Solve right-triangle inradius problems

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Difficulty: Advanced
Estimated time: 24 minutes
Domains: Geometry and Trigonometry
Techniques: Right-triangle-inradiusIncircleEqual-tangent-segmentsSide-ratios

What you’ll learn

Recognize an incircle, a circle inside a triangle that touches all three sides.
Use equal tangent segments to split a right triangle’s sides into pieces.
Show where this formula comes from, and use it:.
Pair the radius with a special right triangle or a trig ratio.
Finish with what the question asks for, such as a leg, the hypotenuse, the radius, or the perimeter.

Prerequisites

Why this matters on the SAT

Turn a circle into side-length bookkeeping

A circle tucked inside a right triangle can look like a whole new topic. It’s really two ideas you already have. The circle gives you pairs of equal tangent segments, and the right triangle gives you a side relationship. Put them together and the math takes a few lines.

Solution to the example

Start with the key circle fact: two tangent segments drawn from the same point to a circle are equal. So in the figure,

AD=AF=x \qquad\text{and}\qquad BE=BF=y.

At the right-angle corner $C$ , both tangent pieces have length $r$ , the circle’s radius. That splits each leg into two pieces:

AC=x+r \qquad\text{and}\qquad BC=y+r.

The hypotenuse gets the other two pieces:

AB=x+y.

Now add the two legs. The $x$ and $y$ inside that sum make up the hypotenuse:

\begin{aligned} 12+9&=(x+r)+(y+r)\\[1.4em] 21&=(x+y)+2r\\[1.4em] 21&=15+2r\\[1.4em] r&=3. \end{aligned}

The answer is B. Equal tangents turned the circle into three simple side sums.

SAT example

Each vertex sends two equal tangent segments to the incircle. The right-angle corner contributes one radius to each leg.

Right triangle $ABC$ has a right angle at $C$ , with $AC=12$ , $BC=9$ , and $AB=15$ . A circle with center $I$ is inscribed in the triangle and is tangent to the sides at $D$ , $E$ , and $F$ , as shown. What is the radius $r$ of the circle?

A
$2$
B
$3$
C
$6$
D
$9$

Build the formula from equal tangents

Why does the corner at $C$ give exactly $r$ to each leg? Look at the small four-sided shape $CDIE$ in the figure.

You need one more circle fact: a radius drawn to the point where the circle touches a side is perpendicular to that side. So $CDIE$ has three right angles. There’s $\angle DCE$ , the triangle’s right angle at $C$ . There’s $\angle CDI$ , where radius $\overline{ID}$ meets side $\overline{AC}$ . And there’s $\angle CEI$ , where radius $\overline{IE}$ meets side $\overline{BC}$ .

The angles of a four-sided shape add up to $360^\circ$ . Three of them already make $270^\circ$ , so the fourth, $\angle DIE$ , is $90^\circ$ too. That makes $CDIE$ a rectangle, and opposite sides of a rectangle are equal:

CD=IE=r \qquad\text{and}\qquad CE=ID=r.

Now let’s do the same bookkeeping for any right triangle. Call the legs $a$ and $b$ and the hypotenuse $c$ , and call the tangent lengths from the other two corners $x$ and $y$ . Then

\begin{aligned} a&=x+r,\\[1.4em] b&=y+r,\\[1.4em] c&=x+y. \end{aligned}

Subtract the hypotenuse from the sum of the legs, and the $x$ and $y$ cancel:

\begin{aligned} a+b-c &=(x+r)+(y+r)-(x+y)\\[1.4em] &=2r. \end{aligned}

Divide by $2$ :

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

In words: add the legs, subtract the hypotenuse, and halve. The formula holds only in a right triangle, because it depends on that rectangle at the right-angle corner.

Check your understanding:

In the opening figure, why is the hypotenuse $x+y$ and not $x+y+r$ ?

Common mistake:

It’s tempting to call whichever side looks longest in the drawing $c$ . But in the formula, $a$ and $b$ must be the legs and $c$ must be the side across from the $90^\circ$ angle. Find the right angle, label the side across from it $c$ , then subtract it from the sum of the two legs.

Choose the shortest setup

Now that you’ve seen why it works, the formula is the quickest way in:

Name the legs $a$ and $b$ and the hypotenuse $c$ .
Turn what the question gives you into side lengths.
Put those lengths into

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

Solve for the scale value, the one letter all three sides share.
Use it to build the side or perimeter the question asks for.

If you already know all three sides, skip step 2. If the question gives you $\tan A$ , a special angle, or another side ratio, write all three sides with one letter, such as $x$ or $k$ , before you use the formula.

When the question asks for the perimeter $P$ , the formula gives you a handy last step. Rearranged, it says

a+b=c+2r,

P=a+b+c=2c+2r.

Use it only once you know $c$ and $r$ , and only in a right triangle. It isn’t a perimeter formula for other triangles.

Check your understanding:

A right triangle has legs $7$ and $24$ and hypotenuse $25$ . What is its inradius?

Example: the radius meets a special triangle

Worked example

In right triangle $ABC$ , $\angle C=90^\circ$ and $\angle A=30^\circ$ . A circle is inscribed in the triangle, and its radius is $6$ . Which choice gives the length of hypotenuse $\overline{AB}$ ?

A
$6(\sqrt3+1)$
B
$6(3+\sqrt3)$
C
$12(\sqrt3+1)$
D
$24(\sqrt3+1)$

Step 1

Write the sides with one letter

With a $90^\circ$ angle and a $30^\circ$ angle, this is a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle. Call the short leg $x$ . Then the three sides are

x,\qquad x\sqrt3,\qquad 2x.

The hypotenuse is $2x$ , because it’s the side across from the right angle.

Step 2

Put the sides into the formula

The radius is $6$ , so put the three sides into the formula:

\begin{aligned} 6 &=\frac{x+x\sqrt3-2x}{2}\\[1.4em] &=\frac{x(\sqrt3-1)}{2}. \end{aligned}

That one equation holds both facts: the circle’s radius and the triangle’s shape.

Step 3

Solve for x

Multiply by $2$ :

12=x(\sqrt3-1).

Divide, then multiply the top and bottom by $\sqrt3+1$ to clear the radical from the bottom:

\begin{aligned} x &=\frac{12}{\sqrt3-1}\\[1.4em] &=\frac{12(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}\\[1.4em] &=\frac{12(\sqrt3+1)}{2}\\[1.4em] &=6(\sqrt3+1). \end{aligned}

Step 4

Answer the question: the hypotenuse

Don’t stop at $x$ . The question asks for $AB$ , which is $2x$ :

\begin{aligned} AB &=2\left[6(\sqrt3+1)\right]\\[1.4em] &=\boxed{12(\sqrt3+1)}. \end{aligned}

The answer is C. Choice A is $x$ itself, where you’d land if you stopped a step early. Keep the answer exact, since the choices use radicals.

Check your understanding:

In the same triangle, how long are the two legs?

Turn a trig ratio into a side ratio

A trig ratio can set up the sides, too. Say

\tan A=\frac34.

Tangent is opposite over adjacent, measured from angle $A$ , so write the legs as $3k$ and $4k$ . The Pythagorean theorem gives the hypotenuse: $\sqrt{(3k)^2+(4k)^2}=\sqrt{25k^2}=5k$ .

If the radius is $5$ , then

\begin{aligned} 5 &=\frac{3k+4k-5k}{2}\\[1.4em] &=k. \end{aligned}

So $k=5$ , and the perimeter is

3k+4k+5k=12k=60.

The perimeter shortcut agrees: the hypotenuse is $5k=25$ , so $2c+2r=2(25)+2(5)=60$ .

Here’s the idea to hold on to: the ratio gives the triangle’s shape, and the radius gives its size.

Common mistake:

If you use $3$ and $4$ as the actual legs, you get $r=\frac{3+4-5}{2}=1$ , not the $5$ the question gave you. A ratio such as $3:4$ only fixes the shape. Write $3k$ and $4k$ , use the Pythagorean theorem for the hypotenuse, and let the radius tell you $k$ .

Practice problems

Do the setup by hand in each one: find the hypotenuse and write all three sides before any arithmetic. That’s the real work in these questions. The calculator is there for the arithmetic afterward, if you want it.

Find the radius from three sides

Practice problem

A right triangle has leg lengths $20$ and $21$ units and hypotenuse length $29$ units. A circle is inscribed in the triangle. What is the radius, in units, of the circle?

Your answer

Calculator loads as you approach

Place the legs and hypotenuse first. Use the calculator only for the arithmetic.

Scale an isosceles right triangle

Practice problem

A circle of radius $5$ is inscribed in a $45^\circ$ - $45^\circ$ - $90^\circ$ triangle. Which choice gives the length of the hypotenuse?

Calculator loads as you approach

Write the sides and the equation by hand, and keep the radical exact.

Find a perimeter from tangent

Practice problem

Right triangle $ABC$ has a right angle at $C$ and satisfies $\tan A=\frac5{12}$ . A circle of radius $4$ is inscribed in the triangle. What is the perimeter of triangle $ABC$ ?

Your answer

Calculator loads as you approach

Turn tan A into three sides and set up the equation first.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

An incircle touches all three sides of a triangle.
The two tangent segments from the same vertex are equal.
In a right triangle, the right-angle corner gives each leg a piece of length $r$ .
Add the legs, subtract the hypotenuse, and halve. With legs $a$ and $b$ and hypotenuse $c$ ,
Given a special angle or a trig ratio, write all three sides with one letter before you use the formula.
The radius fixes that letter. Then go back for what the question asks: a leg, the hypotenuse, the radius, or the perimeter.
Do the setup by hand. Finding the sides and placing the hypotenuse is the real work; save the calculator for the arithmetic afterward.

Next lesson

Use similarity inside right triangles

Use the three similar triangles created by an altitude to a right triangle’s hypotenuse.

Start next lesson

Practice

Practice this lesson

4 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

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