# How to find the length of the hypotenuse

• Nov 21st 2009, 08:04 AM
cvhubb
How to find the length of the hypotenuse
Hi,

Sorry if this a easy problem.... I've been out of school now for 15 years and trying to study up on taking the ACT test to start college in August. There is so much I have forgotten. I could use some help with this problem. Incase you cannot read the question it says....

In the isosceles right triangle below, AB = 10 feet. What is the length, in feet, of AC ?

It has a multiple choice and I'm thinking the answer is E (I believe that this is what my brother said) but I need to know the steps of how he got that for the answer.

http://i241.photobucket.com/albums/f...htTriangle.jpg

Thanks a bunch for your help

Veronica
• Nov 21st 2009, 08:10 AM
skeeter
• Nov 21st 2009, 08:12 AM
Unenlightened
Quote:

Originally Posted by cvhubb
Hi,

Sorry if this a easy problem.... I've been out of school now for 15 years and trying to study up on taking the ACT test to start college in August. There is so much I have forgotten. I could use some help with this problem. Incase you cannot read the question it says....

In the isosceles right triangle below, AB = 10 feet. What is the length, in feet, of AC ?

It has a multiple choice and I'm thinking the answer is E (I believe that this is what my brother said) but I need to know the steps of how he got that for the answer.

http://i241.photobucket.com/albums/f...htTriangle.jpg

Thanks a bunch for your help

Veronica

Note the isoceles, which means two of the sides of the triangle are of the same length.
Then I suggest you look at the wikipedia page for "Pythagoras' theorem" which will explain the formula for getting the length of the hypotenuse of a right-angled triangle. (You can tell the triangle is a right-angled one because of the little square drawn in the corner beside the B. If that weren't there, you could not assume the triangle was right-angled unless stated.)

edit: actually it does say in the question that the triangle is right-angled, apologies..
• Nov 21st 2009, 08:32 AM
cvhubb
Quote:
Thanks skeeter for the link. A lot of good info to help me out. Going now to read some more.

Quote:

Originally Posted by Unenlightened
Note the isoceles, which means two of the sides of the triangle are of the same length.
Then I suggest you look at the wikipedia page for "Pythagoras' theorem" which will explain the formula for getting the length of the hypotenuse of a right-angled triangle. (You can tell the triangle is a right-angled one because of the little square drawn in the corner beside the B. If that weren't there, you could not assume the triangle was right-angled unless stated.)

edit: actually it does say in the question that the triangle is right-angled, apologies..

Thank you too Unenlightened... that is what I thought since it was a right triangle that AB and BC would be the same length, both would be 10. So the forumula I need to use is c = square root of a^2 + b^ , right? When I put this into my calculator I get 14.14213562 . Which I know letter E in the multiple choice is the answer.

One question I have a TI-30X IIS calculator.... is there some way you can change that number to a square root or will I just have to enter in the multiple choice to see which one is right?

Thanks Again,
Veronica
• Nov 21st 2009, 09:21 AM
Unenlightened
Quote:

Originally Posted by cvhubb
Thanks skeeter for the link. A lot of good info to help me out. Going now to read some more.

Thank you too Unenlightened... that is what I thought since it was a right triangle that AB and BC would be the same length, both would be 10. So the forumula I need to use is c = square root of a^2 + b^ , right? When I put this into my calculator I get 14.14213562 . Which I know letter E in the multiple choice is the answer.

One question I have a TI-30X IIS calculator.... is there some way you can change that number to a square root or will I just have to enter in the multiple choice to see which one is right?

Thanks Again,
Veronica

If you leave it as a square root instead of using the calculator, it's probably easier..
ie. (AC)^{2} = (AB)^2+(BC)^{2}
so (AC)^{2}= 100 + 100
(AC)^{2}= 200
AC = $\displaystyle \sqrt{200}$
AC= $\displaystyle \sqrt{100}\sqrt{2}$
AC=$\displaystyle 10\sqrt{2}$
• Nov 21st 2009, 09:25 AM
cvhubb
Quote:

Originally Posted by Unenlightened
If you leave it as a square root instead of using the calculator, it's probably easier..
ie. (AC)^{2} = (AB)^2+(BC)^{2}
so (AC)^{2}= 100 + 100
(AC)^{2}= 200
AC = $\displaystyle \sqrt{200}$
AC= $\displaystyle \sqrt{100}\sqrt{2}$
AC=$\displaystyle 10\sqrt{2}$

OK, gottcha and thanks a bunch