In right triangle , angle is the right angle and the length of the hypotenuse is . The altitude from to meets at point such that . What is the length of the altitude ?

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SAT Lines, Angles & Triangles Question 30 | Free SAT Prep | Aniko

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Question 30·Hard·Lines, Angles, and Triangles

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In right triangle $ABC$ , angle $C$ is the right angle and the length of the hypotenuse $AB$ is $40$ . The altitude from $C$ to $AB$ meets $AB$ at point $D$ such that $AD:DB = 3:2$ . What is the length of the altitude $CD$ ?

When you see a right triangle with an altitude drawn from the right angle to the hypotenuse, immediately think of the special relationships it creates: the altitude is the geometric mean of the two segments of the hypotenuse ( $CD^2 = AD \cdot DB$ ). First, translate any given ratio into actual segment lengths using the total hypotenuse, then plug those into the altitude formula and simplify the square root to match an answer choice. If you forget the formula, you can quickly re-derive it by noting that the two smaller right triangles are similar to the original and setting up proportions.

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Turn the ratio into real lengths

If $AD:DB = 3:2$ and $AD + DB = 40$ , how can you represent $AD$ and $DB$ using a common variable and then solve for that variable?

Use a special property of the altitude to the hypotenuse

When you draw an altitude from the right angle of a right triangle to the hypotenuse, it splits the triangle into two smaller right triangles similar to the original. What equation relates $CD$ , $AD$ , and $DB$ in this situation?

Remember to take a square root and simplify

Once you have an equation involving $CD^2$ , take the square root to find $CD$ , and then factor the number under the root to pull out any perfect-square factors so it matches one of the answer choices.

Desmos Guide

Find the segment lengths from the ratio

In one expression line, type 40/5 to find the common factor $k$ (because $3k + 2k = 40$ ). In the next two lines, type AD = 3*(40/5) and DB = 2*(40/5) so Desmos will display the numerical values of $AD$ and $DB$ .

Use Desmos to compute the altitude length

In a new line, type CD = sqrt(AD * DB). Desmos will show a decimal value for $CD$ ; compare this value to the decimal forms of the answer choices (you can type each choice, like 4*sqrt(6), 8*sqrt(6), etc.) and see which one matches the value of CD.

Step-by-step Explanation

Translate the ratio into actual segment lengths

The ratio $AD:DB = 3:2$ means we can write $AD = 3k$ and $DB = 2k$ for some positive number $k$ .

Because $AB$ is the whole hypotenuse,

AD + DB = AB \Rightarrow 3k + 2k = 40.

So $5k = 40$ , which gives $k = 8$ . Then

$AD = 3k = 3 \cdot 8 = 24$
$DB = 2k = 2 \cdot 8 = 16$ .

Use the altitude-to-hypotenuse property

In a right triangle, the altitude from the right angle to the hypotenuse has a special relationship with the two segments of the hypotenuse it creates.

If $CD$ is the altitude and it splits $AB$ into segments $AD$ and $DB$ , then

CD^2 = AD \cdot DB.

We will use $AD = 24$ and $DB = 16$ from the previous step in this formula.

Write and simplify the expression for the altitude

Substitute $AD = 24$ and $DB = 16$ into the relationship:

CD^2 = 24 \cdot 16.

Compute the product:

24 \cdot 16 = 384.

CD^2 = 384,

which means

CD = \sqrt{384}

(since a length is positive).

Simplify the square root to match an answer choice

Factor 384 to pull out a perfect square:

384 = 64 \cdot 6

because $64 \cdot 6 = 384$ and $64$ is a perfect square.

Then

\sqrt{384} = \sqrt{64 \cdot 6} = \sqrt{64}\,\sqrt{6} = 8\sqrt{6}.

So the length of the altitude $CD$ is $8\sqrt{6}$ , which corresponds to choice B.

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Desmos Guide

Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation