The altitude from the right angle to the hypotenuse is related to the two segments it creates on the hypotenuse. Try writing an equation that uses $CD = 8$ and the product of $AD$ and $DB$, remembering that $AD + DB = 20$.

Once you know the lengths of $AD$ and $DB$, use the fact that each leg is connected to the hypotenuse and one of these segments. Each leg squared equals the product of the hypotenuse and the adjacent segment. Identify which leg (using the smaller segment) must be the shorter one.

When you find the square of the shorter leg, you will get a number like $80$. Factor out the largest perfect square (such as $16$) to simplify $\sqrt{80}$ correctly.

Let the legs be $x$ and $y$. The area from the legs is $\tfrac{1}{2}xy$ and from the base and altitude is $\tfrac{1}{2} \cdot 20 \cdot 8$, so $xy = 160$. By the Pythagorean theorem, $x^2 + y^2 = 400$.

In Desmos, enter y = 160/x on one line and y = sqrt(400 - x^2) on another line. (These are the graphs of $xy = 160$ and the positive branch of $x^2 + y^2 = 400$.)

Locate and tap the intersection point of the two graphs in the first quadrant. The coordinates $(x, y)$ are the two positive leg lengths. Use the smaller of these two values as the shorter leg and match it to the correct answer choice.

• Draw right triangle $\triangle ABC$ with $\angle C$ a right angle and hypotenuse $\overline{AB}$.
• Let $D$ be the foot of the altitude from $C$ to $\overline{AB}$, so $CD$ is perpendicular to $AB$ and $CD = 8$.
• The hypotenuse $AB$ has length $20$, so the two segments on the hypotenuse are $AD$ and $DB$ with $AD + DB = 20$.

In a right triangle, the altitude from the right angle to the hypotenuse is the geometric mean of the two segments of the hypotenuse:

Here $CD = 8$, so

Let $AD = x$, so $DB = 20 - x$. Substitute into the equation:

Rewriting,

Factor:

so $x = 4$ or $x = 16$. Thus, the two hypotenuse segments are $4$ and $16$ (order along $AB$ does not matter).

In a right triangle, each leg is the geometric mean of the whole hypotenuse and the adjacent hypotenuse segment:

• For the leg adjacent to the segment of length $4$:

• For the leg adjacent to the segment of length $16$:

The leg whose square is $80$ is shorter than the leg whose square is $320$, so the shorter leg is the one with length $\sqrt{80}$.

Simplify $\sqrt{80}$ by factoring out a perfect square:

So the shorter leg of $\triangle ABC$ is $4\sqrt{5}$ centimeters.

Correct answer: D) $4\sqrt{5}$.