Question 556·Hard·Math

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In right triangle $DEF$ , $\angle D$ is a right angle and the length of the hypotenuse $EF$ is $14\sqrt{2}$ . If $\cos\angle E = \dfrac{3}{7}$ , what is the area of $\triangle DEF$ ?

A cosine ratio in a right triangle connects the leg beside the named angle to the hypotenuse. Find that leg, then use the Pythagorean theorem for the other leg and the triangle area formula for the result. Desmos can solve the ratio equation and handle the arithmetic, but multiplying the legs without halving gives twice the area.

Hints

Hint 1
Start at $\angle E$ . The adjacent leg touches that angle but is not the hypotenuse. Which side is adjacent to $E$ , and how can you use cosine to relate it to $EF$ ?
Hint 2
Once you know one leg, use the Pythagorean theorem: the squares of the two legs add to the square of the hypotenuse. Subtract the known leg’s square to find the other leg’s square.
Hint 3
The legs meet at the right angle, so they serve as the triangle’s base and perpendicular height. What must you do to their product to get the area of a triangle?

Step-by-step

Find both legs, then the area
Step 1
Turn cosine into a side equation
The hypotenuse is across from the right angle at $D$ , so it’s $EF$ . From $\angle E$ , the adjacent leg is $DE$ . Let $DE=x$ . Cosine means adjacent leg divided by hypotenuse, so:
$\frac{x}{14\sqrt{2}}=\frac{3}{7}.$
Step 2
Solve for the adjacent leg
Type $\frac{x_1}{14\sqrt{2}}\sim\frac37$ in Desmos. The $x_1$ stands for the unknown leg, and $\sim$ tells Desmos to find its value. Under PARAMETERS, Desmos shows $x_1\approx8.485$ , so $DE$ is about $8.485$ .
Step 3
Keep the leg length exact
The choices use radicals, so find the exact length from the ratio. Multiply both sides by $14\sqrt2$ :
$DE=\frac37(14\sqrt2).$
Cancel $14/7$ to get $2$ :
$DE=6\sqrt2.$
Step 4
Set up the other leg
The Pythagorean theorem says the squares of the two legs add to the hypotenuse’s square. Since $D$ is the right angle, $DF$ is the remaining leg, so:
$(6\sqrt2)^2+DF^2=(14\sqrt2)^2.$
Step 5
Find the other leg in Desmos
Subtract the known leg’s square:
$DF^2=(14\sqrt2)^2-(6\sqrt2)^2.$
Take the positive square root, since $DF$ is a length:
$DF=\sqrt{(14\sqrt2)^2-(6\sqrt2)^2}.$
Type that square-root expression in Desmos. It shows about $17.889$ , the length of $DF$ .
Step 6
Keep the other leg exact
For an exact value, square the terms inside:
$DF=\sqrt{392-72}.$
Subtract:
$DF=\sqrt{320}.$
Write $320$ with a perfect-square factor:
$DF=\sqrt{64\cdot5}.$
Take the square root of $64$ :
$DF=8\sqrt5.$
Step 7
Use the legs for area
Use the two legs, not the hypotenuse, as a right triangle’s base and perpendicular height. They meet at $D$ at a right angle. The triangle area formula is half the base times the height, so:
$A=\frac12(DE)(DF)=\frac12(6\sqrt2)(8\sqrt5).$
Step 8
Calculate the area
Type $\frac12(6\sqrt2)(8\sqrt5)$ in Desmos; it shows about $75.895$ for the area. Keep the answer exact by multiplying the radicals:
$A=\frac12(6)(8)\sqrt{2\cdot5}.$
Combine the number factors:
$A=24\sqrt{10}.$
The area of $\triangle DEF$ is $24\sqrt{10}$ . Choice B.

Question 556·Hard·Math

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