Special Right Triangles
Where I would start
There are two right triangles whose sides you never have to work out. In a $45^\circ$-$45^\circ$-$90^\circ$ triangle the sides are $x$, $x$ and $x\sqrt{2}$. In a $30^\circ$-$60^\circ$-$90^\circ$ triangle they are $x$, $x\sqrt{3}$ and $2x$.
Both are on the SAT reference sheet, so you won't be stuck if you forget them. I still want my students to know them by heart, because then they recognize these triangles inside squares, equilateral triangles and trig questions without having to look anything up.
Most of these questions give you one side and expect you to find the other two. Then there is usually one more step: an area, a perimeter, a diagonal, or a sine or cosine.
What you will see on the test
In a $30^\circ$-$60^\circ$-$90^\circ$ triangle, the short leg $x$ is across from the $30^\circ$ angle and the long leg $x\sqrt{3}$ is across from the $60^\circ$ angle.
| You see | What it means |
|---|---|
| A right triangle with two equal legs, or a $45^\circ$ angle | Sides $x$, $x$, $x\sqrt{2}$. Leg to hypotenuse: multiply by $\sqrt{2}$. Hypotenuse to leg: divide by $\sqrt{2}$. |
| A right triangle with a $30^\circ$ or $60^\circ$ angle | Sides $x$, $x\sqrt{3}$, $2x$. The hypotenuse is always twice the short leg. |
| The diagonal of a square with side $s$ | Two $45^\circ$-$45^\circ$-$90^\circ$ triangles. The diagonal is $s\sqrt{2}$. |
| The height of an equilateral triangle with side $s$ | Two $30^\circ$-$60^\circ$-$90^\circ$ triangles. The height is $\dfrac{s\sqrt{3}}{2}$ and the area is $\dfrac{s^2\sqrt{3}}{4}$. |
| Sine, cosine, or tangent of $30^\circ$, $45^\circ$, or $60^\circ$ | Read it from the triangle: $\sin 30^\circ = \dfrac{1}{2}$, $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$, $\sin 45^\circ = \cos 45^\circ = \dfrac{\sqrt{2}}{2}$, $\tan 60^\circ = \sqrt{3}$. |
Four worked examples
Example 1: from the hypotenuse to a leg
A $45^\circ$-$45^\circ$-$90^\circ$ triangle has a hypotenuse of $10$. How long is each leg?
- The hypotenuse is a leg times $\sqrt{2}$, so a leg is the hypotenuse divided by $\sqrt{2}$: $\dfrac{10}{\sqrt{2}}$.
- Multiply the top and bottom by $\sqrt{2}$: $\dfrac{10\sqrt{2}}{2} = 5\sqrt{2}$.
- Each leg is $5\sqrt{2}$, about $7.07$. That is shorter than $10$, as a leg must be.
Example 2: area from the hypotenuse
The hypotenuse of a $30^\circ$-$60^\circ$-$90^\circ$ triangle is $36$ inches. What is its area?
- The hypotenuse is $2x = 36$, so the short leg is $x = 18$.
- The long leg is $x\sqrt{3} = 18\sqrt{3}$.
- The legs are the base and height: $\dfrac{1}{2}(18)(18\sqrt{3}) = 162\sqrt{3}$ square inches.
Example 3: from the long leg back to the others
In a $30^\circ$-$60^\circ$-$90^\circ$ triangle, the side across from the $60^\circ$ angle is $9$. Find the other two sides.
- The side across from $60^\circ$ is the long leg, so $x\sqrt{3} = 9$.
- $x = \dfrac{9}{\sqrt{3}} = \dfrac{9\sqrt{3}}{3} = 3\sqrt{3}$. That is the short leg.
- The hypotenuse is $2x = 6\sqrt{3}$.
Example 4: the triangle hiding in an equilateral triangle
An equilateral triangle has sides of length $8$. What is its area?
- Draw the height. It splits the triangle into two $30^\circ$-$60^\circ$-$90^\circ$ triangles with hypotenuse $8$ and short leg $4$.
- The height is the long leg: $4\sqrt{3}$.
- Area $= \dfrac{1}{2}(8)(4\sqrt{3}) = 16\sqrt{3}$. The shortcut $\dfrac{s^2\sqrt{3}}{4} = \dfrac{64\sqrt{3}}{4}$ gives the same answer.
Where students go wrong
- Putting $\sqrt{3}$ on the wrong leg. The long leg, $x\sqrt{3}$, is across from the $60^\circ$ angle. The short leg is across from the $30^\circ$ angle.
- Multiplying when you should divide. Going from the hypotenuse to a leg in a $45^\circ$-$45^\circ$-$90^\circ$ triangle means dividing by $\sqrt{2}$, not multiplying.
- Mixing up the two triangles. $\sqrt{2}$ belongs only to the $45^\circ$-$45^\circ$-$90^\circ$ triangle. $\sqrt{3}$ and $2x$ belong only to the $30^\circ$-$60^\circ$-$90^\circ$ triangle.
- Making the hypotenuse $x\sqrt{3}$. The hypotenuse is always the longest side. Since $2 > \sqrt{3}$, the hypotenuse is $2x$.
- Missing the triangle inside another shape. A square's diagonal and an equilateral triangle's height both create special right triangles. Draw the line in, and you will usually see it.
Practice questions
Try each one on paper first, then open the solution and compare your steps with mine.
Easy
A $45^\circ$-$45^\circ$-$90^\circ$ triangle has legs of length $7$. What is the length of the hypotenuse?
- A$7$
- B$7\sqrt{2}$
- C$7\sqrt{3}$
- D$14$
Show solution
Answer: B. In a $45^\circ$-$45^\circ$-$90^\circ$ triangle, the hypotenuse is a leg times $\sqrt{2}$, so it is $7\sqrt{2}$, about $9.9$.
Easy
In a $30^\circ$-$60^\circ$-$90^\circ$ triangle, the side across from the $30^\circ$ angle has length $5$. What is the length of the hypotenuse?
- A$5\sqrt{2}$
- B$5\sqrt{3}$
- C$15$
- D$10$
Show solution
Answer: D. The side across from $30^\circ$ is the short leg, $x = 5$. The hypotenuse is $2x = 10$.
Medium
A $45^\circ$-$45^\circ$-$90^\circ$ triangle has a hypotenuse of length $12$. What is the length of each leg?
- A$6$
- B$12\sqrt{2}$
- C$6\sqrt{2}$
- D$6\sqrt{3}$
Show solution
Answer: C. A leg is the hypotenuse divided by $\sqrt{2}$: $\dfrac{12}{\sqrt{2}} = \dfrac{12\sqrt{2}}{2} = 6\sqrt{2}$. Choice B multiplies by $\sqrt{2}$ instead of dividing, which would make a leg longer than the hypotenuse.
Medium
In a $30^\circ$-$60^\circ$-$90^\circ$ triangle, the longer leg has length $12$. What is the length of the shorter leg?
- A$4\sqrt{3}$
- B$6$
- C$6\sqrt{3}$
- D$12\sqrt{3}$
Show solution
Answer: A. The longer leg is $x\sqrt{3}$, so $x\sqrt{3} = 12$ and $x = \dfrac{12}{\sqrt{3}} = \dfrac{12\sqrt{3}}{3} = 4\sqrt{3}$. Choice B halves the long leg, but halving only works on the hypotenuse.
MediumStudent-produced response
A square has a diagonal of length $10$. What is the area of the square?
Enter your answer.
Show solution
Answer: $50$. The diagonal splits the square into two $45^\circ$-$45^\circ$-$90^\circ$ triangles, so the side is $\dfrac{10}{\sqrt{2}} = 5\sqrt{2}$. The area is $(5\sqrt{2})^2 = 25 \cdot 2 = 50$.
Medium
A $20$-foot ladder leans against a vertical wall and makes a $60^\circ$ angle with the level ground. How far up the wall does the top of the ladder reach, in feet?
- A$10$
- B$10\sqrt{2}$
- C$10\sqrt{3}$
- D$20\sqrt{3}$
Show solution
Answer: C. The ladder is the hypotenuse, so $2x = 20$ and $x = 10$. The height on the wall is across from the $60^\circ$ angle, so it is the long leg, $10\sqrt{3}$, about $17.3$ feet. Choice A is the distance from the wall to the foot of the ladder.
Hard
The hypotenuse of a $30^\circ$-$60^\circ$-$90^\circ$ triangle is $36$ inches long. What is the area of the triangle, in square inches?
- A$162$
- B$81\sqrt{3}$
- C$324\sqrt{3}$
- D$162\sqrt{3}$
Show solution
Answer: D. The short leg is $\dfrac{36}{2} = 18$ and the long leg is $18\sqrt{3}$. The area is $\dfrac{1}{2}(18)(18\sqrt{3}) = 162\sqrt{3}$. Choice C forgets the $\dfrac{1}{2}$.
Hard
In right triangle $ABC$, angle $C$ is the right angle, the measure of angle $A$ is $30^\circ$, and $AB = 14$. What is the length of $AC$?
- A$7$
- B$7\sqrt{3}$
- C$7\sqrt{2}$
- D$14\sqrt{3}$
Show solution
Answer: B. $AB$ is across from the right angle, so it is the hypotenuse: $2x = 14$ and $x = 7$. $BC$ is across from the $30^\circ$ angle at $A$, so $BC = 7$. $AC$ is across from the $60^\circ$ angle at $B$, so $AC = 7\sqrt{3}$.
HardStudent-produced response
An equilateral triangle has an area of $25\sqrt{3}$ square units. What is the length of one side of the triangle?
Enter your answer.
Show solution
Answer: $10$. The area of an equilateral triangle with side $s$ is $\dfrac{s^2\sqrt{3}}{4}$. Set $\dfrac{s^2\sqrt{3}}{4} = 25\sqrt{3}$, so $s^2 = 100$ and $s = 10$.
Hard
The hypotenuse of a $30^\circ$-$60^\circ$-$90^\circ$ triangle is $y$ inches long. Which expression gives the area of the triangle, in square inches?
- A$\dfrac{y^2\sqrt{3}}{8}$
- B$\dfrac{y^2\sqrt{3}}{4}$
- C$\dfrac{y^2}{8}$
- D$\dfrac{y^2\sqrt{3}}{2}$
Show solution
Answer: A. The short leg is $\dfrac{y}{2}$ and the long leg is $\dfrac{y\sqrt{3}}{2}$. The area is $\dfrac{1}{2} \cdot \dfrac{y}{2} \cdot \dfrac{y\sqrt{3}}{2} = \dfrac{y^2\sqrt{3}}{8}$. Check with $y = 36$: $\dfrac{1296\sqrt{3}}{8} = 162\sqrt{3}$, the same answer as the earlier question.
Hard
In right triangle $PQR$, angle $R$ is the right angle and $\sin P = \dfrac{\sqrt{3}}{2}$. If $PQ = 18$, what is the length of $QR$?
- A$9$
- B$9\sqrt{2}$
- C$18\sqrt{3}$
- D$9\sqrt{3}$
Show solution
Answer: D. $\sin P = \dfrac{\sqrt{3}}{2}$ means angle $P$ is $60^\circ$. $PQ$ is the hypotenuse, so $2x = 18$ and $x = 9$. $QR$ is across from the $60^\circ$ angle at $P$, so $QR = 9\sqrt{3}$. You can also use $\sin P = \dfrac{QR}{PQ}$ directly: $QR = 18 \cdot \dfrac{\sqrt{3}}{2} = 9\sqrt{3}$.
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