- ErinWeeks

A triangle has side lengths 10, 15, and 7. Is the triangle acute, obtuse, or right? Explain

- schrodinger

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- anonymous

Use Pythagoras.
\[a^2 + b^2 = c^2\]

- anonymous

If they coincide, and relate, the triangle is a right angle.

- ErinWeeks

so it would be 10^2 + 15^2 = what ??

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## More answers

- anonymous

If they don't it's either going to be acute or obtuse. For it to be acute, the hypotenuse of a right angle triangle is longer than your given hypotenuse which is 15 it is acute. If not, then it's obviously left with the only option to be obtuse.

- anonymous

c in that equation is the hypotenuse.

- ErinWeeks

i dont get what numbers to put in which letters
Im SO CONFUSED

- anonymous

the two shortest lengths of the triangle correspond with a^2 and b^2

- anonymous

the longest side which is 15, is the hypotenuse of the triangle.

- ErinWeeks

okay so 7^2 + 10^2 = 149
So Now What

- anonymous

Now c^2= 149, am I right?

- ErinWeeks

No c^2 = 225

- anonymous

Now all you have to do is Find c, which is around 12.2 and the longest length given in your question is 15. So the ideal right angle triangle with the two shortest sides you were given in the question has to have a hypotenuse of 12.2

- anonymous

225?....Are you drinking anything that's giving your problems with your thinking process?

- anonymous

giving you*

- ErinWeeks

well wouldnt it be 15^2 = 225 ? i dont know im confused

- anonymous

You're jumping ships at the moment. Right now c^2 is 149. You're thinking about the longest side given in your question.

- anonymous

Don't do that just yet. Stay at what you're doing right now. Take mathematics as a step by step process.

- anonymous

If you ruin the steps and just jump ahead two steps further, you're going to be in trouble if you do this in exams.

- ErinWeeks

okay so were saying a+b = 149 and c = 149 //

- anonymous

No,
\[a^2 + b^2= 149\]

- anonymous

c^2= 149.
c=? HINT: Use your calculator.

- ErinWeeks

okay i get it

- ErinWeeks

whats next

- anonymous

Okay so, you end up with 12.2 (to the nearest dcp.) So for an ideal right angled triangle, your hypotenuse must have a length of 12.2

- anonymous

that's with the dimensions of 7, 10 that you were given in the question.

- anonymous

Now in the question, you have a triangle that has an hypotenuse of 15 instead of 12.2

- anonymous

That tells you that it's not a right angled triangle. Therefore your only options are that the triangle must be acute or obtuse.

- anonymous

You with me?

- ErinWeeks

yea im with you sorry my nephew is being bad lol . keeep going

- anonymous

Okay. So if your given hypotenuse is less than 12.2 (eg. your dimensions were 7,10 and 11) then the triangle is acute.

- anonymous

But you were given dimensions of 7, 110 and 15.

- anonymous

10*

- ErinWeeks

Okay so its not acute or right. So it would be obtuse ?

- anonymous

so it's not acute, thus it must be obtuse.

- anonymous

Yes. Good.

- ErinWeeks

okay can you explain how tho?

- AravindG

A triangle can aeither be acute,right angled or obtuse
as seen from Aztec's proof it is neither right angled nor zcute so it must definitely be obtuse!!

- ErinWeeks

but i need to explain how it is an obtuse in a shorter term

- anonymous

This question only gave you three options. You can't say it's neither acute, obtuse or right angled because that's not an option. That's an aspect you should also work on is knowing your options.

- anonymous

This question does not need any further analysis because you already solved it. If the question gave you dimensions of 7,10 and 11, you would do the same process. You would then find that the hypotenuse of the triangle (ie. side length of 11) is shorter than the ideal hypotenuse of a right angled triangle which is 12.2. Therefore you say that it's acute because it's shorter than the ideal hypotenuse of 12.2 in a right angled triangle

- anonymous

If you still want to understand more about this. This link gives you everything you need.
http://www.algebra.com/algebra/homework/Triangles/Triangles.faq.question.457725.html

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