Let $t$ be TNYWR.

The area of the triangle formed by the line $\sqrt{k} x+4 y=10$, the $x$-axis and the $y$-axis is $t$. What is the value of $k$ ?
Let $a$ be the $x$-intercept of the line with equation $\sqrt{k} x+4 y=10$ and let $b$ be the $y$-intercept of this line.

![](https://cdn.mathpix.com/cropped/2024_04_30_dfe99ade8da3ee5d11fdg-17.jpg?height=448&width=531&top_left_y=2183&top_left_x=903)

Then the area of the triangle formed by the line and the axes is $\frac{a b}{2}$.

Since $a$ is the $x$-intercept, then $\sqrt{k} a+4(0)=10$ or $a=\frac{10}{\sqrt{k}}$.

Since $b$ is the $y$-intercept, then $\sqrt{k} 0+4(b)=10$ or $b=\frac{10}{4}$.

Since we are given that the area is $t$, then $t=\frac{a b}{2}=\frac{10 \cdot 10}{2 \cdot \sqrt{k} \cdot 4}$.

Thus, $\sqrt{k}=\frac{100}{8 t}=\frac{25}{2 t}$.

Since the answer to (a) is 20 , then $t=20$ and so $\sqrt{k}=\frac{25}{2(20)}=\frac{5}{8}$ and so $k=\frac{25}{64}$.