The last question I have I know I am dumb, but The thing is I'm better at Algebra :/

- anonymous

The last question I have I know I am dumb, but The thing is I'm better at Algebra :/

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- schrodinger

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- anonymous

Idk why I can't set this up!

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- whpalmer4

okay, can you tell me anything about the relationship between the small triangle and the big triangle?

- okdutchman7

The big and small triangles are similar.

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## More answers

- okdutchman7

They would be proportional to each other

- anonymous

Their base, and I guess you would call it a leg is congruent

- whpalmer4

uh, no, the bases are not congruent — the two segments that make up the base of the larger triangle are congruent...

- okdutchman7

Are you a girl?
Cause girls are better at algebra and guys are better at geometry.

- whpalmer4

I'm a counter-example to that theory :-)

- anonymous

Yes I am a girl but my BFF is way better at geometry but she's a lesson behind so she couldn't help!

- okdutchman7

9 times of 10 it's true though : )

- anonymous

I know the theory but I wasn't in class , and my teacher is really hard with her tests

- whpalmer4

|dw:1449196971961:dw|

- anonymous

and she didn't help me when I asked

- whpalmer4

those two segments labeled \(c\) are congruent, so we know that the ratio \(4:c\) is the same as the ratio of the hypotenuse of the big triangle to the base of the big triangle
can you write an equation for that?

- okdutchman7

4c = c(5x+2)
Try to solve it from there

- anonymous

Okay!

- okdutchman7

Sorry that would be
4(2c) = c(5x+2)

- anonymous

Okay so the first part is 8c - 1c = 8=5x+2 subtract two from both sides then you get 7 , so then you would divided by 5 and get 5/7 :/

- whpalmer4

no, that's still not the correct equation...
please, try to write the equation yourself first, you'll need to know how to do that to solve these in the future.

- anonymous

I thought it would be 5x-2 divided by 1/2= 4 , or somethign along that line

- whpalmer4

Here's the ratio of the hypotenuse to the base for the little triangle:
\[\frac{4}{c}\]
What is the ratio of the hypotenuse of the big triangle to the base of the big triangle?
Those two will be equal, so you can set them equal to each other, cross-multiply and solve for \(x\).

- whpalmer4

\(c\) will magically disappear without our ever having needed to know its value :-)

- anonymous

Would I cross mulitply like this X/5 *4/c ?

- whpalmer4

first you write the ratio I asked you for...

- anonymous

I apologize to my medicine is off I can't think clearly!

- whpalmer4

once you're a master at these problems you can start skipping steps, but first you have to become a master :-)

- whpalmer4

what is the hypotenuse of the big triangle?
what is the base of the big triangle?

- okdutchman7

You will need to cross multiply, but rethink what you are cross multiplying

- anonymous

Okay can you explain the ratio to me

- anonymous

5x/2, and 4/c ?

- okdutchman7

5x-2 on the big triangle is proportional to 4 on the little triangle
2c on the big triangle is proportional to c on the little triangle

- anonymous

Oh so it would be 5x-2/2c * c/4?

- whpalmer4

the hypotenuse of the big triangle is 5x-2
the base of the big triangle is the base of the little triangle plus another segment the same length, or c + c = 2c
that gives us:
\[\frac{4}{c} = \frac{5x-2}{2c}\]

- anonymous

Okay so now I cross multiply to get 5x-2*4
/2

- anonymous

would I multiply -2 to 4

- anonymous

Guys I'm sorry I;m so confused now.........

- anonymous

And you left :/

- anonymous

I got the hint ....... okay thanks anyways

- whpalmer4

\[\frac{4}{c} = \frac{5x-2}{2c}\]
Okay, if we cross-multiply, we multiply the numerator of one fraction by the denominator of the other, and vice-versa.
\[4*2c = c*(5x-2)\]\[8c = 5cx - 2c\]
Can you do the rest?

- anonymous

Yes add 2 c to the 8c which would be 10c=5cx and since they have a common variable then I can divided so x= 2?

- whpalmer4

\[8c = 5cx-2c\]\[8c+2c = 5cx-2c+2c\]\[10c = 5cx\]we can divide both sides by \(c\)
\[\frac{10c}{c} = \frac{5cx}{c}\]\[10 = 5x\]\[x=2\]
Now we do the important part, which is to check our answer!
If our hypotenuse on the big triangle is \(5x-2\) and we think \(x=2\), then the hypotenuse is \(5(2)-2 = 10-2=8\). Is that twice the length of the other hypotenuse?
Yes, it is, just like our base is twice the length of the other base. Our answer is correct.

- anonymous

Okay so it is definitely 2 thank you so much for putting up with me!

- whpalmer4

You're welcome!
By the way, I did the ratio as hypotenuse / base, but any way you want to do it is fine, so long as you do it the same way for both triangles.

- anonymous

Okay I'll keep that in mind thank you again!

- okdutchman7

@whpalmer4 you did a great job explaining it!

- anonymous

By the way my subjects I am good at is Biology, and English. The others nope!

- Hero

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- whpalmer4

@Hero you made some mistakes but got the right answer, and I think it is interesting to see why that is.
Small triangle you wrote:
\[a^2+b^2=4\](but it should be \(a^2+b^2=4^2 = 16\))
Large triangle you wrote:
\[4(a^2+b^2)=(5x-2)^2\]\[a^2+b^2=\frac{(5x-2)^2}{4}\]then
Notice: \(4 = \frac{(5x-2)^2}{4}\)
Solve for x
\[64 = (5x-2)^2\]
of course you can only get there from substituting the correct \(a^2+b^2=16\) earlier:
\[(2a)^2+(2b)^2 = (5x-2)^2\]\[4a^2+4b^2=(5x-2)^2\]\[4(a^2+b^2) = (5x-2)^2\]\[4(16) = (5x-2)^2\]\[64 = (5x-2)^2\]\[\sqrt{64} = \sqrt{(5x-2)^2}\]etc. Probably just wrote it down wrong after doing it correctly in your head. That's not the interesting part.
This approach using the Pythagorean theorem assumes that we have a right triangle, but the problem does not specify that it is! The similar triangles approach works for any pair of similar triangles where the legs are doubled in length, not just a right triangle. In fact, it is the natural of similar triangles that makes this work — double the length of the two congruent sides and you double the length of the third side, so the problem is always just \[2*4 = 5x-2\]\[x=2\]and even my approach could have been simplified.
You can write down the work as you did it for triangles which are not right triangles and even though the math is not correct (Pythagorean theorem will not apply), you will still get the same answer! All hail the power of similar triangles! Here is such a pair of triangles. |dw:1449248409607:dw|

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