How do you find the area of this triangle?

- Setsuna-Yuregeshi

How do you find the area of this triangle?

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- schrodinger

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- Setsuna-Yuregeshi

|dw:1438116778704:dw|I need to find the area of this. How do I do that?

- Setsuna-Yuregeshi

|dw:1438116845340:dw| I forgot the right angle

- phi

area is 1/2 base * height

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## More answers

- Setsuna-Yuregeshi

Really? That's it?

- phi

yes

- Setsuna-Yuregeshi

5x3/2?

- phi

yes

- Setsuna-Yuregeshi

But what about:

- Setsuna-Yuregeshi

|dw:1438116970196:dw| What about that part?
WHen they ask find the area which triangle am i finding it for exactly?

- phi

You found the area of the triangle.
we can prove it

- ChillOut

You're finding the are for the full lines one.

- Setsuna-Yuregeshi

I found the area of which triangle?

- ChillOut

area*

- ChillOut

|dw:1438117079817:dw|

- phi

|dw:1438117108451:dw|

- Setsuna-Yuregeshi

Ooh, so the |dw:1438117132170:dw| is giving me the height for the triangle?

- ChillOut

Absolutely.

- Setsuna-Yuregeshi

Ooh, Brainfart!

- Setsuna-Yuregeshi

Since I am taking geometry during summerschool I expected it to be some complicated formula. At first I ruled out that it could be that...

- ChillOut

There are other formulas involving angles and stuff. But leave that for later :)

- phi

here is a proof. We assume we know the area of a rectangle is width * height
|dw:1438117298679:dw|

- Setsuna-Yuregeshi

So if I answer using 1/2BxH on a graded assessment it will be correct?

- phi

now put in a diagonal. the two triangles are congruent, so the area of each is 1/2 of the rectangle's|dw:1438117369207:dw|

- Setsuna-Yuregeshi

So if I answer using 1/2BxH on a graded assessment it will be correct?

- Setsuna-Yuregeshi

They're not expecting a different method?

- phi

now make an obtuse triangle
|dw:1438117452871:dw|

- Setsuna-Yuregeshi

@phi Ooh, I see. That makes sense...

- Setsuna-Yuregeshi

ooh. C is the height of |dw:1438117583062:dw|

- phi

if we subtract of the small triangle on the left we are left with the obtuse triangle
that is
\[ \frac{1}{2} (a+b)\cdot c - \frac{1}{2} b c \\\frac{1}{2} ac + \frac{1}{2} bc - \frac{1}{2} bc \\
=\frac{1}{2} ac
\]

- phi

hopefully you can follow the logic

- Setsuna-Yuregeshi

I can, so you just proved 1/2BxH right?

- phi

yes, that the formula works for obtuse triangles.

- Setsuna-Yuregeshi

Ok thanks. Can I ask another quick question?

- Setsuna-Yuregeshi

How do you find the area of a kite?

- phi

1/2 the product of the diagonals
http://www.mathopenref.com/kitearea.html

- Setsuna-Yuregeshi

Thanks!!!!

- Setsuna-Yuregeshi

Would the diagonals be these? |dw:1438117923899:dw|

- Setsuna-Yuregeshi

@phi

- phi

the "crossbars"
|dw:1438118022805:dw|

- Setsuna-Yuregeshi

Ooh, ok sorry but: |dw:1438118151545:dw|
It would be 1+1 x 2 + 3/2 right? @phi

- Setsuna-Yuregeshi

OOps I mean 1+2 x 3+3/2

- phi

it would be except that would not be a kite (the diagonals of a kite bisect each other)
see
http://www.coolmath.com/reference/kites

- phi

correction: the longer diagonal bisects the shorter diagonal

- Setsuna-Yuregeshi

Ooh, so what i said was right? I have a problem asking for the area where they give me the length of half of both lines.

- Setsuna-Yuregeshi

Like they give me this...

- Setsuna-Yuregeshi

|dw:1438118486114:dw|

- Setsuna-Yuregeshi

I just add the 2's and 1's and use the 1/2bxh formula right?

- phi

yes

- Setsuna-Yuregeshi

Ooh ok. Thanks.

- phi

1/2 d1 * d2
(though you can think of the diagonals as the width and height)

- Setsuna-Yuregeshi

Ooh, thank you so much!!!!!!!!!!!!!!

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