Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (tvinsant):

PLEASE HELP, WILL FAN AND MEDAL. :) Find the length of side x.(Look in the next comment for the picture) a. x = sqrt325 b. x = 15 c. x = sqrt253 d. x = 11

OpenStudy (tvinsant):

Directrix (directrix):

Pythagorean Theorem is needed to do this problem.

Directrix (directrix):

Do you know the theorem? If not, then we need to post it. @tvinsant

OpenStudy (tvinsant):

Yes. I tried but I came up with a different answer :? so I'm kinda stuck. the formula is a^2+b^2+c^2=180

Directrix (directrix):

You have the sum of the angles of a triangle mixed up with the P Theorem

Directrix (directrix):

a^2+b^2 = c^2 where a and b are legs and c is the hypotenuse.

OpenStudy (tvinsant):

OHHHHHH

OpenStudy (tvinsant):

okay... sorry lol

OpenStudy (tvinsant):

so the answer is A.

Directrix (directrix):

Legs are 6 and x.

Directrix (directrix):

so the answer is A. I don't know. That is why we are writing the equation - so that there is no need to guess.

OpenStudy (tvinsant):

because when you square the two and sum it up, the answer is 325. Then you'd have to square it so it'd be x=sqrt325

Directrix (directrix):

I did not get that.

OpenStudy (tvinsant):

lol ummm

OpenStudy (tvinsant):

but is the answer A?

Directrix (directrix):

x^2 + 6 ^2 = 17^2

Directrix (directrix):

>>but is the answer A? I do not know.

OpenStudy (tvinsant):

ohhh nvm is C.

OpenStudy (tvinsant):

its*

Directrix (directrix):

If you don't want to solve this x^2 + 6 ^2 = 17^2 , I'll move along.

OpenStudy (tvinsant):

x^2+36=289 x^2=289-36 x^2=253 sqrtx=sqrt253

Directrix (directrix):

The MaPaw site has a glitch. sqrtx=sqrt253 should be x = sqrt253

OpenStudy (tvinsant):

oh okay

Directrix (directrix):

Now, at long last, I agree with you that C is the correct option.

OpenStudy (tvinsant):

Thank you :) Have a nice day!

Directrix (directrix):

Same to you and you're welcome.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!