Special Right Triangles

📐 Special Right Triangles

Part 1 of 5 — The 45-45-90 Triangle

Topics in This Part

🔑 Key Concept: Two right triangles show up everywhere in geometry, trig, and physics. Their side lengths follow fixed ratios, so you can find every side from just one measurement — no calculator, no trig functions, just a memorized pattern.

Why Are They "Special"?

A special right triangle has angles that produce side lengths in a clean, fixed ratio. Memorize the ratio once and you can solve the triangle instantly.

There are exactly two:

This part covers the $45$-$45$-$90$. Part 2 covers the $30$-$60$-$90$.

💡 You could always fall back on the Pythagorean theorem, but the ratios are faster and give exact answers like $5\sqrt{2}$ instead of decimals.

The $45^\circ$-$45^\circ$-$90^\circ$ Triangle

Because the two acute angles are equal, the triangle is isosceles — the two legs are the same length. Call each leg $x$.

By the Pythagorean theorem the hypotenuse is:

$\text{hyp} = \sqrt{x^2 + x^2} = \sqrt{2x^2} = x\sqrt{2}$

So the sides are always in the ratio:

$\text{leg} : \text{leg} : \text{hyp} \;=\; x : x : x\sqrt{2} \;=\; 1 : 1 : \sqrt{2}$

🔑 Rule: In a $45$-$45$-$90$ triangle, the hypotenuse is a leg times $\sqrt{2}$.

Concept Check 🎯

Finding the Hypotenuse from a Leg

If you know a leg, multiply by $\sqrt{2}$.

Worked Example: legs of length $7$

$\text{hyp} = 7\sqrt{2}$

That's the exact answer. As a decimal, $7\sqrt{2} \approx 7 \times 1.414 \approx 9.90$.

Worked Example: legs of length $3\sqrt{2}$

$\text{hyp} = 3\sqrt{2} \cdot \sqrt{2} = 3 \cdot 2 = 6$

💡 Notice $\sqrt{2}\cdot\sqrt{2} = 2$. When a leg already contains $\sqrt{2}$, the radical often disappears.

Leg → Hypotenuse 🧮

Each triangle is $45$-$45$-$90$. Find the hypotenuse from the given leg. Write radical answers like 5sqrt2 (no spaces) or as a whole number when it simplifies.

1) Leg $= 6$. Hypotenuse $= \,?$ 2) Leg $= 10$. Hypotenuse $= \,?$ 3) Leg $= 5\sqrt{2}$. Hypotenuse $= \,?$ (simplifies to a whole number)

Finding a Leg from the Hypotenuse

Going backward, you divide by $\sqrt{2}$ — then rationalize the denominator.

Worked Example: hypotenuse $= 10$

$\text{leg} = \frac{10}{\sqrt{2}} = \frac{10}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$

⚠️ Common mistake: writing the leg as $10\sqrt{2}$. Multiplying by $\sqrt{2}$ makes a side bigger — but a leg is smaller than the hypotenuse. Going hyp → leg you divide.

Shortcut

Dividing by $\sqrt{2}$ is the same as multiplying by $\dfrac{\sqrt{2}}{2}$:

$\text{leg} = \text{hyp} \cdot \frac{\sqrt{2}}{2}$

Pick the Right Move 🔽

You have a $45$-$45$-$90$ triangle. Choose the correct relationship for each case.

📐 Special Right Triangles

Part 2 of 5 — The 30-60-90 Triangle

🔑 The Idea: Cut an equilateral triangle in half and you get a $30$-$60$-$90$. Its three sides follow the ratio $1 : \sqrt{3} : 2$ — the most useful ratio in all of trigonometry.

Where the $30$-$60$-$90$ Comes From

Start with an equilateral triangle (all sides $2x$, all angles $60^\circ$). Drop an altitude from one vertex straight down. It splits the triangle into two identical right triangles.

Each piece has angles $30^\circ, 60^\circ, 90^\circ$ and these sides:

• Short leg (opposite the $30^\circ$): half the base $= x$
• Hypotenuse (opposite the $90^\circ$): the full side $= 2x$
• Long leg (opposite the $60^\circ$): the altitude

The altitude comes from the Pythagorean theorem:

$\text{long leg} = \sqrt{(2x)^2 - x^2} = \sqrt{4x^2 - x^2} = \sqrt{3x^2} = x\sqrt{3}$

🔑 Ratio: short : long : hyp $= 1 : \sqrt{3} : 2$.

Concept Check 🎯

Solving from the Short Leg

The short leg is the base unit ($x$). If you know it, everything else is easy:

$\text{long leg} = (\text{short leg})\sqrt{3}, \qquad \text{hyp} = 2 \cdot (\text{short leg})$

Worked Example: short leg $= 5$

$\text{long leg} = 5\sqrt{3} \approx 8.66, \qquad \text{hyp} = 2 \cdot 5 = 10$

Worked Example: short leg $= 4\sqrt{3}$

$\text{long leg} = 4\sqrt{3}\cdot\sqrt{3} = 4\cdot 3 = 12, \qquad \text{hyp} = 2 \cdot 4\sqrt{3} = 8\sqrt{3}$

💡 Strategy: always find the short leg first. From the short leg, the other two sides come directly.

From the Short Leg 🧮

Each triangle is $30$-$60$-$90$ with the short leg given. Find the requested side. Use sqrt3 form (e.g. 7sqrt3) or a whole number when it simplifies.

1) Short leg $= 6$. Long leg $= \,?$ 2) Short leg $= 6$. Hypotenuse $= \,?$ 3) Short leg $= 9$. Hypotenuse $= \,?$

Label Every Side at Once

The cleanest way to solve a $30$-$60$-$90$ is to write all three sides as multiples of the short leg $x$, then plug in.

🔑 Find $x$, then read every side off the table. The next check has you fill in the whole triangle.

Match the Side 🔽

A $30$-$60$-$90$ triangle has short leg $= 7$. Fill in each side.

📐 Special Right Triangles

Part 3 of 5 — Working Backward (the hard direction)

🔑 Why it's tricky: When you're given the long leg or the hypotenuse, you have to undo a multiplication — which means dividing by $\sqrt{3}$ (or $2$) and rationalizing. The trick is always to recover the short leg first.

Given the Hypotenuse

Since hyp $= 2x$, the short leg is half the hypotenuse:

$x = \frac{\text{hyp}}{2}, \qquad \text{long leg} = x\sqrt{3}$

Worked Example: hypotenuse $= 14$

$\text{short leg} = \frac{14}{2} = 7, \qquad \text{long leg} = 7\sqrt{3}$

No radicals needed to get the short leg — just halve. The long leg then picks up the $\sqrt{3}$.

💡 Hypotenuse is the friendliest starting point: divide by $2$ and you instantly have the short leg.

Concept Check 🎯

Given the Long Leg

Since long leg $= x\sqrt{3}$, recover the short leg by dividing by $\sqrt{3}$ and rationalizing:

$x = \frac{\text{long leg}}{\sqrt{3}} = \frac{\text{long leg}}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = \frac{(\text{long leg})\sqrt{3}}{3}$

Worked Example: long leg $= 6$

$\text{short leg} = \frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}, \qquad \text{hyp} = 2 \cdot 2\sqrt{3} = 4\sqrt{3}$

✅ Check with Pythagoras: short$^2$ + long$^2 = (2\sqrt{3})^2 + 6^2 = 12 + 36 = 48$, and hyp$^2 = (4\sqrt{3})^2 = 48$ ✓

Recover the Short Leg 🔽

Choose how to find the short leg in each situation.

Both Backward Routes in One Table

⚠️ Dividing by $2$ keeps things rational; dividing by $\sqrt{3}$ needs a rationalize step ($\frac{a}{\sqrt{3}} = \frac{a\sqrt{3}}{3}$). Try both in the next drill.

Work Backward 🧮

Find the short leg of each $30$-$60$-$90$ triangle. Use sqrt3 form (e.g. 5sqrt3) or a whole number.

1) Hypotenuse $= 18$. Short leg $= \,?$ 2) Long leg $= 12$. Short leg $= \,?$ (rationalize; simplifies to 4sqrt3) 3) Hypotenuse $= 7$. Short leg $= \,?$ (decimal is fine)

📐 Special Right Triangles

Part 4 of 5 — Applications: Squares, Hexagons & Word Problems

🔑 Big Payoff: Special triangles are hiding inside squares, equilateral triangles, hexagons, and real-world right-angle problems. Spot the triangle and the rest falls out.

The Diagonal of a Square

A diagonal splits a square into two $45$-$45$-$90$ triangles: the sides of the square are the legs, and the diagonal is the hypotenuse.

$\text{diagonal} = (\text{side})\sqrt{2}$

Worked Example: a square with side $5$

$\text{diagonal} = 5\sqrt{2} \approx 7.07$

Worked Example: a square with diagonal $8$

$\text{side} = \frac{8}{\sqrt{2}} = \frac{8\sqrt{2}}{2} = 4\sqrt{2}$

💡 This is the fastest way to find a square's diagonal — no full Pythagorean setup required.

The Altitude of an Equilateral Triangle

Drop an altitude in an equilateral triangle of side $s$ and you create a $30$-$60$-$90$. The full side $s$ is the hypotenuse, half the base $\left(\frac{s}{2}\right)$ is the short leg, and the altitude is the long leg:

$\text{altitude} = \frac{s}{2}\cdot\sqrt{3} = \frac{s\sqrt{3}}{2}$

Worked Example: equilateral triangle with side $10$

$\text{altitude} = \frac{10\sqrt{3}}{2} = 5\sqrt{3} \approx 8.66$

💡 The area then follows: $A = \frac{1}{2}\,bh = \frac{1}{2}\cdot 10 \cdot 5\sqrt{3} = 25\sqrt{3}$.

Concept Check 🎯

A Word Problem

A ramp rises at a $30^\circ$ angle. The ramp surface (the hypotenuse) is $24$ feet long. How high is the top of the ramp?

The height is opposite the $30^\circ$ angle, so it's the short leg:

$\text{height} = \frac{\text{hyp}}{2} = \frac{24}{2} = 12 \text{ feet}$

⚠️ Watch which side faces which angle. The height is opposite the $30^\circ$, so it's the short leg ($=$ half the hypotenuse), not the long leg. Always label the angle each unknown side is across from.

Spot the Triangle 🔽

Choose the relationship hidden in each figure.

Putting the Shapes Together

💡 The first move in any of these problems is to draw the right triangle and label which side is opposite which angle. The next drill mixes all three figures.

Applications 🧮

Use sqrt2/sqrt3 form (e.g. 7sqrt2) or a whole number.

1) A square has side $11$. Its diagonal $= \,?$ 2) An equilateral triangle has side $8$. Its altitude $= \,?$ (simplifies to 4sqrt3) 3) A $30^\circ$ ramp has a $30$-ft hypotenuse. Its height (short leg, in ft) $= \,?$

📐 Special Right Triangles

Part 5 of 5 — Mixed Practice & Mastery Check

You can now solve both special triangles in either direction and spot them inside squares, equilateral triangles, and word problems. Let's put it all together.

Quick Reference

Recovering the short leg in a $30$-$60$-$90$:

⚠️ Two traps: (1) going hyp → leg you divide (the leg is smaller), and (2) the side opposite $30^\circ$ is the short leg, opposite $60^\circ$ is the long leg. Label angles before you solve.

Mixed Practice 🎯

A Three-Step Habit

Whatever you're given, the same routine solves any special right triangle:

1. Identify the triangle — equal legs $\Rightarrow$ $45$-$45$-$90$; angles $30$/$60$ $\Rightarrow$ $30$-$60$-$90$.
2. Find the base side — a leg for $45$-$45$-$90$, the short leg for $30$-$60$-$90$.
3. Scale up with the ratio ($\times\sqrt{2}$, or $\times\sqrt{3}$ and $\times 2$).

🔑 Run this routine on the drill below, then the Exit Quiz.

Mixed Drill 🧮

Use sqrt2/sqrt3 form or a whole number.

1) $45$-$45$-$90$: leg $= 9$. Hypotenuse $= \,?$ 2) $30$-$60$-$90$: short leg $= 11$. Long leg $= \,?$ 3) $30$-$60$-$90$: hypotenuse $= 30$. Short leg $= \,?$

One Last Reminder Before the Quiz

⚠️ Going from a bigger side to a smaller side, you divide — and dividing by $\sqrt{2}$ or $\sqrt{3}$ means rationalizing ($\frac{a}{\sqrt{2}} = \frac{a\sqrt{2}}{2}$, $\frac{a}{\sqrt{3}} = \frac{a\sqrt{3}}{3}$).

Keep the two ratios in front of you: $1 : 1 : \sqrt{2}$ and $1 : \sqrt{3} : 2$. Ready? Finish with the Exit Quiz.

Exit Quiz ✅

Answer all three to finish the lesson.