Ask your own question, for FREE!
Mathematics 6 Online
OpenStudy (anonymous):

What is the link of the missing side of the right triangle? Sqrt(74) 2sqrt(15) 10 2

OpenStudy (anonymous):

OpenStudy (tkhunny):

Do you have the Pythagorean Theorem?

OpenStudy (anonymous):

Yup. A^2+B^2=C^2 But when I do it I get a awkward number.

OpenStudy (anonymous):

@tkhunny

OpenStudy (tkhunny):

What's wrong with that? Write an exact value that is not awkward. What did you get? Note: The answer is UNIQUE. No matter how you do it, you should get the same value, awkward or not.

OpenStudy (anonymous):

I got -24. And it isn't one of the choices.

OpenStudy (tkhunny):

Excellent. I am glad that you recognized -24 as an indication that something went horribly wrong. 5^2 = 25 7^2 = 49 Why did you subtract them? c^2 = 25+49 = 74

OpenStudy (anonymous):

Ohhh. I thought you would do like 49+b^2=25. That's the part I got lost at.

OpenStudy (tkhunny):

That does explain why you subtracted. It's a convention! a and b are always the short sides - the legs - the sides adjacent to the right angle.

OpenStudy (anonymous):

Okay. So it would be the Sqrt(74)?

OpenStudy (tkhunny):

Can you simplify that?

OpenStudy (anonymous):

8.6. But one of the answers is Sqrt(74)

OpenStudy (tkhunny):

Don't you DARE look at those answers until you have one of your own!! Personally, I would put \(\sqrt{75}\) AND \(5\sqrt{3}\) amongst the answers. I the instructions said "Simplified", I would mark wrong everyone who said \(\sqrt{75}\). I ask again, \(\sqrt{74}\). Can you simplfy that?

OpenStudy (tkhunny):

Just say "no" and we can move on. :-)

OpenStudy (anonymous):

Use Pythagoras Theorem. Given sides are 5 and 7, So, we square them and take the square root of the product as the answer to the measure of the third side. \[\sqrt{5^{2}+7^{2}}=\sqrt{25+49}=\sqrt{74}\]

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!