Mensuration (2D)

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high importance~2 Q in Tier 123 formulas⚡ 12 shortcuts5 subtopics

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Subtopic 1 of 5·Areas of quadrilaterals →

Areas of triangles

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Area of a triangle $=\frac12\times\text{base}\times\text{height}$ . Special cases carry most CGL questions:

Equilateral (side $a$ ): area $\frac{\sqrt3}{4}a^2$ , height $\frac{\sqrt3}{2}a$ .
Heron for three sides $a,b,c$ : semi-perimeter $s=\frac{a+b+c}{2}$ , area $=\sqrt{s(s-a)(s-b)(s-c)}$ .
Right triangle: the two legs are base and height, area $=\frac12\,p\,b$ .

For an equilateral triangle the inradius is $\frac{a}{2\sqrt3}$ and the circumradius is $\frac{a}{\sqrt3}$ .

📖 Notes 🧩 5 question types ⚡ Formulas 🎯 12 practice Q

Detailed notes

The four area routes (pick by what's given)

Base-height: $K = \frac{1}{2} \times \text{base} \times \text{height}$ — any triangle, when the perpendicular height is known.
Equilateral: $K = \frac{\sqrt{3}}{4}a^2$ ; also height $= \frac{\sqrt{3}}{2}a$ . Reverse: from the area, $a^2 = \frac{4K}{\sqrt{3}}$ .
Heron (three sides): $s = \frac{a+b+c}{2}$ , $K = \sqrt{s(s-a)(s-b)(s-c)}$ .
Right triangle: $K = \frac{1}{2} \times \text{leg} \times \text{leg}$ — and this doubles as a Heron bypass whenever the sides form a Pythagorean family.

The Heron bypass (memorised families)

Check for a triplet before touching Heron:

Family	Area
(5,12,13)	30
(9,12,15)	54
(10,24,26)	120
(13,14,15)	84
(7,24,25)	84
(13,20,21)	126

If the sides are a multiple of 3-4-5 or 5-12-13, the triangle is right-angled and the area is half the product of the two smaller sides. (13,14,15) is the classic non-right Heron example — its area 84 is worth memorising outright.

Isosceles triangles

Given equal sides $a$ and base $b$ : the height to the base is $h = \sqrt{a^2 - \left(\frac{b}{2}\right)^2}$ , then $K = \frac{1}{2}bh$ . These are engineered to give Pythagorean halves: (13,13,10) → h 12, K 60; (17,17,16) → h 15, K 120; (25,25,14) → h 24, K 168; (10,10,12) → h 8, K 48.

Area links inside the triangle

Inradius: $K = r \cdot s$ , so $r = \frac{K}{s}$ . For (13,14,15): $r = \frac{84}{21} = 4$ .
A median divides the triangle into two equal areas (same base halves, same height). A centroid therefore creates 6 triangles of equal area.
Altitude to the hypotenuse: $\frac{1}{2}ab = \frac{1}{2}ch$ gives $h = \frac{ab}{c}$ for legs $a,b$ and hypotenuse $c$ — (6,8,10) → $h = 4.8$ .

How the questions combine these

A typical CGL question chains two of the routes: sides given → area via the triplet bypass → then inradius $r = K/s$ ; or isosceles sides → height via Pythagoras → area → then a median split halves it. Read the full question once before computing, note the LAST quantity asked, and work backwards from it to choose the shortest route. Unit discipline matters too: mixed units (m with cm) must be converted before any squaring, and answers in $\text{cm}^2$ asked from metre dimensions need the $\times 10^4$ conversion.

Quick revision

$K = \frac{1}{2}bh$ ; equilateral $\frac{\sqrt3}{4}a^2$ , height $\frac{\sqrt3}{2}a$ .
Heron: $s$ first, then $\sqrt{s(s-a)(s-b)(s-c)}$ ; triplet sides → $\frac12 \times$ legs.
Isosceles: $h = \sqrt{a^2 - (b/2)^2}$ ; know the (13,13,10), (17,17,16), (25,25,14) shapes.
$r = K/s$ ; median halves the area; altitude to hypotenuse $h = \frac{ab}{c}$ .

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: Area from base-height / equilateral formulavery common3 practice Q

How to spot it:

Base and height given directly — or 'equilateral triangle of side a' with area/perimeter/height asked.

K = \frac{1}{2}bh;\quad K_{\text{eq}} = \frac{\sqrt{3}}{4}a^2

General triangle: halve the product of base and perpendicular height.
Equilateral: $\frac{\sqrt3}{4}a^2$ for area, $\frac{\sqrt3}{2}a$ for height, $3a$ for perimeter.
Reverse direction (area given → side) needs $a^2 = \frac{4K}{\sqrt3}$ ; keep the $\sqrt3$ untouched.

Why: the equilateral formula is just base-height with height forced to $\frac{\sqrt3}{2}a$ by the 60-60-60 angles.

Example: Find the area of an equilateral triangle of side 12 cm.

$K = \frac{\sqrt3}{4} \times 12^2 = \frac{\sqrt3}{4} \times 144 = 36\sqrt{3}$ sq cm.

Type 2: Heron's formula (and the triplet bypass)very common2 practice Q

How to spot it:

Three sides given with no height; sides often form a Pythagorean family in disguise.

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

Spot a triplet family first — then the triangle is right-angled and $K = \frac12 \times$ legs.
Otherwise compute $s$ , then the four-factor product, then the root.
(13,14,15) → 84 is the standard non-right Heron answer worth knowing cold.

Why: Heron works on any triangle, but exam sides are chosen so either a triplet bypass or a clean square root appears.

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

$s = 21$ ; $K = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84$ sq cm.

Type 3: Isosceles triangle areacommon2 practice Q

How to spot it:

Equal sides and base given — area asked; the numbers are built for a Pythagorean half-base.

h = \sqrt{a^2 - (b/2)^2},\quad K = \frac{1}{2}bh

Halve the base; it is one leg of the right triangle formed by the height.
Get $h$ from the equal side as hypotenuse — check for (5,12,13), (8,15,17) halves.
Area $= \frac12 \times$ base $\times h$ .

Why: the height to the base of an isosceles triangle is also its median, splitting the base into equal halves.

Example: An isosceles triangle has equal sides of 17 cm and base 16 cm. Its area is:

$h = \sqrt{17^2 - 8^2} = \sqrt{289-64} = 15$ ; $K = \frac12 \times 16 \times 15 = 120$ sq cm.

Type 4: Inradius and median area-splitcommon2 practice Q

How to spot it:

'The inradius of a triangle…' with sides given ( $r = K/s$ ), or a median/centroid with the two area pieces asked.

K = r\,s \ \Rightarrow\ r = \frac{K}{s}

Find the area first (triplet bypass or Heron).
Inradius $r = K/s$ — divide by the SEMI-perimeter, not the perimeter.
Median question: a median halves the area (equal base halves, same height).

Why: the incircle touches all three sides, and joining the incentre to the vertices splits the triangle into three triangles of height $r$ — their areas total $r \cdot s$ .

Example: The sides of a triangle are 13, 14 and 15 cm. Its inradius is:

$K = 84$ (Heron), $s = 21$ : $r = \frac{84}{21} = 4$ cm.

Type 5: Altitude to the hypotenuse (area two ways)occasional2 practice Q

How to spot it:

A right triangle with the perpendicular drawn from the right angle to the hypotenuse; that altitude (or the area) is asked.

\frac{1}{2}ab = \frac{1}{2}ch \ \Rightarrow\ h = \frac{ab}{c}

Compute the area the easy way: half the product of the legs.
The same area equals half × hypotenuse × the altitude to it.
So altitude $h = \frac{ab}{c}$ ; keep it as a fraction if it is not integral.

Why: the two legs and (hypotenuse + altitude) are two base-height pairs of the SAME triangle, so the areas are equal.

Example: In a right triangle with legs 6 cm and 8 cm, find the length of the altitude drawn to the hypotenuse.

Hypotenuse $= 10$ ; area $= \frac12 \cdot 6 \cdot 8 = 24 = \frac12 \cdot 10 \cdot h \Rightarrow h = 4.8$ cm.

Formulas

General triangle

K=\frac12 b h

Equilateral triangle

K=\frac{\sqrt3}{4}a^2,\quad h=\frac{\sqrt3}{2}a

Heron's formula

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

Radii of an equilateral triangle

r_{\text{in}}=\frac{a}{2\sqrt3},\quad R=\frac{a}{\sqrt3}

Shortcut tricks

⚡ Equilateral area from the side table

$\frac{\sqrt3}{4}=0.433$ . Memorise: side 4 → $4\sqrt3$ , side 6 → $9\sqrt3$ , side 8 → $16\sqrt3$ , side 12 → $36\sqrt3$ , side 10 → $25\sqrt3$ . The coefficient is just $\left(\frac{\text{side}}{4}\right)^2$ .

Example: Find the area of an equilateral triangle of side 12 cm.

$\frac{\sqrt3}{4}\times12^2=36\sqrt3$ sq cm.

⚡ Heron with a triplet shortcut

If the sides form a Pythagorean triple the triangle is right-angled: use $\frac12\times$ legs directly. 13-14-15 is the classic non-right Heron case, area 84.

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

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

Where students lose marks

Using $\frac{\sqrt3}{2}a^2$ (the height formula's coefficient) for the equilateral area.
Heron: forgetting the square root at the end.
Taking the longest side as the height in a right triangle — the legs are base and height.

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.