MEDAL AND FAN!!!!! PLEASE HELP!!!!! Referring to the figure, find the area of the quadrilateral shown.

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MEDAL AND FAN!!!!! PLEASE HELP!!!!! Referring to the figure, find the area of the quadrilateral shown.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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|dw:1436931454185:dw|
Would the answer be 12 in. ^2
Yes! You are correct! Actually, you can see that one corner fits perfectly into the other to make a rectangle. |dw:1436932124281:dw| The area is then given by 6in x 2in = 12 in ^2.

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|dw:1436931904321:dw|
Do you know how I can solve this?
Well, if you stare at the picture for a bit, you'll realize that you actually have two triangles. |dw:1436932291925:dw|
Do you know how to find the area of a triangle?
\[base \times height \div 2\]
@jtvatsim Would it be 7 cm x 4cm?
You are close. There are actually a lot of triangles that you could find the area of, after thinking about these these two are easier to work with... |dw:1436932671767:dw| The big triangle on the right has base 4cm and height 7cm. The big triangle on the left has base 4 cm and height 3 cm. Add these two areas together to get: 1/2(4cm)(7cm) + 1/2(4cm)(3cm) = 14cm^2 + 6cm^2 = 20cm^2.
Does that make sense? Any questions? :)
4x3=12 12/2=6
7x4=28 28/2=14 So..... the answer is 20 cm^2
That is correct!

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