In the diagram, each of $\triangle Q P T, \triangle Q T S$ and $\triangle Q S R$ is an isosceles, right-angled triangle, with $\angle Q P T=\angle Q T S=\angle Q S R=90^{\circ}$. The combined area of the three triangles is 56. If $Q P=P T=k$, what is the value of $k$ ?
(A) $\sqrt{2}$
(B) 1
(C) 4
(D) 2
(E) $2 \sqrt{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-028.jpg?height=301&width=548&top_left_y=1216&top_left_x=1122)
By the Pythagorean Theorem,

$$
Q T^{2}=Q P^{2}+P T^{2}=k^{2}+k^{2}=2 k^{2}
$$

Since $Q T>0$, then $Q T=\sqrt{2} k$.

Since $\triangle Q T S$ is isosceles, then $T S=Q T=\sqrt{2} k$.

By the Pythagorean Theorem,

$$
Q S^{2}=Q T^{2}+T S^{2}=(\sqrt{2} k)^{2}+(\sqrt{2} k)^{2}=2 k^{2}+2 k^{2}=4 k^{2}
$$

Since $Q S>0$, then $Q S=2 k$.

Since $\triangle Q S R$ is isosceles, then $S R=Q S=2 k$.

Since $\triangle Q P T$ is right-angled at $P$, its area is $\frac{1}{2}(Q P)(P T)=\frac{1}{2} k^{2}$.

Since $\triangle Q T S$ is right-angled at $T$, its area is $\frac{1}{2}(Q T)(T S)=\frac{1}{2}(\sqrt{2} k)(\sqrt{2} k)=\frac{1}{2}\left(2 k^{2}\right)=k^{2}$.

Since $\triangle Q S R$ is right-angled at $S$, its area is $\frac{1}{2}(Q S)(S R)=\frac{1}{2}(2 k)(2 k)=2 k^{2}$.

Since the sum of the three areas is 56 , then $\frac{1}{2} k^{2}+k^{2}+2 k^{2}=56$ or $\frac{7}{2} k^{2}=56$ which gives $k^{2}=16$. Since $k>0$, then $k=4$.

ANSWER: (C)