Right triangles

How many square units are in the area of triangle APC?

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$AN(x)=-x+6\\ MC(x)=-\frac{3}{12}x+3\\ -x+6=-\frac{1}{4}x+3\\ \color{blue}x=4$

$A_{APC}=A_{ABC}-A_{BCM}-A_{AMP}\\ A_{APC}=\dfrac{12\cdot 6}{2}-\dfrac{12\cdot 3}{2}-\dfrac{3\cdot {\color{blue}4}}{2}\\ \color{blue}A_{APC}=12$

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The area of the total triangle is $6 \times 12 \div 2 = 36$

Draw Median $BO$. All 6 triangles have an area of 6

$\triangle APC$ consists of exactly 2 of these triangles, meaning the area is $2 \times 6 = \color{brown}\boxed{12}$