In right triangle , is the right angle. The altitude from to hypotenuse meets at point . If centimeters and , the length of , in centimeters, can be written in simplest radical form as , where and are positive integers and is square-free. What is the value of ?

Solution available with step-by-step explanation

SAT Right Triangles & Trigonometry Question 33 | Free SAT Prep | Aniko

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Question 33·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 $AB$ meets $AB$ at point $D$ . If $CD = 12$ centimeters and $AD : DB = 1 : 3$ , the length of $AB$ , in centimeters, can be written in simplest radical form as $k\sqrt{d}$ , where $k$ and $d$ are positive integers and $d$ is square-free. What is the value of $d$ ?

Your answer

(Express the answer as an integer)

For right triangles with an altitude drawn to the hypotenuse, look first for the special geometric-mean relationships: $CD^2 = AD \times DB$ , $AC^2 = AD \times AB$ , and $BC^2 = DB \times AB$ . Translate any given ratios on the hypotenuse into algebra using a single variable, then plug into the appropriate relationship to get a simple quadratic in that variable. Solve for the variable, compute the requested length, and always simplify radicals by factoring out perfect squares so you can quickly identify the square-free part when the answer is written as $k\sqrt{d}$ .

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Turn the ratio into actual lengths

You know $AD : DB = 1 : 3$ . Try assigning a variable to $AD$ (for example, $x$ ) and then express $DB$ and $AB$ in terms of that same variable.

Recall the special property of the altitude to the hypotenuse

In a right triangle, the altitude from the right angle to the hypotenuse relates the altitude and the two segments of the hypotenuse. Think about a formula that involves $CD^2$ , $AD$ , and $DB$ .

Set up and solve the equation using CD = 12

Once you remember the relationship $CD^2 = AD \times DB$ , plug in $CD = 12$ and your expressions for $AD$ and $DB$ in terms of $x$ , and solve for $x^2$ .

Simplify the radical and extract d

After you find $x$ , compute $AB$ using your expression from the first hint, write $AB$ in the form $k\sqrt{d}$ , and then make sure the number under the root has no perfect-square factors other than 1.

Desmos Guide

Represent the equation for x in Desmos

In one expression line, type y = 3x^2. In another line, type y = 144. These represent the equation $3x^2 = 144$ that comes from $CD^2 = AD \times DB$ .

Find the positive intersection x

Use the intersection tool (tap the point where the graphs cross) to find the positive x-value where the two graphs meet. This x-value corresponds to the length $AD$ .

Compute the hypotenuse length AB

In a new expression line, type 4 * (x_intersection_value) using the x-value you found (or use a slider for x and link it). This evaluates $AB = 4x$ numerically.

Connect the numeric value to radical form

Compare the numerical value of $AB$ from Desmos with possible radical forms (e.g., by checking that $x^2 = 48$ and then simplifying $\sqrt{48}$ by hand) to determine the simplified $k\sqrt{d}$ form and identify the square-free $d$ . Desmos confirms your numeric result but you must simplify the radical yourself.

Step-by-step Explanation

Translate the ratio into algebra

We know the altitude from $C$ meets $AB$ at $D$ and $AD : DB = 1 : 3$ .

Let $AD = x$ . Then, because the ratio is $1:3$ , we have $DB = 3x$ .

So the full hypotenuse is

AB = AD + DB = x + 3x = 4x.

Use the altitude-to-hypotenuse relationship

In a right triangle, the altitude from the right angle to the hypotenuse has a special property:

CD^2 = AD \times DB.

We are given $CD = 12$ , so

12^2 = AD \times DB.

Substitute $AD = x$ and $DB = 3x$ :

12^2 = x \times 3x.

Solve for the segment length x

First compute $12^2$ :

12^2 = 144.

So the equation becomes

144 = 3x^2.

Divide both sides by $3$ :

x^2 = \frac{144}{3} = 48.

Thus,

x = \sqrt{48}.

(Since $x$ is a length, we only consider the positive root.)

Find AB and identify d in simplest radical form

We already found that $AB = 4x$ and $x = \sqrt{48}$ , so

AB = 4\sqrt{48}.

Now simplify $\sqrt{48}$ :

48 = 16 \times 3,

\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16}\sqrt{3} = 4\sqrt{3}.

Therefore,

AB = 4 \times 4\sqrt{3} = 16\sqrt{3}.

In the form $k\sqrt{d}$ , we have $k = 16$ and $d = 3$ , so the requested value of $d$ is 3.

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Step-by-step Explanation

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