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Mathematics 9 Online
OpenStudy (anonymous):

What is the area of ABC? (Picture below.)

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*

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

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?

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!

OpenStudy (anonymous):

You're welcome.

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