Question 39·Hard·Right Triangles and Trigonometry

You know $\cos \angle E = \dfrac{3}{7}$ and the hypotenuse $EF = 14\sqrt{2}$. Set up a proportion with $\dfrac{ED}{EF}$ to solve for $ED$.

Once you have $ED$, use $ED^2 + DF^2 = EF^2$ to find $DF^2$. Then take a square root and simplify the radical.

After you know both legs, remember that the area of a right triangle is $\tfrac{1}{2}$ times the product of the two legs.

In Desmos, type (3/7)*14*sqrt(2) and note the result; this is the length of $ED$.

In a new line, enter (14*sqrt(2))^2 - ( (3/7)*14*sqrt(2) )^2 to get $DF^2$. Then on the next line, type sqrt( <previous_answer> ) (or retype the expression inside sqrt()), and note the value of $DF$.

Finally, enter 0.5 * ( (3/7)*14*sqrt(2) ) * sqrt( (14*sqrt(2))^2 - ( (3/7)*14*sqrt(2) )^2 ) and read the output; this numeric result is the triangle’s area and should match one of the answer choices.

Cosine of an angle in a right triangle is

Here, $\cos \angle E = \dfrac{3}{7}$ and the hypotenuse $EF = 14\sqrt{2}$.

So the side adjacent to $\angle E$ is $ED$, and

Compute $ED$ by simplifying $\dfrac{3}{7} \cdot 14\sqrt{2}$.

Simplify

First, $\dfrac{14}{7} = 2$, so

So one leg of the triangle is $ED = 6\sqrt{2}$.

Let the other leg be $DF$. In right triangle $DEF$,

Substitute $ED = 6\sqrt{2}$ and $EF = 14\sqrt{2}$:

Compute each square:

• $(6\sqrt{2})^2 = 36 \cdot 2 = 72$
• $(14\sqrt{2})^2 = 196 \cdot 2 = 392$

So

Now take the square root to find $DF$.

We have $DF^2 = 320$, so

In a right triangle, the area is

Here, the legs are $ED = 6\sqrt{2}$ and $DF = 8\sqrt{5}$, so

So the area of $\triangle DEF$ is $24\sqrt{10}$, which matches choice B.