The Areas of A Half-Circle and A Combined Shape

The semicircle with radius $r$ has an area equal to $\dfrac{\pi r^2}{2}$ .

The composite shape has an area equal to the sum of the areas of the semicircle and the rectangle which is $= \dfrac{\pi (r-1)^2}{2} + 2(r-1) \cdot 3 \pi = \dfrac{\pi (r-1)^2}{2} + 6 \pi (r-1)$ .

Equating both areas, we get

$\begin{aligned} & \dfrac{\pi r^2}{2} &=& \dfrac{\pi (r-1)^2}{2} + 6 \pi (r-1) \\ \implies & r^2 &=& (r-1)^2 + 12 (r-1) \\ \implies & r^2 &=& r^2 - 2r + 1 + 12r - 12 \\ \implies & 0 &=& 10r - 11 \\ \implies & r &=& \dfrac{11}{10} = \boxed{1.1} \end{aligned}$

solving for the area of the semi-circle:

$A = \dfrac{1}{2}\pi r^2$

solving for the area of the composite figure:

The base of the rectangle is $(r-1)+(r-1)=2r-2$ . Therefore, the area is $3\pi(2r-2)$

The area of the semi-circle above the rectangle is $\dfrac{1}{2}\pi(r-1)^2$ .

solving for "r":

Equating both areas (because both areas are equal), we have

$\dfrac{1}{2}\pi r^2=3\pi(2r-2)+\dfrac{1}{2}\pi(r-1)^2$

$\pi$ cancels out

$\dfrac{1}{2}r^2=3(2r-2)+\dfrac{1}{2}(r-1)^2$

$\dfrac{1}{2}r^2=3(2r-2)+\dfrac{1}{2}(r^2-2r+1)$

$\dfrac{1}{2}r^2=6r-6+\dfrac{1}{2}r^2-r+\dfrac{1}{2}$

$\dfrac{1}{2}r^2$ cancels out

$0=6r-6-r+\dfrac{1}{2}$

$5.5=5r$

Dividing both sides by $5$ , we have

$1.1=r$

or

$\boxed{r=1.1}$ $\boxed{\text{answer}}$

Since the areas are equal, we have

$\dfrac{\pi}{2}r^2=\dfrac{\pi}{2}(r-1)^2+3\pi(2)(r-1)$

$\dfrac{r^2}{2}=\dfrac{1}{2}(r^2-2r+1)+6r-6$

$\dfrac{r^2}{2}=\dfrac{r^2}{2}-r+\dfrac{1}{2}+6r-6$

$6r-r=6-\dfrac{1}{2}$

$5r=5.5$

$\boxed{r=1.1}$