In right triangle , angle is a right angle and angle measures . A circle is inscribed in triangle (tangent to all three sides), and the radius of the circle is . Which choice is the length of the hypotenuse ?

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Super-Hard SAT Math Question 73 | Free SAT Prep | Aniko

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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 $\overline{AB}$ ?

When a right triangle includes a $30^\circ$ or $45^\circ$ angle, immediately switch to special-triangle side ratios (or equivalent trig values) to express every side in terms of one variable. For an inscribed circle, use the area/semiperimeter relationship (or the right-triangle shortcut $r=\frac{a+b-c}{2}$ ) so you get a single equation in that variable, then solve and simplify (often by rationalizing a denominator).

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Use the special right triangle ratio

Because one acute angle is $30^\circ$ , you can write both legs as multiples of the hypotenuse using a $30$ - $60$ - $90$ triangle.

Connect the inradius to side lengths

For any triangle, area $= r \cdot s$ , where $r$ is the inradius and $s$ is the semiperimeter.

Simplify before solving

After substituting the leg expressions, factor out the hypotenuse so the equation becomes a simple linear equation in $c$ .

Desmos Guide

Define the inradius as a function of the hypotenuse

Enter

r(c) = ( (c/2) + (sqrt(3)/2)*c - c )/2

This matches $r=\frac{a+b-c}{2}$ with $a=\frac{c}{2}$ and $b=\frac{\sqrt{3}}{2}c$ .

Test the answer choices

For each option, define a value for $c$ (for example, c1 = 8*(sqrt(3)+1), c2 = 8*(sqrt(3)-1), etc.). Then evaluate r(c1), r(c2), r(c3), and r(c4).

Match the radius

The correct choice is the one whose computed value makes r(ci) equal to 4 (or extremely close, allowing for rounding).

Step-by-step Explanation

Express the legs in terms of the hypotenuse

Let $c=\overline{AB}$ be the hypotenuse. Since angle $A=30^\circ$ , triangle $ABC$ is a $30$ - $60$ - $90$ triangle.

Side opposite $30^\circ$ is half the hypotenuse: $\overline{BC}=\frac{c}{2}$ .
The other leg is $\overline{AC}=\frac{\sqrt{3}}{2}c$ .

Use the inradius formula for a right triangle

For a right triangle with legs $a$ and $b$ and hypotenuse $c$ , the inradius is

r=\frac{a+b-c}{2}.

Here, $a=\overline{BC}=\frac{c}{2}$ and $b=\overline{AC}=\frac{\sqrt{3}}{2}c$ .

Substitute and solve for the hypotenuse

Substitute into the formula and set $r=4$ :

\begin{aligned} 4 &= \frac{\frac{c}{2}+\frac{\sqrt{3}}{2}c-c}{2} \\ 4 &= \frac{c\left(\frac{1+\sqrt{3}-2}{2}\right)}{2} \\ 4 &= \frac{c(\sqrt{3}-1)}{4} \\ 16 &= c(\sqrt{3}-1). \end{aligned}

c=\frac{16}{\sqrt{3}-1}=\frac{16(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{16(\sqrt{3}+1)}{2}=8(\sqrt{3}+1).

Therefore, the length of $\overline{AB}$ is $8(\sqrt{3}+1)$ .

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