Triangles and their centres — notes, question types & practice · Geometry

01

The angle rules

Inside angles always add to $180^\circ$ . An exterior angle is made by extending one side. It equals the sum of the two remote interior angles, the two far away from it. So the exterior angle at $A$ equals $B+C$ .

Angles given as a ratio $2:3:4$ : write them $2k$ , $3k$ , $4k$ . Then $9k=180^\circ$ , so $k=20^\circ$ . The angles are $40^\circ$ , $60^\circ$ , $80^\circ$ .

An isosceles triangle has two equal sides and two equal base angles. The median to the base is also the height and the bisector: one line, three jobs. Two angles of $40^\circ$ force the third to be $100^\circ$ , because equal angles sit opposite equal sides.

02

The four centres

Centre	Made by	Fact to use
Centroid $G$	medians	cuts each median $2:1$ from the vertex
Incentre $I$	angle bisectors	$\angle BIC=90^\circ+\dfrac{A}{2}$
Circumcentre $O$	perpendicular bisectors	$\angle BOC=2A$
Orthocentre $H$	altitudes	$\angle BHC=180^\circ-A$

Rule: Name the centre from what builds it. Then apply its one formula with the given $\angle A$ .

03

Using the centre facts

Take $\angle A=70^\circ$ :

Incentre: $\angle BIC=90^\circ+35^\circ=125^\circ$
Circumcentre: $\angle BOC=2\times70^\circ=140^\circ$
Orthocentre: $\angle BHC=180^\circ-70^\circ=110^\circ$

The centroid works on lengths, not angles. It cuts every median $2:1$ from the vertex. A median of $15$ cm splits into $10$ cm (vertex side) and $5$ cm (base side). A median of $18$ cm splits into $12$ cm and $6$ cm the same way.

Tip: In a right triangle the orthocentre sits at the right-angle vertex and the circumcentre is the midpoint of the hypotenuse. In an equilateral triangle all four centres are one point.

04

Median lengths

Three facts cover the median questions:

Median to the hypotenuse of a right triangle $=$ half the hypotenuse.
Isosceles with equal sides $13$ and base $10$ : median to base $=\sqrt{13^2-5^2}=\sqrt{169-25}=12$ .
Any triangle: Apollonius, $m_a^2=\dfrac{2b^2+2c^2-a^2}{4}$ , where $m_a$ is the median to side $a$ .

05

Area from three sides

Heron's rule: $s=\dfrac{a+b+c}{2}$ (half the perimeter), then area $=\sqrt{s(s-a)(s-b)(s-c)}$ . Sides $13$ , $14$ , $15$ : $s=21$ , area $=\sqrt{21\times8\times7\times6}=\sqrt{7056}=84$ . Sides $9$ , $12$ , $15$ give $s=18$ and area $\sqrt{18\times9\times6\times3}=54$ by either route.

Tip: If the sides form a Pythagorean triplet such as 9-12-15, skip Heron. The triangle is right-angled, so the area is half the product of the two legs.

06

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Angle at a triangle centre

How to spot it:

The question names I, O or H as incentre, circumcentre or orthocentre (or defines them by bisectors, perpendicular bisectors, altitudes).

\angle BIC=90^\circ+\dfrac{A}{2},\quad \angle BOC=2A,\quad \angle BHC=180^\circ-A

Method

Name the centre from its definition.
Pick its formula and substitute the given $\angle A$ .
Going backwards: from $\angle BIC$ get $A=2(\angle BIC-90^\circ)$ .

Why it works:

Each centre fixes a triangle inside the original, and its angles depend only on $\angle A$ .

Try this

In $\triangle ABC$ , $\angle A=60^\circ$ and $I$ is the incentre. Find $\angle BIC$ .

Show solution

$\angle BIC=90^\circ+\dfrac{A}{2}$ .
$=90^\circ+30^\circ$ .
$=120^\circ$ .

Answer

120°

Type 2very common2 practice Q

Exterior angle of a triangle

How to spot it:

One side of the triangle is extended; the exterior angle is asked, or a remote interior angle is missing.

\text{ext at }A=B+C

Method

Find the two remote interior angles, away from the extended corner.
Add them for the exterior angle.
Subtract instead when one remote angle is missing.

Why it works:

The exterior angle and the corner angle add to $180^\circ$ , and the three interior angles also add to $180^\circ$ .

Try this

An exterior angle of a triangle is $120^\circ$ and one remote interior angle is $50^\circ$ . Find the other remote interior angle.

Show solution

Exterior $=$ sum of remotes.
$120-50=70^\circ$ .
Check: $50+70=120$ .

Answer

70°

Type 3very common2 practice Q

Centroid cuts the median 2 : 1

How to spot it:

A median and the centroid G appear; AG, GD, their sum or difference is asked.

AG=\dfrac{2}{3}AD,\quad GD=\dfrac{1}{3}AD

Method

Mark the median and the centroid on it.
Split it $2:1$ : vertex side twice the base side.
Turn word conditions into parts of the median, e.g. $AG-GD=c$ means $\dfrac{1}{3}AD=c$ .

Why it works:

The centroid is the balance point, and it sits twice as far from the vertex as from the side.

Try this

In $\triangle ABC$ , median $AD=18$ cm and $G$ is the centroid. Find $GD$ .

Show solution

$GD=\dfrac{1}{3}\times18$ .
$GD=6$ cm.
(And $AG=12$ cm.)

Answer

6 cm

Type 4common2 practice Q

Area of a triangle: Heron or triplet

How to spot it:

Three sides given and the area asked, often a triplet family in disguise.

K=\sqrt{s(s-a)(s-b)(s-c)},\quad s=\dfrac{a+b+c}{2}

Method

Check for a triplet: right triangle, so area $=\dfrac{1}{2}\times$ the two legs.
Otherwise compute $s$ , the semi-perimeter.
Multiply $s(s-a)(s-b)(s-c)$ and take the square root.

Why it works:

Heron works for any triangle; triplets just let you skip it.

Try this

Find the area of a triangle with sides 9 cm, 12 cm and 15 cm.

Show solution

9-12-15 is 3-4-5 times 3, so it is right-angled.
Area $=\dfrac{1}{2}\times9\times12$ .
$=54$ sq cm.

Answer

54 sq cm

Type 5very common

Angles given as a ratio

How to spot it:

The three angles of a triangle are in a ratio like 2 : 3 : 4; one angle, or the largest, is asked.

(a+b+c)k=180^\circ

Method

Write the angles as $ak$ , $bk$ , $ck$ .
Add the ratio parts and divide $180^\circ$ by the total.
Multiply each part by $k$ ; read off the angle asked.

Why it works:

The ratio fixes the shares; the $180^\circ$ total fixes their size.

Try this

The angles of a triangle are in the ratio $2:3:4$ . Find the largest angle.

Show solution

$2k+3k+4k=180^\circ$ .
$9k=180^\circ$ , so $k=20^\circ$ .
Largest $=4\times20^\circ=80^\circ$ .

Answer

80°

Type 6common

Isosceles triangle angles

How to spot it:

Two equal sides are stated or drawn; one angle is given and another is asked.

\text{base}+\text{base}+\text{vertex}=180^\circ

Method

Mark the two equal base angles.
Use the $180^\circ$ total with the given angle.
Equal sides sit opposite equal angles.

Why it works:

Equal sides always face equal angles, so two of the three angles match.

Try this

The vertex angle of an isosceles triangle is $40^\circ$ . Find each base angle.

Show solution

Base angles are equal: $2b+40^\circ=180^\circ$ .
$2b=140^\circ$ .
$b=70^\circ$ .

Answer

70°

07

Formula sheet

Angle sum and exterior angle

A+B+C=180^\circ,\quad \text{ext at }A=B+C

Exterior angle = sum of the two remote (far) interior angles.

Centroid division

AG:GD=2:1

G is the centroid on median AD; the vertex piece is twice the base piece.

Incentre angle

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

I = incentre, where the angle bisectors meet.

Circumcentre angle

\angle BOC=2A

O = circumcentre; the angle at O stands on the same arc BC as angle A.

Orthocentre angle

\angle BHC=180^\circ-A

H = orthocentre, where the altitudes meet.

Apollonius (median length)

m_a^2=\dfrac{2b^2+2c^2-a^2}{4}

Median to side a of a triangle with sides a, b, c.

Isosceles median to base

m=\sqrt{a^2-\left(\dfrac{b}{2}\right)^2}

a = equal side, b = base.

Heron's area

K=\sqrt{s(s-a)(s-b)(s-c)},\ s=\dfrac{a+b+c}{2}

s = semi-perimeter (half the perimeter).

08

Shortcuts that save time

⚡ Centre angles from one input

Only $\angle A$ is needed. Read which centre the question names, then use its formula.

Example

In $\triangle ABC$ , $\angle A=70^\circ$ and $H$ is the orthocentre. Find $\angle BHC$ .

Show solution

Orthocentre: $\angle BHC=180^\circ-A$ .
$180^\circ-70^\circ=110^\circ$ .
(Incentre would give $125^\circ$ , circumcentre $140^\circ$ .)

Answer

110°

⚡ Centroid cut in the ratio 2 : 1

Call the median $3$ parts. The centroid gives $2$ parts on the vertex side and $1$ part on the base side.

Example

The median $AD$ of $\triangle ABC$ is 15 cm and $G$ is the centroid. Find $AG$ .

Show solution

$AG=\dfrac{2}{3}\times15$ .
$AG=10$ cm (and $GD=5$ cm).

Answer

10 cm

⚡ Median in an isosceles triangle

Skip Apollonius when two sides are equal. The median to the base is $\sqrt{a^2-(b/2)^2}$ with $a$ the equal side and $b$ the base.

Example

Find the median to the base of an isosceles triangle with equal sides 25 cm and base 14 cm.

Show solution

Half the base $=7$ cm.
$\sqrt{25^2-7^2}=\sqrt{625-49}$ .
$\sqrt{576}=24$ cm.

Answer

24 cm

⚡ Heron families worth memorising

(3,4,5) gives 6, (5,12,13) gives 30, (13,14,15) gives 84, (7,24,25) gives 84, (9,12,15) gives 54, (10,24,26) gives 120.

Example

Find the area of a triangle with sides 13 cm, 14 cm, 15 cm.

Show solution

$s=\dfrac{13+14+15}{2}=21$ .
$\sqrt{21\times8\times7\times6}$ .
$\sqrt{7056}=84$ sq cm.

Answer

84 sq cm

09

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Swapping the centre formulas under time pressure.

Incentre $90^\circ+\dfrac{A}{2}$ , circumcentre $2A$ , orthocentre $180^\circ-A$ . Recite before substituting.

Mistake 02

Taking $GD=\dfrac{2}{3}AD$ at the centroid.

The vertex piece $AG$ is $\dfrac{2}{3}$ . The base piece $GD$ is $\dfrac{1}{3}$ .

Mistake 03

Adding the two near angles for the exterior angle.

The exterior angle equals the two remote interior angles, the ones away from that corner.

Mistake 04

Using the full perimeter inside Heron's formula.

Halve the perimeter first: $s=\dfrac{a+b+c}{2}$ , then use $s-a$ , $s-b$ , $s-c$ .

Mistake 05

Forgetting the median to the hypotenuse is half of it.

In a right triangle that median is always $\dfrac{\text{hypotenuse}}{2}$ .

10

Quick revision

Read this the night before the exam.

Inside angles total $180^\circ$ ; exterior angle $=$ sum of the two remote interior angles.
$\angle BIC=90^\circ+\dfrac{A}{2}$ , $\angle BOC=2A$ , $\angle BHC=180^\circ-A$ .
Centroid cuts each median $2:1$ from the vertex: $AG=\dfrac{2}{3}AD$ .
Median to the hypotenuse $=$ half the hypotenuse.
Isosceles median $=\sqrt{a^2-(b/2)^2}$ ; general median by Apollonius.
Heron: $s$ first, then $\sqrt{s(s-a)(s-b)(s-c)}$ ; triplet sides mean a right triangle.

11

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.