Mensuration (2D)

high importance~2 Q in Tier 126 formulas⚡ 15 shortcuts5 subtopics

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Areas of triangles

⏱ 3 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

Area of a triangle is half of base times height, and any side can be the base. Equilateral, isosceles and right triangles have shortcuts that turn most exam questions into one line. Heron's formula covers everything else using only the three sides.

One master formula

K=\frac{1}{2}\times\text{base}\times\text{height}

The height is the perpendicular dropped onto the chosen base. Pick the side whose height is easiest, then stick with that pair.

Two triangles on the same base and between the same parallels always have equal areas. Half of the same base and the same height forces the same area. A median uses exactly that, which is why it never fails to halve a triangle.

Rule: Any side may be the base, but the height must be measured perpendicular to that exact side.

The three special triangles

Right triangle: the two legs act as base and height. Legs $6$ and $8$ give $K=\dfrac{6\times8}{2}=24$ , hypotenuse $10$ .
Equilateral (side $a$ ): height $=\dfrac{\sqrt3}{2}a$ , area $=\dfrac{\sqrt3}{4}a^2$ . Side $12$ : height $6\sqrt3$ , area $36\sqrt3$ .
Isosceles: the median to the base is the height. Equal sides $17$ , base $16$ : half-base $8$ , height $\sqrt{289-64}=15$ , area $\dfrac{16\times15}{2}=120$ .

Tip: Equilateral numbers stay clean: for $a=6$ the area is $9\sqrt3$ , and every side gives a multiple of $\sqrt3$ .

Heron's formula

With all three sides, take the half-perimeter first:

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

Sides $13$ , $14$ , $15$ : $s=21$ , so $K=\sqrt{21\times8\times7\times6}=\sqrt{7056}=84$ .

Sides $5$ , $12$ , $13$ give $s=15$ and $K=\sqrt{15\times10\times3\times2}=30$ . But $5$ - $12$ - $13$ is a triplet, so $\dfrac{5\times12}{2}=30$ was faster.

Shortcut: Spot a Pythagorean triplet and skip Heron. A right triangle's area is just half the product of its legs.

Altitude to the hypotenuse: two areas

The right triangle with legs $6$ , $8$ has hypotenuse $10$ and area $24$ . Take the hypotenuse as the base instead:

\frac{1}{2}\times10\times h=24 \Rightarrow h=4.8=\frac{24}{5}

Same triangle, same area, second route. That is the whole trick.

Watch: Equate the two area expressions before touching square roots. The altitude rule $h=\dfrac{ab}{c}$ falls out of it.

Radii and the median split

Inradius $r=\dfrac{K}{s}$ . For $13$ - $14$ - $15$ : $r=\dfrac{84}{21}=4$ .
Circumradius $R=\dfrac{abc}{4K}=\dfrac{2730}{336}=\dfrac{65}{8}=8.125$ .
Every median splits a triangle into two equal areas. Two medians cut it into four equal areas.
Equilateral: $R=2r$ , and $a=6$ gives $r=\sqrt3$ , $R=2\sqrt3$ .

Remember: Small radius $r$ divides the area by the half-perimeter; big radius $R$ multiplies the sides. $R=2r$ only in the equilateral case.

Question types you will see

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

Type 1very common3 practice Q

Area from base-height / equilateral formula

How to spot it:

A triangle with one length given, or an equilateral with its side.

Method

Pick the base with the easiest height.
Right triangle: legs are base and height.
Equilateral: use $K=\dfrac{\sqrt3}{4}a^2$ directly.
Multiply and simplify.

Why it works:

The direct formula is a one-liner, and equilateral sides keep the answer a clean multiple of $\sqrt3$ .

Try this

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

Show solution

$K=\dfrac{\sqrt3}{4}\times12^2$ .
$=\dfrac{\sqrt3}{4}\times144=36\sqrt3$ sq cm.

Answer

36*sqrt(3) sq cm (about 62.35 sq cm)

Type 2very common2 practice Q

Heron's formula (and the triplet bypass)

How to spot it:

Three sides, no right angle, no height.

Method

Compute $s=\dfrac{a+b+c}{2}$ .
Form the product $s(s-a)(s-b)(s-c)$ .
Take the square root.
Check for a triplet first to skip the work.

Why it works:

Heron works from sides alone, and most exam values are perfect squares under the root.

Try this

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

Show solution

$s=\dfrac{42}{2}=21$ .
$K=\sqrt{21\times8\times7\times6}=\sqrt{7056}=84$ .

Answer

84 sq cm

Type 3common2 practice Q

Isosceles triangle area

How to spot it:

Two equal sides and the base, no height.

Method

Halve the base.
Height $=\sqrt{(\text{equal side})^2-(\text{half-base})^2}$ .
$K=\dfrac{1}{2}\times\text{base}\times\text{height}$ .

Why it works:

The median to the base doubles as the height, so Pythagoras finishes the job in two steps.

Try this

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

Show solution

Half-base $=8$ ; height $=\sqrt{17^2-8^2}=15$ .
$K=\dfrac{16\times15}{2}=120$ sq cm.

Answer

120 sq cm

Type 4common2 practice Q

Inradius and median area-split

How to spot it:

Sides given, question asks the inradius; or a median asks for an area ratio.

Method

Find the area first (Heron or a triplet).
$r=\dfrac{K}{s}$ with the half-perimeter.
For medians: each one splits the area in half.

Why it works:

Both facts reuse the area you already computed, so nothing extra is needed.

Try this

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

Show solution

$K=84$ , $s=21$ .
$r=\dfrac{84}{21}=4$ cm.

Answer

4 cm

Type 5occasional2 practice Q

Altitude to the hypotenuse (area two ways)

How to spot it:

A right triangle, question asks the height on the hypotenuse.

Method

Compute the area from the two legs.
Set $\dfrac{1}{2}\times\text{hypotenuse}\times h$ equal to it.
Solve for $h$ .

Why it works:

One triangle, one area, two base choices. Equating them avoids the similar-triangle mess.

Try this

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

Show solution

$K=\dfrac{6\times8}{2}=24$ ; hypotenuse $=10$ .
$\dfrac{1}{2}\times10\times h=24$ .
$h=4.8$ cm.

Answer

4.8 cm

Formula sheet

Triangle area

K=\frac{1}{2}bh

b = any side, h = perpendicular height on it.

Equilateral triangle

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

a = side; height is root-three over two of the side.

Heron's formula

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

s = half the perimeter.

Altitude to hypotenuse

h=\frac{ab}{c}=\frac{2K}{c}

a, b legs, c hypotenuse; from equating two areas.

Inradius / circumradius

r=\frac{K}{s},\quad R=\frac{abc}{4K}

K = area, s = half-perimeter.

Median split

\text{median}\Rightarrow\text{two equal areas}

Each median halves the area of a triangle.

Shortcuts that save time

⚡ Triplet beats Heron

Before Heron, check for a triplet. A right triangle needs only half the product of its legs.

Example

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

Show solution

$9^2+12^2=81+144=225=15^2$ , so it is right-angled.
$K=\dfrac{9\times12}{2}$ .
$K=54$ sq cm.

Answer

54 sq cm

⚡ Isosceles: half it first

The median to the base is the height. Halve the base, then use Pythagoras with an equal side.

Example

An isosceles triangle has equal sides of 17 cm and base 16 cm. Find its area.

Show solution

Half-base $=8$ cm.
Height $=\sqrt{17^2-8^2}=\sqrt{225}=15$ cm.
$K=\dfrac{16\times15}{2}=120$ sq cm.

Answer

120 sq cm

⚡ Two radii from Heron

After Heron gives the area, both radii are one division away: r divides by s, R uses abc over 4K.

Example

The sides of a triangle are 13, 14 and 15 cm. Find its inradius.

Show solution

$s=21$ , $K=84$ .
$r=\dfrac{K}{s}=\dfrac{84}{21}$ .
$r=4$ cm.

Answer

4 cm

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using the slant side as the height.

The height is perpendicular to the base. In the 17-17-16 triangle it is 15, not 17.

Mistake 02

Forgetting to halve the base in an isosceles triangle.

Pythagoras needs the half-base 8, so the height is $\sqrt{289-64}=15$ .

Mistake 03

Running Heron on a right triangle.

A triplet like 6-8-10 needs only $\dfrac{6\times8}{2}=24$ .

Mistake 04

Using the full perimeter as $s$ .

$s$ is the half-perimeter. For 13-14-15, $s=21$ , not 42.

Mistake 05

Mixing the legs with the hypotenuse in $h=\dfrac{ab}{c}$ .

$c$ is the hypotenuse. Legs 6 and 8 over 10 give $h=4.8$ .

Quick revision

Read this the night before the exam.

Master formula: half of base times perpendicular height, any side as base.
Equilateral side $a$ : height $\dfrac{\sqrt3}{2}a$ , area $\dfrac{\sqrt3}{4}a^2$ .
Isosceles: halve the base, Pythagoras with an equal side gives the height.
Heron: half-perimeter first; a triplet triangle skips Heron entirely.
Altitude to the hypotenuse equals $\dfrac{ab}{c}$ , from two area expressions.
$r=\dfrac{K}{s}$ , $R=\dfrac{abc}{4K}$ , and every median halves the area.

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.

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