Geometry

high importance~4 Q in Tier 136 formulas⚡ 19 shortcuts6 subtopics

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Subtopic 2 of 6·← Lines and angles·Congruence, similarity and BPT →

Triangles and their centres

Angle facts: the interior angles add to $180^\circ$ ; an exterior angle equals the sum of the two remote interior angles. In an isosceles triangle the median to the base is also the altitude and the bisector.

The four centres (memorise the angle results):

Centre	Defined by	Key result
Centroid G	medians	divides each median $2:1$ from the vertex
Incentre I	angle bisectors	$\angle BIC=90^\circ+\frac{A}{2}$
Circumcentre O	perpendicular bisectors	$\angle BOC=2\angle A$
Orthocentre H	altitudes	$\angle BHC=180^\circ-\angle A$

For a right triangle the orthocentre is the right-angle vertex and the circumcentre is the midpoint of the hypotenuse; in an equilateral triangle all four centres coincide.

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 12 practice Q

Detailed notes

The angle machinery of a triangle

Interior angles always add to $180^\circ$ . An exterior angle (one side produced) equals the sum of the two remote interior angles — the two not touching that vertex. So exterior at $A = B + C$ . Ratio questions: let the angles be $ak, bk, ck$ with $k(a+b+c)=180^\circ$ .

An isosceles triangle (two equal sides) has two equal base angles, and the median to the base is also the altitude and the angle bisector — one line does three jobs. Equilateral: all $60^\circ$ ; every median is an altitude.

The four centres — memorise three angle results

Centre	Defined by	The result asked in exams
Incentre $I$	angle bisectors	$\angle BIC = 90^\circ + \frac{A}{2}$
Circumcentre $O$	perpendicular bisectors	$\angle BOC = 2A$ (same arc $BC$ )
Orthocentre $H$	altitudes	$\angle BHC = 180^\circ - A$

Each gives the asked angle from $\angle A$ alone — identify the centre from its definition, then read the formula. The centroid $G$ (medians) is about lengths, not angles: it divides every median $2:1$ from the vertex, so $AG = \frac{2}{3}AD$ and $GD = \frac{1}{3}AD$ . Conditions like " $AG$ exceeds $GD$ by 3" translate to $\frac{1}{3}AD = 3$ .

Special positions worth remembering: in a right triangle the orthocentre is the right-angle vertex and the circumcentre is the midpoint of the hypotenuse (so the median to the hypotenuse equals half of it). In an equilateral triangle all four centres coincide.

Median lengths

Apollonius: median to side $a$ : $m_a^2 = \frac{2b^2+2c^2-a^2}{4}$ .
Isosceles shortcut: median to the base $= \sqrt{a^2 - \left(\frac{\text{base}}{2}\right)^2}$ — the altitude-right-triangle relation, usually a Pythagorean triplet (13, 13, 10 → height 12).
Median to the hypotenuse $= \frac{h}{2}$ .

Area of a triangle (three sides)

Heron: semi-perimeter $s = \frac{a+b+c}{2}$ , area $= \sqrt{s(s-a)(s-b)(s-c)}$ . Memorise the standard families: (13,14,15)→84, (5,12,13)→30, (9,12,15)→54, (10,24,26)→120, (7,24,25)→84. If the sides are a Pythagorean family, skip Heron — the triangle is right-angled and area $= \frac{1}{2}\times$ legs.

Quick revision

$A+B+C=180^\circ$ ; exterior $=$ sum of remote interiors.
$\angle BIC = 90^\circ+\frac{A}{2}$ , $\angle BOC = 2A$ , $\angle BHC = 180^\circ-A$ .
Centroid: $AG:GD = 2:1$ on every median; $AG-GD = \frac{1}{3}AD$ .
Median: Apollonius, or isosceles $\sqrt{a^2-(b/2)^2}$ ; median to hypotenuse $=\frac{h}{2}$ .
Heron: $s$ , then $\sqrt{s(s-a)(s-b)(s-c)}$ ; triplet sides → $\frac12 \times$ legs.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Angle at a triangle centre (I, O or H)very common2 practice Q

How to spot it:

$I$ / $O$ / $H$ is named as incentre/circumcentre/orthocentre (or defined by bisectors/perpendicular bisectors/altitudes); $\angle A$ is given or asked.

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

Identify the centre from its definition (bisectors → I, perpendicular bisectors → O, altitudes → H).
Apply the matching formula with the given $\angle A$ .
Reverse direction: from $\angle BIC$ get $A = 2(\angle BIC - 90^\circ)$ ; from $\angle BHC$ get $A = 180^\circ - \angle BHC$ .

Why: each centre fixes a triangle whose angles are forced by $A$ alone.

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

$\angle BIC = 90^\circ + \frac{60^\circ}{2} = 120^\circ$ . (Check the centre type first — $O$ would also give $2 \times 60^\circ$ here.)

Type 2: Exterior angle of a trianglevery common2 practice Q

How to spot it:

One side of the triangle is produced; the exterior angle is asked from the two remote interiors, or one interior is missing from the exterior angle.

\text{ext at } A = B + C

Spot which two interior angles are REMOTE from the produced side.
Add them for the exterior angle; subtract to recover a missing remote interior.
The interior angle at the same vertex is $180^\circ -$ exterior — read carefully which is asked.

Why: the exterior and the adjacent interior are supplementary, and the three interiors total $180^\circ$ — the same fact written twice.

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

$120 - 50 = 70^\circ$ . Check: $50+70=120$ ✓ (the interior at the produced vertex is $60^\circ$ — not asked).

Type 3: Centroid cuts the median 2 : 1very common2 practice Q

How to spot it:

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

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

Name the median $AD$ with $G$ on it; the vertex-side piece is twice the base-side piece.
From $AD$ : $AG = \frac{2}{3}AD$ , $GD = \frac{1}{3}AD$ .
Conditions like $AG - GD = c$ mean $\frac{1}{3}AD = c$ ; solve for the asked piece.

Why: the three medians balance at $G$ , and on each median the balance point sits twice as far from the vertex as from the opposite side's midpoint.

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

$GD = \frac{1}{3} \times 18 = 6$ cm (and $AG = 12$ cm — the $2:1$ split).

Type 4: Area of a triangle: Heron and isosceles heightcommon2 practice Q

How to spot it:

Three sides given (a triplet family) — area asked; or an isosceles triangle with equal sides and base — area asked.

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

Triplet sides? Use right-triangle area $= \frac12 \times$ legs directly.
Otherwise Heron: halve the perimeter, multiply the four factors, take the root.
Isosceles: height $= \sqrt{a^2 - (b/2)^2}$ , then area $= \frac12 b\,h$ .

Why: Heron is the general version of the right-triangle shortcut; both run on memorised families.

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

$9\text{-}12\text{-}15$ is $3\text{-}4\text{-}5 \times 3$ → right-angled: area $= \frac12 \times 9 \times 12 = 54$ sq cm. (Heron agrees: $s=18$ , $\sqrt{18 \cdot 9 \cdot 6 \cdot 3} = 54$ .)

Formulas

Angle sum & exterior angle

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

Centroid division

AG:GD=2:1\ \text{on median }AD

Incentre angle

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

Circumcentre angle

\angle BOC=2A\ \text{(minor arc BC)}

Orthocentre angle

\angle BHC=180^\circ-A

Apollonius (median length)

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

Isosceles median

m_{\text{base}}=\sqrt{a^2-\left(\frac{\text{base}}{2}\right)^2}

Heron's area

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

Shortcut tricks

⚡ Centre angles from one input

The asked angle involves $\angle A$ only. Read off which centre is drawn: bisectors → $90+\frac A2$ , perpendicular bisectors → $2A$ , altitudes → $180-A$ .

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

$\angle BHC=180^\circ-70^\circ=110^\circ$ . (For the incentre it would be $90+35=125^\circ$ ; for the circumcentre $140^\circ$ .)

⚡ Centroid ratios without coordinates

The centroid cuts each median in $2:1$ from the vertex. If the median is $3m$ long, the pieces are $2m$ and $m$ .

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

$AG=\frac23\times15=10$ cm, $GD=5$ cm.

⚡ Median by Apollonius or symmetry

For an isosceles triangle skip the formula: the median to the base is $\sqrt{a^2-(b/2)^2}$ . Otherwise use $m_a^2=\frac{2b^2+2c^2-a^2}{4}$ .

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

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

⚡ Heron with a standard semi-perimeter

For 13-14-15, $s=21$ , area $=\sqrt{21\cdot8\cdot7\cdot6}=84$ . Memorise the common Heron families: (13,14,15)→84, (3,4,5)→6, (5,12,13)→30, (6,8,10)→24, (7,24,25)→84, (9,12,15)→54, (10,24,26)→120.

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

$s=21$ : $K=\sqrt{21\times8\times7\times6}=\sqrt{7056}=84$ sq cm.

Where students lose marks

$\angle BIC=90^\circ+\frac A2$ vs $\angle BOC=2A$ vs $\angle BHC=180^\circ-A$ get swapped under exam pressure.
Centroid ratio read the wrong way: $GD=\frac13 AD$ , not $\frac23$ .
Exterior angle taken at the wrong vertex — it equals the sum of the two remote interior angles.
Heron: forgetting to halve the perimeter before subtracting.

Practice sets — 12 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 12 questions

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