The Formulas
For a right triangle with legs $a$ and $b$, hypotenuse $c$ (the side opposite the right angle, always the longest side), and the two acute angles $\angle A$ and $\angle B$:
$$\boxed{;\begin{aligned}\text{Pythagorean theorem:}\quad & a^2 + b^2 = c^2\ \text{Area:}\quad & \tfrac{1}{2},b,a\ \text{Perimeter:}\quad & a + b + c\ \sin A = \tfrac{\text{opposite}}{\text{hypotenuse}},\quad & \cos A = \tfrac{\text{adjacent}}{\text{hypotenuse}},\quad \tan A = \tfrac{\text{opposite}}{\text{adjacent}}\end{aligned};}$$
Each variable points to a part of the figure above. $a$ and $b$ are the two legs — the sides that meet at the right angle. $c$ is the hypotenuse — opposite the right angle, and always the longest side. $\angle A$ and $\angle B$ are the two acute angles, and they always add to $90°$, because all three angles of any triangle sum to $180°$ and one of them is already $90°$.
The trig memory aid is SOHCAHTOA: Sin = Opposite over Hypotenuse, Cos = Adjacent over Hypotenuse, Tan = Opposite over Adjacent. "Opposite" and "adjacent" are read relative to whichever acute angle you are working with — the side touching the angle (other than the hypotenuse) is adjacent, the side across from it is opposite.
How the Pythagorean Theorem Is Derived
The Pythagorean theorem is not a rule you memorise on faith. It comes from a rearrangement of area.
Take four identical copies of the right triangle, each with legs $a$ and $b$ and hypotenuse $c$. Arrange them inside a large square of side $(a + b)$ so their hypotenuses form a tilted inner square of side $c$.
The big square's area can be counted two ways. As one square: $(a + b)^2 = a^2 + 2ab + b^2$. As four triangles plus the inner square: $4 \cdot \tfrac{1}{2}ab + c^2 = 2ab + c^2$. Set them equal:
$$a^2 + 2ab + b^2 = 2ab + c^2.$$
The $2ab$ cancels from both sides, leaving:
$$a^2 + b^2 = c^2.$$
That is the whole proof — no trigonometry, just area counted twice. The law of cosines is the same idea generalised to any triangle; for a right triangle the cosine term vanishes and you are back to Pythagoras. (We cross-link the full proof and properties in our concept guide to the right-angled triangle.)
How Do You Find the Area of a Right Triangle?
In a right triangle the two legs are already perpendicular, so one leg is the base and the other is the height — no separate altitude to hunt for. That is why the area formula is simply
$$\text{Area} = \frac{1}{2} \cdot b \cdot a,$$
half the product of the two legs. For a non-right triangle you would have to drop a perpendicular first; the right angle hands you the height for free.
Examples of Right Triangle Formulas
Example 1
A right triangle has legs of 6 cm and 8 cm. Find the hypotenuse.
By the Pythagorean theorem:
$$c^2 = 6^2 + 8^2 = 36 + 64 = 100,$$
so $c = \sqrt{100} = 10$ cm.
Final answer: $c = 10$ cm.
Example 2
A right triangle has hypotenuse 13 m and one leg 5 m. Find the other leg.
Here the unknown is a leg, not the hypotenuse, so the missing side is not the square root of a sum — it is the square root of a difference. Rearrange Pythagoras:
$$a^2 = c^2 - b^2 = 13^2 - 5^2 = 169 - 25 = 144,$$
so $a = \sqrt{144} = 12$ m.
Final answer: $a = 12$ m.
Example 3
A right triangle has legs 9 in and 12 in. Find its area and perimeter.
The legs are perpendicular, so they serve as base and height:
$$\text{Area} = \frac{1}{2} \cdot 9 \cdot 12 = 54 \text{ in}^2.$$
For the perimeter, find the hypotenuse first: $c = \sqrt{9^2 + 12^2} = \sqrt{225} = 15$ in. Then
$$\text{Perimeter} = 9 + 12 + 15 = 36 \text{ in}.$$
Final answer: Area $= 54 \text{ in}^2$, perimeter $= 36$ in.
Example 4
In a right triangle, the angle $\angle A$ faces a side of 7 cm and the hypotenuse is 25 cm. Find $\sin A$, then the angle.
The side facing $\angle A$ is the opposite side, so
$$\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{7}{25} = 0.28.$$
Taking the inverse sine, $\angle A = \sin^{-1}(0.28) \approx 16.3°$.
Final answer: $\sin A = 0.28$, $\angle A \approx 16.3°$.
Example 5
A 45-45-90 triangle has legs of 5 units each. Find the hypotenuse and the area.
A 45-45-90 triangle is an isosceles right triangle — two equal legs, two $45°$ angles. Its sides follow the fixed ratio $1 : 1 : \sqrt{2}$, so the hypotenuse is a leg times $\sqrt{2}$:
$$c = 5\sqrt{2} \approx 7.07 \text{ units}, \qquad \text{Area} = \frac{1}{2} \cdot 5 \cdot 5 = 12.5 \text{ units}^2.$$
This same shape sits inside our guide to the isosceles right triangle.
Final answer: $c = 5\sqrt{2} \approx 7.07$ units, area $= 12.5$ units².
Example 6
A ramp must rise 1.5 m over a horizontal run of 6 m. Find the ramp's length and the angle it makes with the ground.
The rise and run are the two legs of a right triangle; the ramp itself is the hypotenuse:
$$c = \sqrt{1.5^2 + 6^2} = \sqrt{2.25 + 36} = \sqrt{38.25} \approx 6.18 \text{ m}.$$
The angle of inclination uses tangent, since rise and run are opposite and adjacent:
$$\tan \theta = \frac{1.5}{6} = 0.25, \qquad \theta = \tan^{-1}(0.25) \approx 14°.$$
Final answer: Ramp length $\approx 6.18$ m at about $14°$ to the ground.
Where Right Triangle Formulas Show Up
The right triangle is the workhorse of applied geometry, because any sloped or angled measurement can be split into a horizontal leg and a vertical leg.
Construction and roofing. Roof pitch is rise over run — a tangent ratio. Rafter length is the hypotenuse, computed straight from Pythagoras.
Navigation and surveying. Bearings and distances resolve into right triangles; the trigonometric ratios turn an angle and one distance into the others.
Screens and pixels. A "55-inch TV" quotes the diagonal — the hypotenuse of the right triangle formed by the screen's width and height.
Physics. Any force or velocity at an angle is broken into perpendicular components using $\sin$ and $\cos$ — the same two legs of a right triangle.
For a Grade 9 or 10 student, the most-met context is the coordinate-geometry distance formula, which is just Pythagoras applied to the horizontal and vertical gaps between two points. The right triangle is hiding inside almost every "find the distance" or "find the angle" problem you will ever solve.
Special Right Triangles — The Two Ratios Worth Memorising
Two right triangles appear so often that their side ratios are worth knowing cold, so you never re-derive them mid-problem.
45-45-90 (isosceles right triangle). Two equal legs, sides in ratio $1 : 1 : \sqrt{2}$. The hypotenuse is always a leg times $\sqrt{2}$.
30-60-90. Sides in ratio $1 : \sqrt{3} : 2$. The side opposite $30°$ is the shortest; the side opposite $60°$ is $\sqrt{3}$ times it; the hypotenuse is twice the shortest side.
Where Students Lose the Mark on Right Triangle Formulas
Mistake 1: Treating a leg like the hypotenuse
Where it slips in: When the unknown side is a leg, not the hypotenuse, but the student reflexively adds the two known squares.
Don't do this: Write $c = \sqrt{13^2 + 5^2}$ when 13 is actually the hypotenuse and 5 is a leg. That gives $\sqrt{194} \approx 13.9$, which is longer than the hypotenuse — impossible, since the hypotenuse is the longest side.
The correct way: Identify the hypotenuse first (it is opposite the right angle, and longest). If the hypotenuse is known and a leg is missing, subtract: $a^2 = c^2 - b^2$.
Mistake 2: Forgetting the square root
Where it slips in: After correctly computing $a^2 + b^2$, the student reports that value as the answer and stops.
Don't do this: Write $c = 100$ for legs 6 and 8. You found $c^2 = 100$, not $c$.
The correct way: Take the square root at the end: $c = \sqrt{100} = 10$. The memorizer who recites "a-squared plus b-squared equals c-squared" is the one most likely to stop a step early — the formula gives $c^2$, and the question almost always wants $c$.
Mistake 3: Mixing up opposite and adjacent
Where it slips in: Setting up a trig ratio without anchoring "opposite" and "adjacent" to a specific angle.
Don't do this: Use the same side as "opposite" for both acute angles. The side opposite $\angle A$ is adjacent to $\angle B$ — they swap when you switch reference angles.
The correct way: Pick the angle, then read the figure: the side across from it is opposite, the leg touching it is adjacent, and the longest side is always the hypotenuse. SOHCAHTOA only works once you have fixed which angle you mean.
Key Takeaways
The right triangle formulas start from the Pythagorean theorem $a^2 + b^2 = c^2$, which comes from counting one square's area two ways.
Area is $\frac{1}{2} \times \text{leg} \times \text{leg}$, because the two legs are already perpendicular.
The trig ratios sin, cos, and tan turn any acute angle plus one side into all the others (SOHCAHTOA).
The 45-45-90 ($1:1:\sqrt{2}$) and 30-60-90 ($1:\sqrt{3}:2$) ratios are the two special cases worth memorising.
The costliest mistake is treating a leg as the hypotenuse — always identify the longest, right-angle-facing side first.
Practice These Problems to Solidify Your Understanding
A right triangle has legs 20 cm and 21 cm. Find the hypotenuse, then the area.
A right triangle has hypotenuse 17 m and one leg 8 m. Find the other leg.
In a 30-60-90 triangle the shortest side is 4 cm. Find the other two sides.
Answer to Question 1: hypotenuse $= 29$ cm, area $= 210$ cm². Answer to Question 2: other leg $= 15$ m. Answer to Question 3: $4\sqrt{3} \approx 6.93$ cm and $8$ cm. If Question 2 gave you a number larger than 17, you added the squares instead of subtracting — return to Mistake 1.
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