Mathematics
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OpenStudy (anonymous):
What is the area of ABC?
(Picture below.)
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OpenStudy (anonymous):
OpenStudy (anonymous):
First find the area of the triangle on the left, what do you get?
OpenStudy (anonymous):
a=1/2bh or Pythagorean theorem?
OpenStudy (anonymous):
a=i/2bh
OpenStudy (anonymous):
sorry, 1/2bh*
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OpenStudy (anonymous):
a=1/2(7)(12)
a=42
OpenStudy (anonymous):
Good, and add the area of the other triangle to that also
OpenStudy (anonymous):
a=1/2(9)(7)
a=63/2
OpenStudy (mathstudent55):
@zaynahf Be careful, 12 and 7 are not base and height
OpenStudy (anonymous):
True.. then use pythagorean theorem
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OpenStudy (mathstudent55):
The height is an altitude, which has to be perpendicular to the base.
OpenStudy (anonymous):
In that case I have to use Pythagorean theorem right?
OpenStudy (anonymous):
oh ok
OpenStudy (anonymous):
a^2+b^2=c^2
OpenStudy (anonymous):
what numbers do i plug in?
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OpenStudy (anonymous):
@mathstudent55 @zaynahf
OpenStudy (anonymous):
9^2+12^2=225
OpenStudy (anonymous):
is that right? @mathstudent55
OpenStudy (anonymous):
\[\text{Area} =\frac{1}{2}\left(\left(\sqrt{9^2-7^2}\right)+\left(\sqrt{12^2-7^2}\right)\right)*7=\frac{7}{2} \left(4 \sqrt{2}+\sqrt{95}\right)=53.9128 \]
OpenStudy (anonymous):
Thank you so much!
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OpenStudy (anonymous):
You're welcome.