Thread: How to find the length of the hypotenuse

1. 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.

Thanks a bunch for your help

Veronica

2. 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.

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..

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

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

4. 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=$\displaystyle 10\sqrt{2}$

5. 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