Question

Can someone please help me with some Special Right Triangles? I will post them in a link below.

Answer 1

http://gyazo.com/89545e2b6570d8819ed399807c220b26 The first one.

Answer 2

Do you know the special right triangles?

Answer 3

http://www.mathplanet.com/media/44806/45-45-90___30-60-90.png

Answer 4

I don't, sorry..

Answer 5

@whpalmer4 Could you maybe help me again?

Answer 6

On a 45-45-90 triangle, both the legs have the same value, while the hypotenuse has same value but with a square root of 2 value.. coming from
\[\LARGE a^2+b^2=c^2\]

Answer 7

Okay..

Answer 8

So...??

Answer 9

whpalmer is here o:

Answer 10

So plug in the values and solve it :-)

You know that one side is 8. Let that side be \(a\). You know that the other side is equal, because you have an isosceles triangle (2 angles are congruent). Let that side be \(b\) which also equals 8. That gives you:
\[a^2+b^2=c^2\]\[8^2+8^2=c^2\]
Can you solve that for \(c\)?

Answer 11

So would it be 16?

Answer 12

No...

\[8^2 = 8*8 = \]

Answer 13

Uh..I got a big number from that but it's not there on my answers list?

Answer 14

What is 8*8? Don't jump ahead to thinking you have the answer!

Answer 15

64. So it would be 128.

Answer 16

\[8^2 + 8^2 = c^2\]\[128 = c^2\]What does \(c=\)

Answer 17

Put another way:\[c*c = 128\]\[c = \]

Answer 18

128 right? I'm so confused, sorry.

Answer 19

128*128 = 128?

Answer 20

And if I'm allowed to jump it..
\[\LARGE \sqrt{64*2}=c\]c:

Answer 21

Oh! I thought I did that, oops.

Answer 22

Yes, @Luigi0210 you are always welcome to jump in!

Answer 23

16384

Answer 24

yes, 128*128 = 16384.

We want a number,\(c\), such that \(c*c = 128\). You've shown that \(c\) is definitely not 128 :-)

Answer 25

@Luigi0210 Thank you for your help too!!!

Answer 26

Okay..

Answer 27

If \[c^2 = 128\]we take the square root of both sides:
\[\sqrt{c^2} = \sqrt{128}\]What is the square root of \(c^2\), assuming that \(c>0\)?

Answer 28

Hint: \[\sqrt{c^2} = \sqrt{c*c}\]

Answer 29

0?

Answer 30

perhaps you don't understand the concept of a square root?

The square root of some number \(x\) is another number, \(a\), such that \(a*a = x\).

The square root of 4 is 2, because 2*2 = 4. The square root of 9 is 3, because 3*3 = 9. Can you tell me what the square root of 64 is?

Answer 31

8?

Answer 32

I understand the concept, it just looks like soup to me.

Answer 33

Sigh.

4*4 =
5*5 =
6*6 =
7*7 =
8*8 =
9*9 =
10*10 =

What number times itself gives you 64? The answer is in the table you just made.

Answer 34

(Sorry if it looks like I'm gone, bad internet connection.)

Answer 35

I just said 8.

Answer 36

Sorry.

Answer 37

Okay. So \[\sqrt{64} = 8\]However, we need to find the square root of 128, or\[\sqrt{128} = \sqrt{64*2}\]as @Luigi0210 helpfully pointed out. By the properties of square roots, we can write \[\sqrt{128} = \sqrt{64*2} = \sqrt{64}*\sqrt{2}\]We know that \(\sqrt{64} = 8\), so that becomes
\[\sqrt{128} = \sqrt{64*2} = \sqrt{64}*\sqrt{2} = 8*\sqrt{2}\]

Is there an integer that we can square to get 2? If not, we're done.

Answer 38

Unless four would work, no?

Answer 39

Does 4*4 = 2?

You're confusing squaring with taking a square root.

No, there is no rational number we can square (multiply by itself) to get 2.

\[1*1=1\]\[2*2=4\]If such a number existed, it would be somewhere between 1 and 2.

Our final answer is \[c =\sqrt{128} = \sqrt{64}*\sqrt{2} = 8*\sqrt{2} = 8\sqrt{2}\]

What @Luigi0210 was trying to tell you with the reference to special triangles is that in an isosceles right triangle, the ratio of the length of the hypotenuse to the length of either side is always \(\sqrt{2}:1\). So if you know that the side is \(8\), you can immediately conclude that the hypotenuse is \(8\sqrt{2}\). If the side is 10, the hypotenuse is \(10\sqrt{2}\). It also works the other way, of course: if you know what the hypotenuse is, you can find the side by dividing by \(\sqrt{2}\). This particular relationship is ONLY true if you have a 45/45/90 triangle.

I would encourage you to watch this video: https://www.khanacademy.org/math/arithmetic/exponents-radicals/radical-radicals/v/understanding-square-roots

You seem to be on rather unsteady ground when it comes to square roots. Maybe the video can help clear it up. If not, ask your teacher for a review.