Triangles — scalene, isosceles, equilateral; acute, right, obtuse: Worked examples

Before we start, one tool we lean on repeatedly, built from zero:

Definition The "longest-side-versus-corner" test

Take a triangle with sides $a \leq b \leq c$ (so $c$ is the longest). Compare $c^{2}$ with $a^{2} + b^{2}$ :

If $c^{2} = a^{2} + b^{2}$ → the corner opposite $c$ is exactly $90°$ → right triangle. (This is the Pythagorean Theorem read backwards.)
If $c^{2} < a^{2} + b^{2}$ → that corner is less than $90°$ , and since it was the biggest corner, all corners are → acute.
If $c^{2} > a^{2} + b^{2}$ → that corner is more than $90°$ → obtuse.

Why does this work? Picture the two short sides $a, b$ hinged at the corner like a pair of scissors. If you open the scissors wider than a right angle, the far ends spread apart, so the gap $c$ grows — that makes $c^{2}$ overshoot $a^{2} + b^{2}$ . Squeeze the scissors tighter than $90°$ and the ends come close, $c$ shrinks, $c^{2}$ falls short. The right angle is the exact break-even point. We only ever test the longest side because the biggest corner is always opposite it, and only the biggest corner can reach or pass $90°$ .

The scenario matrix

Cell	What it tests	Example
A — Scalene / acute	all sides different, all angles $< 90°$	Ex 1
B — Scalene / right	different sides, one $90°$ (the $3, 4, 5$ )	Ex 2
C — Scalene / obtuse	different sides, one angle $> 90°$	Ex 3
D — Isosceles / apex geometry	drop altitude, find base angles + area	Ex 4
E — Isosceles right (45-45-90)	two equal legs, $90°$ apex	Ex 5
F — Equilateral / area from side	all equal, use $\frac{a ^{2} 3}{4}$	Ex 6
G — Degenerate / does-not-exist	Triangle Inequality fails or is tight	Ex 7
H — Real-world word problem	ladder against wall, hidden right triangle	Ex 8
I — Exam twist / classify by angles only	given angles, name side + angle class	Ex 9

Example 1 — Cell A: Scalene, acute

Forecast: all sides look different (scalene, yes) — but is it acute or does that $8$ sneak past $90°$ ? Guess before reading.

Order the sides. $6 \leq 7 \leq 8$ , so the longest is $c = 8$ . Why this step? The test above only needs the longest side; ordering finds it in one glance.
Compute $c^{2}$ and $a^{2} + b^{2}$ . $c^{2} = 8^{2} = 64$ . And $a^{2} + b^{2} = 6^{2} + 7^{2} = 36 + 49 = 85$ . Why this step? This is the scissors comparison — it decides the biggest angle in one subtraction.
Compare. $64 < 85$ , so $c^{2} < a^{2} + b^{2}$ → biggest angle is under $90°$ → acute. Why this step? Since the largest angle is acute, every smaller angle is too.
Side class. $6, 7, 8$ all different → scalene.

Verify: all three sides distinct ✓. Gap $85 - 64 = 21 > 0$ means genuinely acute, not a knife-edge right triangle. Units are cm throughout, consistent. → scalene acute.

Example 2 — Cell B: Scalene, right

Forecast: you've probably met $3, 4, 5$ . Right triangle? And the area — guess it before dividing.

Figure — Triangles — scalene, isosceles, equilateral; acute, right, obtuse

Order + test. $c = 5$ , $c^{2} = 25$ ; $a^{2} + b^{2} = 9 + 16 = 25$ . Equal! → exactly $90°$ → right. Why this step? Equality is the break-even point of the scissors: neither open nor squeezed.
Identify the hypotenuse. The side opposite the $90°$ corner is $5$ — the longest, as it must be. Why this step? Naming the hypotenuse tells us which two sides are the legs (the ones forming the right angle).
Area = half base × height. The two legs $3$ and $4$ meet at the right angle, so one is the base and the other is the height: $A = \frac{1}{2} \times 3 \times 4 = 6 cm^{2} .$ Why this step? When the corner is $90°$ , the two legs are already perpendicular — no need to hunt for an altitude, the legs are base and height.

Verify: $3, 4, 5$ all different → scalene ✓. $c^{2} = a^{2} + b^{2}$ exactly → right ✓. Area $6 cm^{2}$ , units area-squared ✓. → scalene right, area $6 cm^{2}$ .

Example 3 — Cell C: Scalene, obtuse

Forecast: that $8$ is much bigger than the other two. Will the scissors open past $90°$ ?

Order + test. $c = 8$ , $c^{2} = 64$ ; $a^{2} + b^{2} = 16 + 25 = 41$ . Why this step? Longest side is the only candidate for a big angle.
Compare. $64 > 41$ → $c^{2} > a^{2} + b^{2}$ → biggest angle exceeds $90°$ → obtuse. Why this step? Overshoot means the scissors opened wide — an obtuse corner opposite side $8$ .
Side class. $4, 5, 8$ distinct → scalene.

Verify: overshoot $64 - 41 = 23 > 0$ confirms obtuse (not right). Also check it's even a valid triangle: Triangle Inequality needs $4 + 5 = 9 > 8$ ✓ — it exists. → scalene obtuse.

Example 4 — Cell D: Isosceles apex geometry

Forecast: the two equal sides lean in like a tent. Guess the height (hint: it's a whole number).

Drop an altitude from the apex. Because the two sides are equal, the triangle has a mirror line straight down the middle (see Symmetry in Geometry). That altitude splits the base $6$ exactly in half → each half is $3$ . Why this step? Symmetry hands us two identical right triangles for free — much easier than the slanted whole.
Pythagorean theorem on one half. The half-triangle has hypotenuse $5$ (the equal side), one leg $3$ (half base), the other leg $h$ (height): $h^{2} + 3^{2} = 5^{2} \Rightarrow h^{2} = 25 - 9 = 16 \Rightarrow h = 4.$ Why this step? The altitude created a right angle, unlocking Pythagoras — our only clean tool for a right triangle's missing side.
Base angle $α$ . In the right half-triangle, the base angle sits opposite the height $4$ with adjacent side $3$ . Using a trig ratio, $tan α = \frac{opposite}{adjacent} = \frac{4}{3}$ , so $α = arctan \frac{4}{3} \approx 53.13°$ . Why $tan$ ? We know the two legs (opposite and adjacent to $α$ ) but not the hypotenuse's role in the angle — $tan$ is exactly "opposite over adjacent," the ratio that these two known sides form.
Area. Base $6$ , height $4$ : $A = \frac{1}{2} \times 6 \times 4 = 12 cm^{2} .$

Verify: two sides $5$ , base $6$ , all not equal → genuinely isosceles ✓. Apex angle $= 180° - 2 (53.13°) \approx 73.74°$ , all angles under $90°$ so it's also acute. Height $4$ cm, area $12 cm^{2}$ ✓.

Example 5 — Cell E: Isosceles right (45-45-90)

Forecast: two equal legs + one right angle. Guess the other two angles before computing.

Hypotenuse by Pythagoras. Legs $7, 7$ : $c^{2} = 7^{2} + 7^{2} = 49 + 49 = 98 \Rightarrow c = 98 = 72 \approx 9.90.$ Why this step? The two legs meet at the $90°$ corner, so they are the perpendicular pair Pythagoras needs.
Base angles. Equal legs face equal angles (angles opposite equal sides are equal). Call each $θ$ : $θ + θ + 90° = 180° \Rightarrow 2 θ = 90° \Rightarrow θ = 45°.$ Why this step? Isosceles symmetry forces the two non-right angles to match, and the $180°$ total pins their value.

Verify: angles $45 + 45 + 90 = 180$ ✓. Hypotenuse $72 \approx 9.90 >$ each leg ✓ (longest side, as required). Side pattern $a, a, a 2$ matches the parent's 45-45-90 result ✓. → isosceles right.

Example 6 — Cell F: Equilateral area from side

Forecast: all sides $10$ , all angles $60°$ . The area is not $50$ — guess how much less.

Use the derived formula. From the parent note, an equilateral triangle of side $a$ has $A = \frac{a ^{2} 3}{4} .$ Why this step? We already proved this by dropping an altitude and using Pythagoras; no need to redo it.
Substitute $a = 10$ . $A = \frac{1 0 ^{2} 3}{4} = \frac{100 3}{4} = 253 \approx 43.30 cm^{2} .$ Why this step? Direct plug-in — the whole point of having a formula.

Verify: all angles $60° < 90°$ so it's also acute ✓. $253 \approx 43.3$ , comfortably less than $\frac{1}{2} \cdot 10 \cdot 10 = 50$ (which would need a full right angle between the two sides) ✓. → equilateral acute, $253 cm^{2}$ .

Example 7 — Cell G: Degenerate / cannot exist

Forecast: guess yes or no for each before reading.

State the Triangle Inequality. For three sides to close into a triangle, each side must be shorter than the sum of the other two. Why this step? If one side is as long as (or longer than) the other two laid end-to-end, they can't reach across to meet — the "triangle" flattens into a line or fails entirely.
Test $2, 3, 5$ . Longest is $5$ ; the other two sum to $2 + 3 = 5$ . Here $5 = 5$ — exactly equal. Why this step? Equality is the knife-edge: the two short sides lie flat along the long one, all three points on a straight line. This is a degenerate triangle — zero height, zero area.
Test $2, 3, 6$ . Longest is $6$ ; the other two sum to $2 + 3 = 5 < 6$ . The short sides can't even reach → no triangle exists. Why this step? Sum falls short of the longest side — the ends never touch.

Verify: $2, 3, 5$ : $2 + 3 = 5$ , area $= 0$ (collapsed) ✓. $2, 3, 6$ : $5 < 6$ , impossible ✓. A valid triangle needs a strict inequality, e.g. $2, 3, 4$ works since $2 + 3 = 5 > 4$ .

Example 8 — Cell H: Real-world word problem

Forecast: the ladder, wall, and ground make a triangle. Guess the height (it's a whole number).

Find the hidden right angle. The wall is vertical, the ground horizontal — they meet at $90°$ . So the ground ( $5$ m), wall (height $h$ ), and ladder ( $13$ m) form a right triangle with the ladder as hypotenuse. Why this step? Spotting the right angle is what lets us use Pythagoras at all; the ladder, being opposite that corner, must be the longest side.
Pythagoras for the missing leg. $h^{2} + 5^{2} = 1 3^{2} \Rightarrow h^{2} = 169 - 25 = 144 \Rightarrow h = 12 m .$ Why this step? We know the hypotenuse and one leg; Pythagoras solves for the other leg.
Classify. Sides $5, 12, 13$ all different → scalene; one $90°$ → right.

Verify: $5^{2} + 1 2^{2} = 25 + 144 = 169 = 1 3^{2}$ ✓ (a valid right triangle). Height $12$ m is below the ladder length $13$ m, physically sensible ✓. Units metres throughout ✓. → the ladder reaches $12$ m; scalene right triangle.

Example 9 — Cell I: Exam twist, classify from angles only

Forecast: you're given only angles. Guess the third angle and the two classifications.

Find the third angle. Angles sum to $180°$ (Interior Angles of Polygons for $n = 3$ ): $180° - 35° - 35° = 110°.$ Why this step? The third angle is forced — this single subtraction unlocks everything.
Angle class. One angle $110° > 90°$ → obtuse. Why this step? Any angle over $90°$ makes the whole triangle obtuse (and only one can be).
Side class. Two equal angles ( $35°, 35°$ ) sit opposite two equal sides → isosceles. Why this step? "Angles opposite equal sides are equal" runs both ways: equal angles ⇒ equal opposite sides.
Equilateral? Equilateral needs every angle $= 60°$ . Here we have $35°$ and $110°$ — impossible. Why this step? Equilateral is fixed by the $60$ - $60$ - $60$ pattern; any other angle set rules it out.

Verify: $35 + 35 + 110 = 180$ ✓. Largest angle $110°$ obtuse ✓. Two equal base angles ⇒ isosceles ✓. Not $60$ - $60$ - $60$ ⇒ never equilateral ✓. → isosceles obtuse.

Recall Quick self-test

Sides $6, 7, 8$ : acute, right or obtuse? ::: Acute — $64 < 85$ . Sides $4, 5, 8$ : which class by angle? ::: Obtuse — $64 > 41$ . Can sides $2, 3, 5$ form a triangle? ::: No (degenerate) — $2 + 3 = 5$ , it collapses to a line. Isosceles triangle, sides $5, 5, 6$ : its height to the base? ::: $4$ — since $3^{2} + 4^{2} = 5^{2}$ . Ladder $13$ m, foot $5$ m out: height reached? ::: $12$ m — since $5^{2} + 1 2^{2} = 1 3^{2}$ . Angles $35°, 35°, ?$ : name it. ::: Isosceles obtuse (third angle $110°$ ).