, if you
think about it, the square root of 100 is equal to 10. And the square root of
121 is equal to 11. So this value right here is
going to be someplace in between 10 and 11, which
makes sense if you think about it visually.