Answer
$\color{blue}{\dfrac{49}{3} \approx 16.33}$ square units
Work Step by Step
For each fixed $Y=y \in[6,10]$, the length of each leg of the isosceles right triangle is $a = y/\sqrt{2}$.
Consequently, the area of the triangle is ${\large\frac{1}{2}}a^2 = {\large\frac{1}{2}}(y/\sqrt{2})^2 = \dfrac{1}{4}y^2.$
Now,
\begin{align*}
E\left(\frac{1}{4}Y^2\right) &= \frac{1}{4}E(Y^2) & [\ \text{Corollary 3.5.1, p. 148}\ ] \\
&= \frac{1}{4}\left( \int_{\Omega_Y} y^2\cdot f_Y(y)\ dy \right) \\
&= \frac{1}{4}\left( \int_{y\in [6,10]} y^2\cdot \frac{1}{4}\ dy \right) & Y\sim \text{Uniform}[6,10] \implies \\
& & f_Y(y) = \frac{1}{10-6} = \frac{1}{4},\ y\in[6,10] \\
&= \frac{1}{4}\left( \int_6^{10} y^2\cdot \frac{1}{4}\ dy \right) \\
&= \frac{1}{4}\cdot \left( \frac{1}{4}\cdot \frac{y^3}{3}\ \right\vert_6^{10} \\
&= \frac{1}{(4)(4)(3)} \biggl( y^3\ \biggr\vert_6^{10} \\
&= \frac{1}{(4)(4)(3)} \left( 10^3-6^3 \right) \\
&= \frac{1}{(4)(4)(3)}(1,\!000-216) \\
&= \frac{1}{(4)(4)(3)}(784) \\
&= \frac{1}{(4)(4)(3)}(4)(4)(49) \\
\color{blue}{E\left(\frac{1}{4}Y\right)}\ &\color{blue}{= \frac{49}{3} \;\approx\; 16.33 }
\end{align*}
Thus, the expected value of such a triangle is $\color{blue}{\dfrac{49}{3} \approx 16.33}$ square units.
