Composite area

Let $r$ be the radius of semicircle $n$ . Then

$A_{ABC}=\dfrac{1}{2}bh$

$32=\dfrac{1}{2}(2r)(2r)$

$64=4r^2$

$16=4r$

$r=4$

It follows that the diameter of semicircle $n$ is $3(4)=12$ and $CE=BD=12$ and $DE=BC=8$ .

So the area of semicircle $n$ is $\dfrac{1}{2}\pi (4^2)=8\pi$ . The area of semicircle $m$ is $\dfrac{1}{2}\pi (6^2)=18 \pi$ . The area of rectangle $BCED$ is $8(12)=96$ . The area of equilateral triangle $FBD$ is $\dfrac{\sqrt{3}}{4}(12^2)=36\sqrt{3}$ .

Finally, the area of the whole figure is

$32+96+36\sqrt{3}+8\pi + 18 \pi=$ $\boxed{128+36\sqrt{3}+26\pi}$

Formulas used:

$A=\pi r^2$ $\implies$ Area of a circle where $r$ is the radius. When looking for the area of a semicircle, we need to divide it by $2$ .

$A=\dfrac{\sqrt{3}}{4}x^2$ $\implies$ Area of an equilateral triangle where $x$ is the side length. It is a derived formula.

$A=lw$ $\implies$ Area of a rectangle where $l$ is the length and $w$ is the width. It can also be $base~\times~height$ or $length~\times~breadth$