Question 73·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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In right triangle $ABC$ , angle $C$ is a right angle and angle $A$ measures $30^\circ$ . A circle is inscribed in triangle $ABC$ (tangent to all three sides), and the radius of the circle is $4$ .

Which choice is the length of the hypotenuse $\overline{AB}$ ?

A right triangle with a $30^\circ$ angle has a 30-60-90 side ratio, so you can write all three sides using one short-leg length. For a circle inscribed in a right triangle, relate its radius to the two legs and the hypotenuse, then solve for that length in Desmos. The radius sets the triangle’s scale; it is not itself a leg. Finish with the side the question asks for.

Hints

Hint 1
In a 30-60-90 triangle, the short leg, long leg, and hypotenuse have lengths $x$ , $x\sqrt3$ , and $2x$ . Which of these is opposite the right angle?
Hint 2
For a right triangle with legs $a$ and $b$ and hypotenuse $c$ , the inradius is $r=\frac{a+b-c}{2}$ . The hypotenuse is subtracted, not added.
Hint 3
Put the three side expressions into the radius formula to find $x$ . That gives the short leg; what must you do with it to find the hypotenuse?

Step-by-step

Use the side ratio and inradius
Step 1
Name the three sides
The right angle at $C$ and $30^\circ$ angle at $A$ make a 30-60-90 triangle. Call the leg opposite $30^\circ$ $x$ . The side ratio gives legs $x$ and $x\sqrt3$ , and hypotenuse $AB=2x$ . $AB$ is opposite the right angle, not the $30^\circ$ angle.
Step 2
Relate the circle to the sides
The circle’s radius is the inradius $r$ . For legs $a,b$ and hypotenuse $c$ in a right triangle, $r=\frac{a+b-c}{2}$ . Here’s why: tangent pieces from each vertex match. The two legs together contain the hypotenuse’s pieces plus two pieces of length $r$ at the right-angle corner.
Step 3
Turn the radius into an equation
The given radius is $4$ , so put $4$ in for $r$ and the three side expressions into the formula: $4=\frac{x+x\sqrt3-2x}{2}$ . The $2x$ is subtracted because it is the hypotenuse.
Step 4
Find the short leg in Desmos
Type $4\sim\frac{x_1+x_1\sqrt3-2x_1}{2}$ . Use $x_1$ and $\sim$ so Desmos solves for the unknown instead of graphing $x$ . Under PARAMETERS, it shows $x_1\approx10.9282$ . That is the short leg, not yet $AB$ .
Step 5
Match the hypotenuse to an exact choice
Since $AB=2x$ , type $2x_1$ , then type the answer expressions to compare. Desmos shows $2x_1\approx21.8564$ ; $8(\sqrt3+1)$ matches that value. The other expressions give about $5.8564$ , $43.7128$ , and $37.8564$ . So the hypotenuse $AB$ is $8(\sqrt3+1)$ . Choice D.

Question 73·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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