Question 127·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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In triangle $ABC$ , angle $C$ is a right angle. Point $D$ lies on $\overline{AB}$ , point $E$ lies on $\overline{BC}$ , and $\overline{DE}$ is parallel to $\overline{AC}$ . The length of $\overline{AB}$ is 25 units, the length of $\overline{DE}$ is 6 units, the length of $\overline{AC}$ is greater than the length of $\overline{BC}$ , and the area of triangle $ABC$ is 150 square units.

Which choice gives the length of $\overline{BE}$ , in units?

A right triangle’s area and hypotenuse can determine its two legs. Write the area and Pythagorean equations, then graph them in Desmos; a condition about which leg is longer picks the right intersection. A parallel segment creates similar triangles, so match sides before scaling. The tempting slip is pairing that segment with the hypotenuse instead of its corresponding leg.

Hints

Hint 1
The area of a right triangle is half the product of the two sides that meet at the right angle. Call $AC$ and $BC$ the legs. What equation does the given area make?
Hint 2
The hypotenuse is the side opposite the right angle. Use its length in the Pythagorean theorem, then compare the two positive Desmos intersections. Which one makes $AC$ longer than $BC$ ?
Hint 3
A segment parallel to one side of a triangle makes a smaller, similar triangle. Match $DE$ with $AC$ first. Which side of the larger triangle then matches the requested $BE$ ?

Step-by-step

Graph the legs, then use similarity
Step 1
Write the area equation
Let $x=AC$ and $y=BC$ . These are the legs, the sides that meet at the right angle $C$ . A triangle’s area is half its base times its perpendicular height, so the given area gives
$\frac12xy=150.$
Step 2
Write the right-triangle equation
$AB=25$ is opposite the right angle, so it’s the hypotenuse. The Pythagorean theorem says the squares of the two legs add to the hypotenuse’s square:
$x^2+y^2=25^2.$
Step 3
Choose the intersection that fits the lengths
Type both equations in Desmos and click their intersections with positive coordinates: $(20,15)$ and $(15,20)$ . The problem says $AC>BC$ , or $x>y$ , so choose $(20,15)$ . This means $AC=20$ and $BC=15$ ; the other point reverses the legs.
Step 4
Match the sides of the similar triangles
Triangles $BDE$ and $BAC$ share angle $B$ . Because $DE$ is parallel to $AC$ , their angles at $D$ and $A$ are equal too. By angle-angle similarity, two matching angles make the triangles the same shape. So match $DE$ with $AC$ and $BE$ with $BC$ .
Step 5
Set up the side proportion
In similar triangles, matching sides have the same small-to-large ratio. Use $DE=6$ with $AC=20$ , then $BE$ with $BC=15$ : $\frac{BE}{15}=\frac6{20}$ . Don’t pair $DE$ with $AB=25$ : $AB$ is the hypotenuse, not the side matching $DE$ .
Step 6
Find the requested length
Type $x_1/15\sim6/20$ , with $x_1$ standing for $BE$ . The $\sim$ tells Desmos to find the unknown; under PARAMETERS, it shows $x_1=4.5$ . Type $x_1$ on a new line and use its fraction button to see $\frac92$ . So $BE$ is $\frac92$ units. Choice C.

Question 127·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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