In , the side lengths are , , and . Point is the midpoint of . What is the length of ?

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SAT Lines, Angles & Triangles Question 18 | Free SAT Prep | Aniko

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Question 18·Hard·Lines, Angles, and Triangles

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In $\triangle ABC$ , the side lengths are $AB = 13$ , $BC = 14$ , and $AC = 15$ . Point $D$ is the midpoint of $\overline{BC}$ . What is the length of $\overline{AD}$ ?

When a triangle gives all three side lengths and asks for a segment to a midpoint, stay SAT-native: add an altitude, use two Pythagorean relationships, and only then use one final right triangle. This is cleaner than reaching for named theorems that are not part of normal SAT math.

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Add a helpful altitude

Draw a perpendicular from $A$ to $\overline{BC}$ and call the foot $E$ . Then $\triangle ABE$ and $\triangle ACE$ are both right triangles.

Use the side lengths to split BC

Let $BE=x$ . Then $CE=14-x$ . Use $AB=13$ and $AC=15$ in the two right triangles to compare the squares.

Use the midpoint last

Once you know where the altitude lands, remember that $D$ is the midpoint of $BC$ , so the horizontal distance from that foot to $D$ is small and easy to compute.

Desmos Guide

Find point A with two circles

Type $x^2+y^2=169$ and $(x-14)^2+y^2=225$ on separate lines. These are circles of radius $13$ and $15$ centered at $B=(0,0)$ and $C=(14,0)$ . Click the upper intersection — Desmos shows $(5,12)$ . This is point $A$ .

Compute the distance AD

The midpoint of $\overline{BC}$ is $D=(7,0)$ . On a new line, type $\sqrt{(5-7)^2+12^2}$ . Desmos returns approximately $12.17$ .

Verify against the answer choices

On another line, type $2\sqrt{37}$ . Desmos also returns approximately $12.17$ , confirming $AD=2\sqrt{37}$ (choice C).

Step-by-step Explanation

Drop an altitude and name the segments

Draw altitude $AE$ to $\overline{BC}$ , and let $BE=x$ . Then $CE=14-x$ . Because $AB=13$ and $AC=15$ , the right triangles give $AE^2 + x^2 = 13^2$ and $AE^2 + (14-x)^2 = 15^2$ .

Solve for the split of BC

Subtract the equations: $(14-x)^2 - x^2 = 15^2 - 13^2 = 56$ . Since $(14-x)^2 - x^2 = (14-2x)(14)$ , this simplifies to $14(14-2x)=56$ , so $14-2x=4$ and $x=5$ . Therefore $BE=5$ and $CE=9$ .

Find the altitude

Use $\triangle ABE$ : $AE^2 = 13^2 - 5^2 = 169-25 = 144$ , so $AE=12$ .

Use the midpoint to get AD

Since $D$ is the midpoint of $BC$ , $BD=DC=7$ . The foot $E$ is $5$ units from $B$ , so $ED=2$ . In right triangle $AED$ , $AD^2 = 12^2 + 2^2 = 148$ , so $AD = 2\sqrt{37}$ , which matches choice C.

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Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation