In triangle , angle is a right angle. Point lies on , point lies on , and is parallel to . The length of is 25 units, the length of is 6 units, the length of is greater than the length of , and the area of triangle is 150 square units. Which choice gives the length of , in units?

Solution available with step-by-step explanation

Super-Hard SAT Math Question 127 | Free SAT Prep | Aniko

00:00

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

0 / 200

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. Аnіko АІ Т‍utοr

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

When a segment inside a triangle is parallel to a side, immediately look for similar triangles and write a ratio using corresponding sides (here, $\frac{BE}{BC}=\frac{DE}{AC}$ ). If the large triangle is right and you are given its area and hypotenuse, solve for the legs using $AC^2+BC^2=AB^2$ and $\frac{1}{2}(AC)(BC)=\text{area}$ ; computing $(AC+BC)^2$ is often the fastest path. Then apply the similarity scale factor to get the requested segment. Prepar‌еd‍ bу Ani⁠кο‍.аі

Hints

Click me!

Use the parallel line

Since $\overline{DE} \parallel \overline{AC}$ , identify two triangles that must be similar and write a ratio involving $BE$ .

Translate the area into an equation

Because $\angle C$ is a right angle, the area can be written as $\frac{1}{2}(AC)(BC)$ . Turn that into an equation for $AC\cdot BC$ . aniк‌o.аі

Combine area with the hypotenuse

You have $AC^2+BC^2=AB^2$ and $AC\cdot BC$ from the area. Try computing $(AC+BC)^2$ to find $AC+BC$ .

Desmos Guide

Graph the two relationships between the legs

Let $x=AC$ and $y=BC$ . Enter these equations:

$y=300/x$
$y=\sqrt{625-x^2}$

The intersection(s) give the positive leg lengths.

Read the legs and assign $AC$ and $BC$

Click the intersection in the first quadrant. You should see $(x,y)=(20,15)$ (or swapped). Use the condition $AC>BC$ to take $AC=20$ and $BC=15$ . а‌n‌і‌kο.ai SАТ Que‍st‍iοn⁠ Ва‌nк

Compute $BE$ with the similarity ratio

Enter the expression $15*(6/20)$ . The resulting value is the length of $BE$ .

Step-by-step Explanation

Find the legs of right triangle $ABC$

Let $AC=x$ and $BC=y$ . Since $\angle C$ is a right angle,

\begin{aligned} \text{(Pythagorean)}\quad x^2+y^2&=25^2=625 \\ \text{(Area)}\quad \frac{1}{2}xy&=150 \;\Rightarrow\; xy=300 \end{aligned}

Now compute $(x+y)^2$ :

(x+y)^2=x^2+y^2+2xy=625+600=1225

So $x+y=35$ (lengths are positive).

Use the sum and product to get $AC$ and $BC$

The numbers $x$ and $y$ satisfy both Aniko Queѕtіon Вan⁠k

$x+y=35$
$xy=300$

So $x$ and $y$ are the roots of $t^2-35t+300=0$ . The discriminant is

35^2-4(300)=1225-1200=25

t=\frac{35\pm 5}{2} \Rightarrow 20,\;15

Given $AC>BC$ , it follows that $AC=20$ and $BC=15$ .

Apply similarity to relate $BE$ to the whole triangle

Because $\overline{DE} \parallel \overline{AC}$ , triangles $BDE$ and $BAC$ are similar, with corresponding sides:

$DE \leftrightarrow AC$
$BE \leftrightarrow BC$

Thus,

\frac{BE}{BC}=\frac{DE}{AC}

Compute $BE$

Substitute $BC=15$ , $DE=6$ , and $AC=20$ :

BE=BC\cdot \frac{DE}{AC}=15\cdot\frac{6}{20}=15\cdot\frac{3}{10}=\frac{9}{2}

So the correct choice is $\frac{9}{2}$ .

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation