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Mathematics 6 Online
OpenStudy (misssmartiez):

Find the area on this irregular figure. Does anyone know how to find it? Medal, Fan, etc...!

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.

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: \(\sf A = \dfrac{1}{2}bh\)

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.

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?

jimthompson5910 (jim_thompson5910):

I guess another way is to do it like this |dw:1455240527138:dw| but that seems like more work than necessary

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