In right triangle , is the right angle. The altitude from to hypotenuse has length units. If is twice as long as , what is the length of ?

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SAT Right Triangles & Trigonometry Question 3 | Free SAT Prep | Aniko

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Question 3·Hard·Right Triangles and Trigonometry

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In right triangle $\triangle ABC$ , $\angle C$ is the right angle. The altitude from $C$ to hypotenuse $\overline{AB}$ has length $12$ units. If $AC$ is twice as long as $BC$ , what is the length of $AC$ ?

When you see a right triangle with an altitude to the hypotenuse and a ratio between the legs, first introduce a variable for the smaller leg and express all sides (including the hypotenuse via the Pythagorean theorem) in terms of that variable. Then, instead of memorizing special altitude formulas, use the fact that the triangle’s area can be written both as $\tfrac12(\text{leg})(\text{other leg})$ and as $\tfrac12(\text{hypotenuse})(\text{altitude})$ to create a simple equation. Solve for the variable, carefully identify which side the problem asks for, and only then match your result to the answer choices so you don’t pick an intermediate value by mistake.

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Turn the side relationship into algebra

If $AC$ is twice as long as $BC$ , try letting $BC = x$ . How can you write $AC$ in terms of $x$ ?

Use the Pythagorean theorem

Once you have $AC$ and $BC$ in terms of $x$ , use $AB^2 = AC^2 + BC^2$ to express the hypotenuse $AB$ in terms of $x$ as well.

Use the altitude via area

You know the altitude from $C$ to $AB$ is $12$ . Write the area of the triangle once using $AC$ and $BC$ as base and height, and once using $AB$ as the base and $12$ as the height. Set these two area expressions equal to get an equation in $x$ .

Finish solving and answer the question asked

After you solve for $x$ (which is $BC$ ), remember that the problem asks for $AC$ , which is twice $BC$ . Multiply appropriately before picking your answer.

Desmos Guide

Set up the equation in Desmos

In Desmos, type the equation representing the area relationship from the work: x^2 = 6x*sqrt(5) (or equivalently x^2 - 6x*sqrt(5) = 0).

Find the positive solution for x

If you entered y = x^2 - 6x*sqrt(5), look at the graph and use the intercept tool to find the positive $x$ -intercept; that value is $BC$ . If you entered x^2 = 6x*sqrt(5), use the solver or inspect the value of $x$ that makes both sides equal.

Compute AC from x

In a new expression line, enter AC = 2x and substitute your $x$ value from the previous step; the resulting output is the length of $AC$ , which should match one of the answer choices.

Step-by-step Explanation

Represent the side lengths with a variable

We are told that $AC$ is twice as long as $BC$ .

Let $BC = x$ .
Then $AC = 2x$ .

We also know the altitude from $C$ to hypotenuse $AB$ has length $12$ .

Find the hypotenuse in terms of x

Use the Pythagorean theorem in right triangle $\triangle ABC$ with legs $AC$ and $BC$ and hypotenuse $AB$ .

AB^2 = AC^2 + BC^2 \\[0.7em] AB^2 = (2x)^2 + x^2 = 4x^2 + x^2 = 5x^2

AB = x\sqrt{5}

(using only the positive root because side lengths are positive).

Relate the altitude to the sides using the area

The area of $\triangle ABC$ can be computed in two ways:

Using the two legs $AC$ and $BC$ as base and height:

\text{Area} = \frac12 (AC)(BC) = \frac12 (2x)(x)

Using the hypotenuse $AB$ as the base and the altitude from $C$ (which is $12$ ) as the height:

\text{Area} = \frac12 (AB)(12) = \frac12 (x\sqrt{5})(12)

Set these equal:

\frac12 (2x)(x) = \frac12 (x\sqrt{5})(12) \\[0.7em] x^2 = 6x\sqrt{5}

Since $x>0$ , divide both sides by $x$ :

x = 6\sqrt{5}

This is the length of $BC$ . To answer the question, we still need $AC$ .

Find AC and choose the matching option

We defined $AC$ to be twice $BC$ :

AC = 2x = 2(6\sqrt{5}) = 12\sqrt{5}

So the length of $AC$ is $12\sqrt{5}$ , which corresponds to choice C.

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Step-by-step Explanation

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