Special Right Triangles
Two triangle shapes appear so often the SAT expects you to know their side ratios cold.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
A 45-45-90 triangle has sides $s$, $s$, $s\sqrt{2}$. A 30-60-90 has sides $x$, $x\sqrt{3}$, $2x$. Match the side you know to the ratio.
How to solve one, step by step
Example: a 45-45-90 triangle with legs $5$.
- Legs are equal at $5$.
- Hypotenuse: $5\sqrt{2}$.
The mistakes that cost points
- Mixing up the two ratios. 30-60-90 and 45-45-90 have different side relationships.
- Multiplying when you should divide by √3. Track which side is which in the 30-60-90 ratio.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
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$324\sqrt{3}$
- B$648$
- C$324$
- D$162\sqrt{3}$
Show solution
Answer: D, $162\sqrt{3}$. Step 1: Apply the 30-60-90 ratio $1 : \sqrt{3} : 2$. Hypotenuse $= 2x = 36$, so $x = 18$.
Step 2: Short leg $= 18$, long leg $= 18\sqrt{3}$.
Step 3: Area $= \frac{1}{2} \cdot 18 \cdot 18\sqrt{3} = 162\sqrt{3}$ square inches.
Medium
In a $45°$-$45°$-$90°$ triangle, the hypotenuse has length $k$ and one leg has length $9$. What is the value of $k$?
- A$9\sqrt{2}$
- B$18$
- C$9\sqrt{3}$
- D$\frac{9}{\sqrt{2}}$
Show solution
Answer: A, $9\sqrt{2}$. In a $45°$-$45°$-$90°$ triangle, hypotenuse $= \text{leg} \times \sqrt{2} = 9\sqrt{2}$. So $k = 9\sqrt{2}$.
Hard
The hypotenuse of a $30^\circ$, $60^\circ$, $90^\circ$ triangle is $y$ inches long. What is the area of the triangle, in square inches?
- A$\frac{y^2\sqrt{3}}{8}$
- B$\frac{y^2}{2}$
- C$\frac{y^2}{4}$
- D$\frac{y^2\sqrt{3}}{4}$
Show solution
Answer: A, $\frac{y^2\sqrt{3}}{8}$. Step 1: Apply the 30-60-90 ratio $1 : \sqrt{3} : 2$. Hypotenuse $= 2x = y$, so $x = \frac{y}{2}$.
Step 2: Short leg $= \frac{y}{2}$, long leg $= \frac{y\sqrt{3}}{2}$.
Step 3: Area $= \frac{1}{2} \cdot \frac{y}{2} \cdot \frac{y\sqrt{3}}{2} = \frac{y^2\sqrt{3}}{8}$ square inches.
Find your exact gaps
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