Without using a calculator, find the hypotenuse of a right triangle that has legs with lengths and .

for the extra challange you dont have to but if you want to use a calculator you can.

P.S. Since im looking fo a quick answer you dont have to write the explanation, but again if you want to you can.

Guest Oct 28, 2017

#1**+2 **

How are you going to know how to do these on a test if you don't know the explanation??

We can use the Pythagorean theorem to do this problem. The Pythagorean theorem says

a^{2} + b^{2} = c^{2} , where " a " and " b " are legs of a right triangle, and " c " is the hypotenuse.

They tell us the legs are 264 and 495 . So...

264^{2} + 495^{2} = c^{2} Multiply 264 * 264 to get 69,696 and 495 * 495 to get 245,025 .

69,696 + 245,025 = c^{2} Add them

314,721 = c^{2} Take the positive square root of both sides.

561 = c

hectictar Oct 28, 2017

#2**+2 **

Well, if I was truly not allowed to use a calculator during any step throughout this problem, I would do the following:

\(264^2=(200+60+4)(200+60+4)=200(200+60+4)+60(200+60+4)+4(200+60+4)\)

There are a lot of friendly numbers here, do you agree? Now, let's continue the expansion.

\(200(200+60+4)+60(200+60+4)+4(200+60+4)=40000+12000+800+12000+3600+240+800+240+16\)

WOAH! That's a lot of terms! I would first rearrange the numbers such that the largest are added to the smallest.

\(40000+12000+12000+3600+800+800+240+240+16\)

You can do addition in any order you'd like. The colors indicate an addition where the result will be friendly and will end in another 0.

Now, just do the addition.

\(40000+12000+12000+3600+800+800+240+240+16=69696\)

I wouldn't do 495 the same way, however. I would break that into \((500-5)(500-5)\). Now, do the same thing like before.

\(500(500-5)-5(500-5)\)

\(250000-2500-2500+25\)

\(245025\)

The square root is where I would hit a barrier, though...

TheXSquaredFactor Oct 28, 2017