In right triangle , is a right angle. The altitude from to the hypotenuse has length . If , what is the length of side ?

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SAT Right Triangles & Trigonometry Question 45 | Free SAT Prep | Aniko

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Question 45·Hard·Right Triangles and Trigonometry

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In right triangle $DEF$ , $\angle F$ is a right angle. The altitude from $\angle F$ to the hypotenuse $\overline{DE}$ has length $12$ . If $\sin(D)=\dfrac{4}{5}$ , what is the length of side $\overline{DE}$ ?

Your answer

(Express the answer as an integer)

For right-triangle problems involving a trigonometric ratio and an altitude to the hypotenuse, first interpret the trig ratio (like $\sin$ ) as a ratio of specific sides and, when possible, match it to a known Pythagorean triple (such as 3-4-5) with a scale factor. Next, use geometric relationships—most efficiently, the area identity $\tfrac{1}{2}ab = \tfrac{1}{2}ch$ so the altitude equals $(ab)/c$ —to connect the altitude to the legs and hypotenuse. Solve the resulting simple equation for the scale factor or hypotenuse, and only then plug back to get the specific side length, being careful not to confuse altitudes with legs.

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Relate sine to triangle sides

For angle $D$ in right triangle $DEF$ , which side is opposite angle $D$ , and which side is the hypotenuse? Use $\sin(D) = \dfrac{\text{opposite}}{\text{hypotenuse}}$ to set up a ratio.

Look for a Pythagorean triple

If one leg and the hypotenuse are in the ratio $4:5$ , what is the simplest integer value the other leg can have so that the Pythagorean theorem still holds? Think of the common 3-4-5 right triangle.

Use area to connect the altitude and the sides

The altitude from the right angle to the hypotenuse can be related to the legs and hypotenuse using the area formula. Write the area once using the two legs as base and height, and once using the hypotenuse and the altitude as base and height, then set them equal.

Solve for the hypotenuse

After you have an expression for the altitude in terms of the scale factor (or hypotenuse), set it equal to $12$ and solve. Then use that scale factor to find the hypotenuse length.

Desmos Guide

Set up an equation for the altitude

From the geometry steps, you should have that the legs are $3x$ and $4x$ and the hypotenuse is $5x$ , and that the altitude from the right angle to the hypotenuse is $h = \dfrac{(3x)(4x)}{5x} = \dfrac{12}{5}x$ . Since the altitude is given as $12$ , the equation to solve is $\dfrac{12}{5}x = 12$ .

Graph both sides in Desmos

In Desmos, enter two expressions on separate lines:

y = (12/5)x
y = 12 Then find the point where the two graphs intersect. The $x$ -coordinate of this intersection is the value of $x$ .

Use the scale factor to find the hypotenuse

On a new line, type 5 * x and substitute the $x$ -value you found from the intersection. The resulting output is the length of $DE$ , the hypotenuse.

Step-by-step Explanation

Use the sine information to relate the sides

In right triangle $DEF$ with right angle at $F$ , side $\overline{DE}$ is the hypotenuse and side $\overline{EF}$ is opposite angle $D$ .

By definition of sine:

\sin(D) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{EF}{DE} = \frac{4}{5}.

So $EF : DE = 4 : 5$ .

Recognize or build the 3-4-5 triangle pattern

In a right triangle, if one leg and the hypotenuse are in the ratio $4:5$ , the other leg must be in ratio $3$ to make a 3-4-5 Pythagorean triple.

Let a common scale factor be $x$ :

$DF = 3x$
$EF = 4x$
$DE = 5x$

You can confirm this with the Pythagorean theorem:

(3x)^2 + (4x)^2 = 9x^2 + 16x^2 = 25x^2 = (5x)^2.

Write a formula for the altitude to the hypotenuse

Let $h$ be the altitude from $F$ to hypotenuse $DE$ . The area of the triangle can be written two ways:

Using legs $DF$ and $EF$ as base and height:

\text{Area} = \frac{1}{2} (DF)(EF) = \frac{1}{2} (3x)(4x).

Using hypotenuse $DE$ as the base and altitude $h$ as the height:

\text{Area} = \frac{1}{2} (DE)(h) = \frac{1}{2} (5x)(h).

Set these equal:

\frac{1}{2} (3x)(4x) = \frac{1}{2} (5x)(h).

The $\tfrac{1}{2}$ cancels, and so does one factor of $x$ :

12x^2 = 5xh \implies 12x = 5h.

Thus the altitude is

h = \frac{12}{5}x.

Use the given altitude to solve for the scale factor

You are told the altitude from $F$ to $DE$ is $12$ , so set $h = 12$ :

\frac{12}{5}x = 12.

Solve for $x$ :

\begin{aligned} \frac{12}{5}x &= 12 \\ 12x &= 60 \\ x &= 5. \end{aligned}

Find the hypotenuse length

Now use $x = 5$ to find $DE$ :

DE = 5x = 5 \cdot 5 = 25.

So the length of side $\overline{DE}$ is $\boxed{25}$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation