SAT Question 694b7fce: Lines, Angles, and Triangles answer and explanation with Desmos solution

It's given in the figure that angle $ACB$ and angle $AED$ are right angles. It follows that angle $ACB$ is congruent to angle $AED$. It's also given that angle $BAC$ and angle $DAE$ are the same angle. It follows that angle $BAC$ is congruent to angle $DAE$. Since triangles $ABC$ and $ADE$ have two pairs of congruent angles, the triangles are similar. Sides $AB$ and $AC$ in triangle $ABC$ correspond to sides $AD$ and $AE$, respectively, in triangle $ADE$. Corresponding sides in similar triangles are proportional. Therefore, $\frac{AD}{AB}=\frac{AE}{AC}$. It's given that $AC=3$ units and $CE=21$ units. Therefore, $AE=24$ units. It's also given that $AB=\sqrt{34}$ units. Substituting $3$ for $AC$, $24$ for $AE$, and $\sqrt{34}$ for $AB$ in the equation $\frac{AD}{AB}=\frac{AE}{AC}$ yields $\frac{AD}{\sqrt{34}}=\frac{24}{3}$, or $\frac{AD}{\sqrt{34}}=8$. Multiplying each side of this equation by $\sqrt{34}$ yields $AD=8\sqrt{34}$. By the Pythagorean theorem, if a right triangle has a hypotenuse with length $c$ and legs with lengths $a$ and $b$, then $a^2+b^2=c^2$. Since triangle $ADE$ is a right triangle, it follows that $AD$ represents the length of the hypotenuse, $c$, and $DE$ and $AE$ represent the lengths of the legs, $a$ and $b$. Substituting $24$ for $b$ and $8\sqrt{34}$ for $c$ in the equation $a^2+b^2=c^2$ yields $a^2+{\left(24\right)}^2={\left(8\sqrt{34}\right)}^2$, which is equivalent to $a^2+576=64\left(34\right)$, or $a^2+576=2,176$. Subtracting $576$ from both sides of this equation yields $a^2=1,600$. Taking the square root of both sides of this equation yields $a=\pm 40$. Since $a$ represents a length, which must be positive, the value of $a$ is $40$. Therefore, $DE=40$. Since $DE$ and $AE$ represent the lengths of the legs of triangle $ADE$, it follows that $DE$ and $AE$ can be used to calculate the area, in square units, of the triangle as $\frac{1}{2}\left(40\right)\left(24\right)$, or $480$. Therefore, the area, in square units, of triangle $ADE$ is $480$.