In the right-angled triangle $P Q R, P Q=Q R$. The segments $Q S, T U$ and $V W$ are perpendicular to $P R$, and the segments $S T$ and $U V$ are perpendicular to $Q R$, as shown. What fraction of $\triangle P Q R$ is shaded?
(A) $\frac{3}{16}$
(B) $\frac{3}{8}$
(D) $\frac{5}{32}$
(E) $\frac{7}{32}$
(C) $\frac{5}{16}$

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-046.jpg?height=377&width=358&top_left_y=1012&top_left_x=1320)
Since $\triangle P Q R$ is isosceles with $P Q=Q R$ and $\angle P Q R=90^{\circ}$, then $\angle Q P R=\angle Q R S=45^{\circ}$. Also in $\triangle P Q R$, altitude $Q S$ bisects $P R(P S=S R)$ forming two identical triangles, $S Q P$ and $S Q R$.

Since these two triangles are identical, each has $\frac{1}{2}$ of the area of $\triangle P Q R$.

In $\triangle S Q R, \angle Q S R=90^{\circ}, \angle Q R S=45^{\circ}$, and so $\angle S Q R=45^{\circ}$.

Thus, $\triangle S Q R$ is also isosceles with $S Q=S R$.

Then similarly, altitude $S T$ bisects $Q R(Q T=T R)$ forming two identical triangles, $S Q T$ and $S R T$.

Since these two triangles are identical, each has $\frac{1}{2}$ of the area of $\triangle S Q R$ or $\frac{1}{4}$ of the area of $\triangle P Q R$.

Continuing in this way, altitude $T U$ divides $\triangle S T R$ into two identical triangles, $S T U$ and $R T U$. Each of these two triangles has $\frac{1}{2}$ of $\frac{1}{4}$ or $\frac{1}{8}$ of the area of $\triangle P Q R$.

Continuing, altitude $U V$ divides $\triangle R T U$ into two identical triangles, $R U V$ and $T U V$.

Each of these two triangles has $\frac{1}{2}$ of $\frac{1}{8}$ or $\frac{1}{16}$ of the area of $\triangle P Q R$.

Finally, altitude $V W$ divides $\triangle R U V$ into two identical triangles, $U V W$ and $R V W$.

Each of these two triangles has $\frac{1}{2}$ of $\frac{1}{16}$ or $\frac{1}{32}$ of the area of $\triangle P Q R$.

Since the area of $\triangle S T U$ is $\frac{1}{8}$ of the area of $\triangle P Q R$, and the area of $\triangle U V W$ is $\frac{1}{32}$ of the area of $\triangle P Q R$, then the total fraction of $\triangle P Q R$ that is shaded is $\frac{1}{8}+\frac{1}{32}=\frac{4+1}{32}$ or $\frac{5}{32}$.