SSC CGL Maths · Triangles

Centres of a Triangle

The centroid, incentre, circumcentre and orthocentre are where the medians, angle bisectors, perpendicular bisectors and altitudes meet; the in- and circumradius each have a short formula.

All Triangles notes

Why this matters

Fifteen questions in the bank, 14 EASY and 1 MODERATE, and nine of them come from the 2025 shifts. Four ask where a centre lies, three ask the angle at the incentre, and eight ask for a radius, nearly always of a right or an equilateral triangle. Know the right-triangle and equilateral formulas cold and these are ten-second questions.

Concept 1 of 4: The four centres of a triangle and where each one lies

Each centre is where three special lines cross. Two of them always stay inside the triangle; the other two move outside when one angle opens past 90°.

Definition

Centroid (G): where the three medians meet. It divides each median 2 : 1.
Incentre (I): where the three angle bisectors meet. It is the same distance from all three sides, so it is the centre of the inscribed circle (the incircle).
Circumcentre (O): where the perpendicular bisectors of the sides meet. It is the same distance from all three corners, so it is the centre of the circle through them (the circumcircle).
Orthocentre (H): where the three altitudes meet.
In an equilateral triangle all four are the same point.

Centre	Lines that meet there	Acute triangle	Right triangle	Obtuse triangle
Centroid	the three medians	inside	inside	inside
Incentre	the three angle bisectors	inside	inside	inside
Circumcentre	the perpendicular bisectors of the sides	inside	at the midpoint of the hypotenuse	outside, beyond the longest side The circumradius of a right triangle is half its hypotenuse.
Orthocentre	the three altitudes	inside	at the corner with the right angle	outside, beyond the obtuse corner The two legs of a right triangle are altitudes of each other, so they meet at the right angle.

Centroid and incentre never leave the triangle; circumcentre and orthocentre do.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 24 Sep 2025, Shift 3 · Q73Easy

Example 1 · Triangles · Centres of a Triangle

If a triangle has an obtuse angle, where is its Circumcenter located?

Swapping the circumcentre and the orthocentre of a right triangle

In a right triangle the circumcentre is the midpoint of the hypotenuse and the orthocentre is the right-angle corner.

Making the incentre equidistant from the corners

The incentre is the same distance from the three sides. The point the same distance from the three corners is the circumcentre.

Drill 3 more on the four centres of a triangle and where each one lies

Concept 2 of 4: The angle at the incentre between two angle bisectors

Triangle BIC keeps only half of angles B and C. Half of B + C is 90° minus half of A, so the angle at I is 90° plus half of A.

Definition

The bisectors of $\angle B$ and $\angle C$ meet at the incentre I. Then $\angle BIC = 90^\circ + \frac{\angle A}{2}$ .
Why: in triangle BIC, $\angle BIC = 180^\circ - \frac{\angle B + \angle C}{2} = 180^\circ - \frac{180^\circ - \angle A}{2}$ .
Working back: $\angle A = 2(\angle BIC - 90^\circ)$ .
The angle a side makes at the incentre is always more than 90°, by half of the angle opposite that side.

Angle at the incentre

\angle BIC = 90^\circ + \frac{\angle A}{2}

Ithe incentre, where the bisectors of angles B and C meet

Worked example

In triangle ABC, the bisectors of

\angle B

and

\angle C

meet at I, and

\angle A = 64^\circ

. Find

\angle BIC

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 9 Sep 2024, Shift 3 · Q63Easy

Example 2 · Triangles · Centres of a Triangle

In a

\Delta PQR

, the bisectors of

\angle Q

and

\angle R

meet at point O, inside the triangle. If

\angle QOR = 107^\circ

, then the measure of

\angle P

is:

Adding the whole of the opposite angle

The angle at the incentre is 90° plus half the opposite angle. With ∠A = 60° it is 120°, not 150°.

Using the wrong angle of the triangle

The formula for ∠BIC uses ∠A, the corner whose bisector is not involved. Angles B and C have already been halved inside triangle BIC.

Drill 2 more on the angle at the incentre between two angle bisectors

Concept 3 of 4: Inradius: the area divided by the semi-perimeter

Join the incentre to the three corners and the triangle splits into three triangles, each with height r on one side as base. Their areas add to r times half the perimeter.

Definition

The inradius r is the distance from the incentre to any side.
Area $= r \times s$ , where $s = \frac{a + b + c}{2}$ is the semi-perimeter. So $r = \frac{\text{Area}}{s} = \frac{2 \times \text{Area}}{a + b + c}$ .
Right triangle with legs $a$ , $b$ and hypotenuse $c$ : $r = \frac{a + b - c}{2}$ . This comes from the equal tangents from each corner.
Equilateral triangle of side $a$ : $r = \frac{a}{2\sqrt{3}}$ , one third of the height.
Working back: with r and one leg known in a right triangle, write the hypotenuse in terms of the other leg and use Pythagoras.

Inradius

r = \frac{\text{Area}}{s} \qquad s = \frac{a + b + c}{2} \qquad r_{\text{right}} = \frac{a + b - c}{2}

sthe semi-perimeter
cthe hypotenuse, in the right-triangle form

Worked example

Find the inradius of a right triangle with legs 8 cm and 15 cm.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 19 Sep 2025, Shift 1 · Q74Easy

Example 3 · Triangles · Centres of a Triangle

A circle is inscribed in a right triangle with legs of length 5 and 12. What is the radius of the circle?

Dividing the area by the full perimeter

The inradius is the area divided by the semi-perimeter. Dividing by the whole perimeter gives half the radius.

Adding the hypotenuse in the right-triangle form

For a right triangle r = (a + b − c) ÷ 2: the two legs minus the hypotenuse. Adding all three sides gives the semi-perimeter instead.

Drill 4 more on inradius: the area divided by the semi-perimeter

Concept 4 of 4: Circumradius of a right, an equilateral and any triangle

The circle through the three corners of a right triangle has the hypotenuse as a diameter, so its radius is half the hypotenuse. For an equilateral triangle the centre is the centroid, two thirds of the way down a median.

Definition

The circumradius R is the distance from the circumcentre to each corner.
Right triangle: $R = \frac{\text{hypotenuse}}{2}$ . Check for a Pythagorean triple first.
Equilateral triangle of side $a$ : $R = \frac{a}{\sqrt{3}}$ , two thirds of the height, and $r = \frac{a}{2\sqrt{3}}$ , so R : r = 2 : 1.
Any triangle: $R = \frac{abc}{4 \times \text{Area}}$ .

Circumradius

R = \frac{abc}{4 \times \text{Area}} \qquad R_{\text{right}} = \frac{\text{hypotenuse}}{2} \qquad R_{\text{equilateral}} = \frac{a}{\sqrt{3}}

a, b, cthe three sides

Worked example

Find the circumradius of a triangle with sides 20 cm, 21 cm and 29 cm.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 17 Sep 2025, Shift 2 · Q67Easy

Example 4 · Triangles · Centres of a Triangle

In an equilateral triangle with side length 'a', what is the ratio of the circumradius to the inradius?

Halving a leg instead of the hypotenuse

The circumradius of a right triangle is half the hypotenuse, the longest side. Half of a leg is not a radius of anything here.

Swapping R and r in an equilateral triangle

The circumradius a ÷ √3 is twice the inradius a ÷ (2√3). The circle through the corners is the bigger one.

Drill 2 more on circumradius of a right, an equilateral and any triangle

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The angle at the incentre between two angle bisectors
Angle at the incentre
$\angle BIC = 90^\circ + \frac{\angle A}{2}$
Inradius: the area divided by the semi-perimeter
Inradius
$r = \frac{\text{Area}}{s} \qquad s = \frac{a + b + c}{2} \qquad r_{\text{right}} = \frac{a + b - c}{2}$
Circumradius of a right, an equilateral and any triangle
Circumradius
$R = \frac{abc}{4 \times \text{Area}} \qquad R_{\text{right}} = \frac{\text{hypotenuse}}{2} \qquad R_{\text{equilateral}} = \frac{a}{\sqrt{3}}$

Reference tables (1)

The four centres of a triangle and where each one lies4 rows

Centre	Lines that meet there	Acute triangle	Right triangle	Obtuse triangle
Centroid	the three medians	inside	inside	inside
Incentre	the three angle bisectors	inside	inside	inside
Circumcentre	the perpendicular bisectors of the sides	inside	at the midpoint of the hypotenuse	outside, beyond the longest side The circumradius of a right triangle is half its hypotenuse.
Orthocentre	the three altitudes	inside	at the corner with the right angle	outside, beyond the obtuse corner The two legs of a right triangle are altitudes of each other, so they meet at the right angle.

Centroid and incentre never leave the triangle; circumcentre and orthocentre do.

Watch out for (8)

Swapping the circumcentre and the orthocentre of a right triangle
→ The four centres of a triangle and where each one lies
Making the incentre equidistant from the corners
→ The four centres of a triangle and where each one lies
Adding the whole of the opposite angle
→ The angle at the incentre between two angle bisectors
Using the wrong angle of the triangle
→ The angle at the incentre between two angle bisectors
Dividing the area by the full perimeter
→ Inradius: the area divided by the semi-perimeter
Adding the hypotenuse in the right-triangle form
→ Inradius: the area divided by the semi-perimeter
Halving a leg instead of the hypotenuse
→ Circumradius of a right, an equilateral and any triangle
Swapping R and r in an equilateral triangle
→ Circumradius of a right, an equilateral and any triangle

Test yourself on Triangles

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