Question

The sides and classification of a triangle are given. The length of the longest side is the integer given. What value(s) of x make the triangle?

Answer 1

1. x, x, 6 acute

Answer 2

Are you studying the Triangle Inequality Theorem: One side of a triangle is shorter than the sum of the other two?

Answer 3

1. x, x, 6 --> triangle with 4 sides lol

Answer 4

Wait how is it 4 sided?

Answer 5

1, x, x, 6 ?

Answer 6

1 as in #1 lol

Answer 7

Look at the attachment. (You did not answer my question about the Triangle Inequality Theorem).

Answer 8

No. But I'm pretty sure it's on her handout. I just haven't gotten the chance to look through them.

Answer 9

6 ^2 < x^2 + x^2 if triangle is to be acute.

36 < 2 x^2

18 < x^2

Answer 10

If we do not know the theorems, we cannot do the problems. The theorems are the map we need.

Answer 11

Yup, but you get there, so how would you find x?

Answer 12

Get the theorem list out.

Answer 13

She numbers her theorems like theorem 7

Answer 14

We have to solve this before we "get there" without a map or with a map. lol

18 < x^2

Answer 15

So I don't exactly know if it is the theorem you're talking about

Answer 16

This is algebra.

18 < x^2 or x^2 >18

x^2 - 18 > 0. Algebra solves that. You could try taking square roots of both sides of x^2 >18.

Answer 17

Since the triangle you're looking for is isosceles and acute, you know that the sum of the squares of the two equal side are greater than the square of the non-equal side. In other words, \[x^2+x^2 > 6^2\]

Answer 18

I thought Directrix summed it up pretty well.

Answer 19

So couldn't x be 4 or 3?

Answer 20

x^2 + x^2 < 36, I think.

Answer 21

No, x can't be 4 or 3.

Answer 22

I get the rest of the steps, just don't understand how to find x.

Answer 23

Algebra Time: x^2 + x^2 < 36

Answer 24

That should be greater than, not less than.

Answer 25

\(x^2+x^2 > 36\)

Answer 26

So \(2x^2 > 36 \Longrightarrow x^2 > 18\) Since \(36/2 = 18\)

Answer 27

Then, we only have to take the square root of both sides, so \(\sqrt{x^2} > \sqrt{18} \Longrightarrow x > 3\sqrt2\)

Answer 28

Did this make more sense?

Answer 29

Oh! I see now. I think it was the > sign that threw me off a bit.

Answer 30

It's all good. Do you think you could do a similar problem in the future?

Answer 31

Yes, in fact, I have one right now. After I finish, will you tell me if I'm correct?

Answer 32

of course