In the diagram, $\triangle P Q R$ is an isosceles triangle with $P Q=P R$. Semi-circles with diameters $P Q, Q R$ and $P R$ are drawn. The sum of the areas of these three semicircles is equal to 5 times the area of the semi-circle with diameter $Q R$. The value of $\cos (\angle P Q R)$ is
(A) $\frac{1}{3}$
(B) $\frac{1}{\sqrt{8}}$
(C) $\frac{1}{\sqrt{12}}$
(D) $\frac{1}{\sqrt{15}}$
(E) $\frac{1}{\sqrt{10}}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-016.jpg?height=345&width=323&top_left_y=234&top_left_x=1302)
Suppose that $P Q=P R=2 x$ and $Q R=2 y$.

The semi-circles with diameters $P Q$ and $P R$ thus have radii $x$ and the radius of the semi-circle with diameter $Q R$ is $y$.

The area of each semi-circle with radius $x$ is $\frac{1}{2} \pi x^{2}$ and the area of the semi-circle with radius $y$ is $\frac{1}{2} \pi y^{2}$.

Since the sum of the areas of the three semi-circles equals 5 times the area of the semi-circle with diameter $Q R$, then

$$
\frac{1}{2} \pi x^{2}+\frac{1}{2} \pi x^{2}+\frac{1}{2} \pi y^{2}=5 \cdot \frac{1}{2} \pi y^{2}
$$

which gives $x^{2}+x^{2}+y^{2}=5 y^{2}$ and so $2 x^{2}=4 y^{2}$ which gives $x^{2}=2 y^{2}$ and so $x=\sqrt{2} y$. Suppose that $M$ is the midpoint of $Q R$ and that $P$ is joined to $M$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-031.jpg?height=344&width=323&top_left_y=744&top_left_x=955)

Since $\triangle P Q R$ is isosceles with $P Q=P R$, then $P M$ is perpendicular to $Q R$. In other words, $\triangle P M Q$ is right-angled at $M$.

Therefore, $\cos (\angle P Q R)=\cos (\angle P Q M)=\frac{Q M}{P Q}=\frac{\frac{1}{2} Q R}{P Q}=\frac{y}{2 x}=\frac{y}{2 \sqrt{2} y}=\frac{1}{2 \sqrt{2}}=\frac{1}{\sqrt{4 \cdot 2}}=\frac{1}{\sqrt{8}}$.

ANsWer: (B)