Mathematics
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OpenStudy (misssmartiez):
Find the area on this irregular figure. Does anyone know how to find it? Medal, Fan, etc...!
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OpenStudy (misssmartiez):
OpenStudy (igreen):
I have to go after this.
OpenStudy (misssmartiez):
Alright.
OpenStudy (igreen):
We can cut it into 4 triangles.
OpenStudy (triciaal):
one approach
split it into regular shapes. find the areas of the regular shapes then add all the areas together.
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OpenStudy (triciaal):
or 2 trapezoids
OpenStudy (igreen):
|dw:1455240165667:dw|
OpenStudy (misssmartiez):
That is the problem, I don't know how to find the formula, I know how to cut it up, but not add it all up.
OpenStudy (igreen):
|dw:1455240239946:dw|
OpenStudy (igreen):
The formula of the area of a triangle is:
\(\sf A = \dfrac{1}{2}bh\)
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OpenStudy (igreen):
That's what Google is for ;)
OpenStudy (misssmartiez):
I did that, but thanks
OpenStudy (misssmartiez):
XD thanks
OpenStudy (misssmartiez):
So I?...
OpenStudy (igreen):
Hm. 4 triangles don't cut it, because we can't find the height of the biggest triangle.
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OpenStudy (igreen):
doesn't*
jimthompson5910 (jim_thompson5910):
I would split it into 2 trapezoids
|dw:1455240333992:dw|
jimthompson5910 (jim_thompson5910):
then use the area of a trapezoid formula
\[\Large A = \frac{h(b_1+b_2)}{2}\]
OpenStudy (igreen):
Yeah, that's the only approach I see.
OpenStudy (misssmartiez):
Jim, can you help me still?
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jimthompson5910 (jim_thompson5910):
I guess another way is to do it like this
|dw:1455240527138:dw|
but that seems like more work than necessary