A right triangle is the one shape whose area needs no construction at all. The two legs meet at a right angle, so each is already the perpendicular height above the other, and the area is half their product. Every other triangle makes you find a height first; this one hands it to you. Pythagoras th...
A right triangle is the one case where the base and the height are handed to you. The two legs meet at a right angle, so each is already the perpendicular height above the other and the area is half their product — nothing to construct, nothing to measure twice. Pythagoras then gives whichever side you are missing, exactly: legs of 1 and 1 have hypotenuse √2, not 1.414. The animation below is the proof of that theorem rather than a picture of it.
Any two facts about the triangle. Whole numbers, decimals and fractions all work.
LEG a
LEG b
AREA
6
Exactly 6. All three sides are whole numbers, which makes this a Pythagorean triple — uncommon enough to be worth noticing.
| Leg a | 3 | 3 |
| Leg b | 4 | 4 |
| Hypotenuse | 5 | 5 |
| Area | 6 | 6 |
THE ANSWER PUT BACK
a² + b² = 9 + 16 = 25 ✓
the altitude 12/5 × c ÷ 2 → 6 ✓
HYPOTENUSE
5
the side opposite the right angle
ALTITUDE TO IT
12/5
ab ÷ c — shorter than either leg
INRADIUS
1
(a + b − c) ÷ 2, for a right triangle
PYTHAGOREAN TRIPLE
yes
all three sides are whole numbers
WHAT IS HAPPENING
A = 3 × 4 ÷ 2 = 6
In a right triangle no construction is needed. The two legs meet at a right angle, so each one is already the perpendicular height above the other — which is why this is the one triangle whose area needs no extra measurement.
c = √(3² + 4²) = √25 = 5
The sum of the two squares is itself a square, so the hypotenuse is a whole number and these three sides form a Pythagorean triple. That is uncommon.
Why that rearrangement is a proof. The outer square has side a + b and never changes. Inside it sit four copies of this triangle, which slide but are never stretched, turned over or resized. So the area they cover is the same at the start and the end, and therefore the area they leave uncovered is the same too. At the start that uncovered gap is a single tilted square whose side is the hypotenuse — area c². At the end it is two upright squares of sides a and b — area a² + b². Two names for one quantity, which is the theorem.
Exact values throughout · surds kept as surds, never rounded away
Choose which two facts you have — two legs, a leg with the hypotenuse, a leg with an acute angle, or the area with a leg — and fill in the boxes.
Read the decimal first and the exact value beneath it. Where the hypotenuse is a surd, that surd is the answer and the decimal is a rounding.
Press replay and watch the four pieces slide. They never turn or change size, so the gap they leave cannot change either — that is the whole proof of a² + b² = c².
Check the extras below the fold: the altitude to the hypotenuse, the inradius and the circumradius all take simpler forms here than in any other triangle.
Take legs of 3 and 4. The area needs nothing extra: A = 3 × 4 ÷ 2 = 6. The legs meet at a right angle, so 3 is the base and 4 is the height above it — or the other way round, which gives the same product. Pythagoras gives the third side: c = √(3² + 4²) = √(9 + 16) = √25 = 5. Twenty-five happens to be a square number, so the hypotenuse is a whole number and 3-4-5 is a Pythagorean triple. That is the exception rather than the rule. Change one leg and it stops being tidy. Legs of 2 and 3 give c = √13, which is irrational: 3.6056 is a rounding, and √13 is the answer. Legs of 1 and 1 give c = √2, the number whose irrationality broke the Pythagorean school's own philosophy. Now backwards. Given a leg of 3 and a hypotenuse of 5, rearrange: b = √(5² − 3²) = √16 = 4, and the area is 6 again. Give a hypotenuse of 3 with a leg of 5 instead and there is no triangle at all — the subtraction under the root turns negative, because the hypotenuse is always the longest side. The extras, on the 3-4-5. The altitude from the right angle to the hypotenuse is ab ÷ c = 12 ÷ 5 = 2.4, shorter than either leg as it always is. The inradius is (3 + 4 − 5) ÷ 2 = 1, a formula that works for right triangles and no others. The circumradius is 5 ÷ 2 = 2.5, half the hypotenuse, because the hypotenuse is a diameter of the circle through all three corners — which is Thales' theorem read backwards. One warning if you ever code this yourself. Computing c as √(a² + b²) squares the inputs first, and the squares overflow long before the answer would. At legs of 10¹⁶⁰ the squares become Infinity in double precision and the formula returns Infinity, while the true hypotenuse is about 1.4142 × 10¹⁶⁰ and perfectly representable. Across a set built to test that, the textbook arrangement failed on 94% of the pairs.
| Rule | What it says | Why |
|---|---|---|
| Area | A = ab ÷ 2 | The two legs are the base and the height. No perpendicular has to be constructed. |
| Why no construction | the legs meet at 90° | Each leg is already the perpendicular height above the other. |
| Pythagoras | a² + b² = c² | Any two sides give the third. The animation above is the proof. |
| Finding a leg | b = √(c² − a²) | A negative under the root means the hypotenuse was not the longest side. |
| From an acute angle | b = a·tan θ | Exact only at 30°, 45° and 60°; a decimal everywhere else. |
| Altitude to the hypotenuse | h = ab ÷ c | Shorter than either leg, always. Rational in only 1.86% of whole-number cases. |
| Inradius | r = (a + b − c) ÷ 2 | This simple form holds for right triangles alone. |
| Circumradius | R = c ÷ 2 | The hypotenuse is a diameter of the circle through all three corners. |
| Pythagorean triples | 1.25% of leg pairs | Whole-number legs give a whole-number hypotenuse surprisingly rarely. |
| The smallest few | 3-4-5, 5-12-13, 8-15-17 | 6-8-10 is 3-4-5 doubled, so it is not a new shape. |
| The 45° case | c = a√2 | Equal legs. The hypotenuse is never a whole number when the legs are. |
| The 30-60 case | sides 1, √3, 2 | The half of an equilateral triangle, and the source of those exact tangents. |
| Squaring overflows first | √(a² + b²) fails early | At legs of 1e160 the squares overflow while the answer is representable. |
| Units | legs in cm give cm² | An area answer without a squared unit means the wrong formula was used. |
Multiply the two legs and halve the result: A = ab ÷ 2. The legs are the two sides that meet at the right angle, and because they are perpendicular each one is already the height above the other. No extra measurement or construction is needed, which is not true of any other triangle.
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Last updated: September 1, 2026 · Exact values throughout, surds kept as surds · In a right triangle the two legs are already the base and the height.