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CCSS.Math:

say we have a right triangle so 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 this sides length right here is 14 this sides 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 we can always figure out 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 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 plus 9 squared is going to be equal to 14 squared 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 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 one is four plus one is five 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 let's subtract 81 from both sides on the left hand side we're going to be left with just the a squared these two guys cancel out whole point of subtracting 81 so left with a squared is equal to 196 minus 81 196 minus 81 what is that see 190 so that if you just subtract 195 you subtract 80 it'd be 115 if I'm doing that right right it would be 115 and then to solve for a we just take the square root of both sides the principle 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 a is equal to the square root of 115 now let's see if we can break down 115 any further so let's see it's clearly divisible by 5 it's a fit if you factor it out it's 5 and then what is it / - 1 15 23 times 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 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 if you think about it visually