NMTC 2014(Junior level)

Let non-shaded area of the semicircle be $C$ . Therefore, the area of the semicircle $- C = B$ and if the radius of the semicircle is $r = 21$ .

$\quad \Rightarrow \frac {1}{2}\pi r^2 - C = B\quad ...(1)$

We also note that, the area of $\triangle PQR - C = A$ and if $QR = h$ , then:

$\quad \Rightarrow \frac{1}{2}(2r)h - C = A \quad \Rightarrow rh - C = A\quad ...(2)$

Equation 2 - Equation 1:

$\quad \Rightarrow rh - \frac {1}{2} \pi r^2 = A-B=357\quad \Rightarrow h = \dfrac {357} {r} + \dfrac {\pi r}{2}$ $\quad \Rightarrow h = \dfrac {357} {21} + \dfrac {22}{7} \times \dfrac {21}{2} = 17 + 33 = \boxed{50}$

Let $T$ be the area of $\triangle RQP$ and $C$ be the area of the unshaded region of the semicircle.

$T=\dfrac{1}{2}(RQ)(42)=21RQ$ $\color{#D61F06}(1)$

$A=T-C$ $\color{#D61F06}(2)$

$C=\dfrac{1}{2}\pi(21^2)-B=\dfrac{1}{2}\left(\dfrac{22}{7}\right)(441)-B=693-B$ $\color{#D61F06}(3)$

Substitute $\color{#D61F06}(1)$ and $\color{#D61F06}(3)$ in $\color{#D61F06}(2)$ .

$A=21RQ-(693-B)=21RQ-693+B$

$A-B+693=21RQ$

$357+693=21RQ$

$1050=21RQ$

$\boxed{50=RQ}$

The area of triangle $PQR$ is $A$ plus the semicircle minus $B$ . Since $A-B=357$ , then the area of $PQR$ is the area of the semicircle plus $357$ . The area of the semicircle is $\frac{21^2\pi}{2}=\frac{441}{2}\pi$

Let $QR=x$ . Then the area of $PQR$ is also equal to $\frac{42x}{2}=21x$ . We can set the two areas equal to get the equation $\frac{441}{2}\pi+357=21x$ Dividing both sides by 21 gives us $\frac{21}{2}\pi+17=x$ Since $\pi=\frac{22}{7}$ , then $x=\frac{21}{2}\times\frac{22}{7}+7=3\times11+17=\boxed{50}$