What Is an Obtuse Triangle?
An obtuse triangle (also called an obtuse-angled triangle) is a triangle in which one interior angle measures more than 90° but less than 180°. The other two angles are acute — each less than 90° — because the three interior angles of any triangle add to 180°. So if one angle is, say, 110°, the remaining two must share only 70° between them.
That single rule explains a lot. A triangle can hold at most one obtuse angle. Two angles above 90° would already exceed 180° on their own, which no triangle can do. By the same logic, an obtuse triangle can never also be a right triangle.
Obtuse triangles are one branch of the wider types of triangle, classified by their angles alongside acute and right triangles.
How Do You Know If a Triangle Is Obtuse?
There are two ways to tell, depending on what you are given.
By its angles. Check the largest angle. If it is more than 90°, the triangle is obtuse. If exactly 90°, it is right; if all three are below 90°, it is acute.
By its sides. When you only know the three side lengths, square them. Take the longest side as $c$ and the two shorter sides as $a$ and $b$. Then compare:
$$ \begin{aligned} a^2 + b^2 > c^2 &\quad\Rightarrow\quad \text{acute} \ a^2 + b^2 = c^2 &\quad\Rightarrow\quad \text{right} \ a^2 + b^2 < c^2 &\quad\Rightarrow\quad \text{obtuse} \end{aligned} $$
The obtuse case, written the other way round, is $c^2 > a^2 + b^2$. This is the Pythagoras theorem used as an inequality — the longest side has "overshot" what a right triangle would allow, so the angle facing it has opened past 90°.
Properties of an Obtuse Triangle
A short opener before the list: every property below follows from the one obtuse angle and the 180° angle sum.
One obtuse angle only. Exactly one angle exceeds 90°; the other two are acute.
Longest side faces the obtuse angle. The side opposite the obtuse vertex is the longest in the triangle — larger angle, longer opposite side.
The side test. For the longest side $c$, the triangle is obtuse precisely when $c^2 > a^2 + b^2$.
Angle sum holds. As in any triangle, the three interior angles total 180°, so the two acute angles add to less than 90°.
Circumcenter and orthocenter sit outside. Unlike an acute triangle, both the circumcenter (centre of the circle through all three vertices) and the orthocenter (where the altitudes meet) fall outside an obtuse triangle.
It can be scalene or isosceles, never equilateral. An equilateral triangle has three 60° angles, so none can be obtuse.
Two named cases come up often. An isosceles obtuse triangle has two equal sides with the obtuse angle wedged between them, while an acute scalene triangle is the opposite extreme — no equal sides and no obtuse angle at all. (A right scalene triangle is the right-angled cousin.)
Obtuse Triangle Formulas
The two everyday formulas are perimeter and area, and they read the same as for any triangle.
Perimeter. Add the three sides:
$$P = a + b + c$$
where $a$, $b$, $c$ are the three side lengths (in the same unit).
Area, base-and-height form.
$$\text{Area} = \frac{1}{2} \times b \times h$$
Here $b$ is the chosen base and $h$ is the perpendicular height to that base. This is where obtuse triangles surprise students: drop a perpendicular from the obtuse vertex and it lands inside the base, but drop one from an acute vertex onto a short side and the foot can land outside the triangle, on an extension of that side. The height still exists; it just sits beyond the edge.
Area, Heron's formula (when you know all three sides but no height). First compute the semi-perimeter $s$:
$$s = \frac{a + b + c}{2}$$
then
$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$
Heron's formula, attributed to Heron of Alexandria, needs no angle and no height — only the three sides — which makes it the natural tool for an obtuse triangle whose height runs outside the figure.
Examples of Obtuse Triangle
Example 1
Is a triangle with angles 30°, 40°, and 110° obtuse?
Find the largest angle: 110°.
Since 110° is greater than 90° and less than 180°, the triangle has one obtuse angle.
Check the sum: 30° + 40° + 110° = 180°. Valid.
Final answer: Yes, it is an obtuse triangle.
Example 2 (a tempting shortcut that fails)
Do the sides 6 cm, 8 cm, and 9 cm form an obtuse triangle?
Wrong attempt. A quick reflex is to add the two smaller squares and compare to the largest side rather than its square: $6^2 + 8^2 = 36 + 64 = 100$, and "100 is bigger than 9," so the triangle "must be acute." That comparison is meaningless — it pits an area-sized number (100) against a length (9).
Why it breaks. The test compares squares with squares. You must square the longest side too.
Correct. Longest side is 9, so $c^2 = 81$. The other two give $a^2 + b^2 = 36 + 64 = 100$. Compare: $100 > 81$, i.e. $a^2 + b^2 > c^2$.
Final answer: The triangle is acute, not obtuse — the side test only works once every term is squared.
Example 3
Do the sides 4 cm, 5 cm, and 7 cm form an obtuse triangle?
Longest side $c = 7$, so $c^2 = 49$.
Other two: $a^2 + b^2 = 4^2 + 5^2 = 16 + 25 = 41$.
Compare: $49 > 41$, so $c^2 > a^2 + b^2$.
Final answer: Yes — the triangle is obtuse.
Example 4
An obtuse triangle has a base of 8 in and a perpendicular height of 5 in to that base. Find its area.
$$\text{Area} = \frac{1}{2} \times b \times h$$
Substitute:
$$\text{Area} = \frac{1}{2} \times 8 \times 5$$
$$\text{Area} = 20 \text{ in}^2$$
Final answer: 20 square inches.
Example 5
Find the area of an obtuse triangle with sides 5 cm, 6 cm, and 9 cm using Heron's formula.
Semi-perimeter:
$$s = \frac{5 + 6 + 9}{2} = \frac{20}{2} = 10$$
Apply Heron's formula:
$$\text{Area} = \sqrt{10(10-5)(10-6)(10-9)}$$
$$\text{Area} = \sqrt{10 \times 5 \times 4 \times 1}$$
$$\text{Area} = \sqrt{200} \approx 14.14 \text{ cm}^2$$
Final answer: about 14.14 cm² (and the side test, $81 > 61$, confirms it is obtuse).
Example 6
Two angles of an obtuse triangle are 25° and 35°. Find the third angle and confirm the triangle is obtuse.
Subtract from 180°:
$$\angle 3 = 180° - 25° - 35° = 120°$$
Since 120° is greater than 90°, the triangle has one obtuse angle.
Final answer: The third angle is 120°, so the triangle is obtuse.
Why the Obtuse Case Earns Its Own Name
"The square on the side opposite the obtuse angle is greater."
That line is essentially Proposition 12 of Book II of Euclid's Elements — the obtuse-triangle companion to the Pythagorean theorem. The Greeks needed it because real triangles rarely arrive with a right angle, and surveyors, builders, and astronomers had to reason about the awkward, splayed-open ones too.
Where the obtuse case shows up:
Structural bracing. A widely-splayed truss spreads load across a long span; engineers compute its area and centroid using exactly the Heron and base-height tools above. Because the orthocenter lies outside the figure, the geometry of the bracing forces behaves differently from a compact acute frame.
Navigation and surveying. When a sightline between two landmarks subtends a wide angle from a third point, the triangle formed is obtuse, and the triangle sum theorem lets the surveyor recover the missing angle from the two measured ones.
The law of cosines. For an obtuse angle, $\cos$ is negative, so $c^2 = a^2 + b^2 - 2ab\cos C$ becomes $c^2 = a^2 + b^2 + (\text{a positive amount})$ — which is precisely why $c^2 > a^2 + b^2$. The side test isn't a separate rule; it is the law of cosines in disguise.
This last point connects the obtuse triangle to the full set of properties of a triangle — the same angle-and-side relationships, read in the obtuse regime.
Where Students Trip Up on Obtuse Triangles
Mistake 1: Comparing a side to a sum of squares
Where it slips in: Using the side test with all three side lengths in hand.
Don't do this: Compare $a^2 + b^2$ to the longest side $c$ instead of to $c^2$.
The correct way: Square every side. Compare $a^2 + b^2$ with $c^2$ — squares against squares, never a square against a raw length.
A specific habit fixes this: write all three squares in a row first ($a^2$, $b^2$, $c^2$), and only then compare. Students who skip straight to the inequality are the ones who quietly drop a square. The first-instinct error here is comparing $a^2 + b^2$ to $c$ rather than $c^2$ — once the three squares are written out, the mismatch is obvious.
Mistake 2: Assuming the height is always inside the triangle
Where it slips in: Finding the area of an obtuse triangle from a base and height.
Don't do this: Insist the perpendicular height must land between the two base vertices.
The correct way: Allow the foot of the altitude to fall outside the base, on its extension, when the height is dropped from an acute vertex. The height is still the perpendicular distance; it just lives outside the figure.
This is the point I had to slow down on myself when I first taught it — the picture fights the intuition. The memorizer who learned "height goes straight down inside the triangle" is exactly the student who freezes on an obtuse figure, because that mental rule was never the real definition.
Mistake 3: Thinking a triangle can have two obtuse (or an obtuse plus a right) angle
Where it slips in: Building or checking a triangle from its angles.
Don't do this: Allow two angles above 90°, or one obtuse and one right angle.
The correct way: Remember the 180° cap. One obtuse angle (say 100°) already leaves under 80° for the other two combined — a second 90°-plus angle is impossible.
Key Takeaways
An obtuse triangle has exactly one angle greater than 90°, and the longest side sits opposite it.
The side test identifies one fast: a triangle is obtuse when $c^2 > a^2 + b^2$ for the longest side $c$.
Area comes from $\tfrac{1}{2}bh$ or, when only the three sides are known, from Heron's formula.
The most common error is comparing a side to a sum of squares — always square every term first.
Unlike acute triangles, the circumcenter and orthocenter of an obtuse triangle lie outside the figure.
A Practical Next Step
Practice these problems to solidify your understanding. Take three lengths — try 7, 8, and 12 — and run the side test; then try 5, 12, 13 and notice it lands exactly on a right angle. If you get stuck on which side is $c$, return to "How Do You Know If a Triangle Is Obtuse?" and pick the longest side first, every time.
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