Special Right Triangles

A 45-45-90 triangle is exactly half of a square cut by its diagonal. If the square has side $s$, the diagonal splits it into two congruent isosceles right triangles.

The Ratio: By the Pythagorean theorem, $s^2 + s^2 = d^2$, so $d = s\sqrt{2}$. The side lengths are always in the ratio $1 : 1 : \sqrt{2}$.

SAT Use: Whenever you see a square, immediately draw the diagonal to create these triangles. If the problem gives you the diagonal, divide by $\sqrt{2}$ to get the side; if it gives the side, multiply by $\sqrt{2}$ to get the diagonal.

Worked Example: A square has a diagonal of $10$. The side $s$ satisfies $s\sqrt{2} = 10$, so $s = \frac{10}{\sqrt{2}} = 5\sqrt{2}$.

A 30-60-90 triangle is half of an equilateral triangle cut by its altitude. If the equilateral triangle has side $2s$, the altitude bisects the base, creating a side of length $s$.

The Ratio: By the Pythagorean theorem, $s^2 + h^2 = (2s)^2$, so $h^2 = 3s^2$ and $h = s\sqrt{3}$. The ratio is $1 : \sqrt{3} : 2$ (short leg : long leg : hypotenuse).

SAT Use: The most common trap is misidentifying the legs. Always locate the $30^\circ$ angle first; the side opposite it is the shortest side ($s$), and the side opposite $60^\circ$ is the middle side ($s\sqrt{3}$).

Worked Example: If the long leg is $6$, then $s\sqrt{3} = 6$, so $s = \frac{6}{\sqrt{3}} = 2\sqrt{3}$. The hypotenuse is $2s = 4\sqrt{3}$.

To solve these triangles efficiently, treat the ratio as a multiplier. If you have the 'base' side ($s$), you simply multiply by the ratio factor. If you have the 'hypotenuse' or 'long leg', you divide.

Rationalization: The SAT often leaves answers in radical form. Remember that $\frac{k}{\sqrt{n}} = \frac{k\sqrt{n}}{n}$.

SAT Use: When you are given a side with a radical, don't panic. The radicals often cancel out during the division process, leaving clean integers or simple fractions.

Worked Example: Given a 45-45-90 triangle with hypotenuse $8$. Side $s = \frac{8}{\sqrt{2}} = \frac{8\sqrt{2}}{2} = 4\sqrt{2}$.

While you can derive these, memorizing the standard formulas for squares and equilateral triangles saves critical time on the Digital SAT.

• Square Diagonal: $d = s\sqrt{2}$
• Equilateral Altitude: $h = \frac{s\sqrt{3}}{2}$

SAT Use: These appear frequently in coordinate geometry and area problems. If a problem asks for the area of an equilateral triangle with side $s$, use $Area = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times s \times \frac{s\sqrt{3}}{2} = \frac{s^2\sqrt{3}}{4}$.

Worked Example: An equilateral triangle has side $6$. The altitude is $\frac{6\sqrt{3}}{2} = 3\sqrt{3}$.

The SAT rarely gives you a naked 30-60-90 triangle. Instead, it hides them inside rectangles, trapezoids, or composite figures.

Recognition: Look for angles of $30^\circ, 45^\circ, 60^\circ, 120^\circ,$ or $135^\circ$. These are 'dead giveaways' that you should drop an altitude to create a special right triangle.

SAT Use: In a trapezoid with a $60^\circ$ base angle, dropping an altitude creates a 30-60-90 triangle. This allows you to find the height and the missing base segments.

Worked Example: A rectangle has a diagonal of $10$ and an angle of $30^\circ$ between the diagonal and a side. The sides are $10 \cos(30^\circ) = 5\sqrt{3}$ and $10 \sin(30^\circ) = 5$.

The values of $\sin, \cos,$ and $\tan$ for $30^\circ, 45^\circ,$ and $60^\circ$ are not arbitrary; they are the ratios of the sides of these specific triangles.

• $\sin(30^\circ) = \frac{1}{2}$, $\cos(30^\circ) = \frac{\sqrt{3}}{2}$
• $\sin(45^\circ) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$, $\cos(45^\circ) = \frac{\sqrt{2}}{2}$
• $\sin(60^\circ) = \frac{\sqrt{3}}{2}$, $\cos(60^\circ) = \frac{1}{2}$

SAT Use: If you forget these values, draw the triangle on your scratch paper and use SOH CAH TOA. It takes 5 seconds and guarantees accuracy.

Worked Example: $\tan(60^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{s\sqrt{3}}{s} = \sqrt{3}$.