SSC CGL Maths · Mensuration 2D

Areas of Triangles

A triangle's area is half its base times its height; when the height is not given, use two sides and the angle between them, or all three sides.

All Mensuration 2D notes

Why this matters

Twenty-one questions in the bank, nineteen EASY and two MODERATE, ten of them from the 2025 shifts. Equilateral triangles and regular hexagons come up again and again, so their formulas are worth knowing by heart. The scaling rule on this page also settles every 'side doubled' or 'area up by 44%' question in the chapter.

Concept 1 of 4: Half base times height, and two sides with the angle between them

Every triangle is half of a parallelogram with the same base and height, so its area is half of base times height. When the height is not given but two sides and the angle between them are, the sine of that angle supplies the height.

Definition

Area $= \frac{1}{2} \times \text{base} \times \text{height}$ , where the height is drawn perpendicular to that base. Any side can be the base if you use the height to it.
In a right triangle, the two legs are a base and its height: area $= \frac{1}{2} \times$ leg $\times$ leg.
A right isosceles triangle with legs $a$ has area $\frac{a^2}{2}$ and hypotenuse $a\sqrt{2}$ .
Two sides and the included angle: area $= \frac{1}{2}ab\sin C$ . Values to know: $\sin 30^\circ = \sin 150^\circ = \frac{1}{2}$ , $\sin 60^\circ = \sin 120^\circ = \frac{\sqrt{3}}{2}$ , $\sin 90^\circ = 1$ .
Comparing triangles: area ratio $=$ base ratio $\times$ height ratio. With equal heights, areas are in the ratio of the bases; with equal areas, bases and heights are in inverse ratio.
Fast route: when two triangles share a height, skip the area and divide the bases straight in the area ratio. Safe only when the heights really are equal.

Area of a triangle

A = \frac{1}{2}bh \qquad A = \frac{1}{2}ab\sin C

b, ha base and the perpendicular height to it
a, b, Ctwo sides and the angle between them

Worked example

A triangle has a base of 18 cm and an area of 108 cm². A second triangle has the same height and an area of 72 cm². Find the height, and the base of the second triangle.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 21 Sep 2025, Shift 2 · Q59Easy

Example 1 · Mensuration 2D · Areas of Triangles

A triangular flag has base and height in the ratio 5:4. If the area is 160

\text{cm}^2

, what are the base and height?

Pairing a base with the wrong height

The height must be perpendicular to the side you call the base. A slanting side is not a height unless the triangle has a right angle there.

Reading an area ratio as a base ratio

Area ratio is base ratio times height ratio. Areas

5 : 3

with heights

5 : 7

give bases

\frac{5}{3} \div \frac{5}{7} = 7 : 3

, not

5 : 3

Drill 6 more on half base times height, and two sides with the angle between them

Concept 2 of 4: Three sides: semi-perimeter, Heron's formula and the inradius

Three sides fix a triangle completely, so they must fix its area too. Heron's formula does it with the semi-perimeter alone. The same semi-perimeter links the area to the circle that fits inside the triangle.

Definition

Semi-perimeter $s = \frac{a + b + c}{2}$ , half the perimeter.
Heron's formula: area $= \sqrt{s(s - a)(s - b)(s - c)}$ .
Check for a right angle first. If $a^2 + b^2 = c^2$ (as with 8, 15, 17), the area is simply half the product of the two shorter sides. This fast route is safe only when the squares really add up.
Isosceles triangle with equal sides $a$ and base $b$ : the height bisects the base, so height $= \sqrt{a^2 - \frac{b^2}{4}}$ and area $= \frac{b}{4}\sqrt{4a^2 - b^2}$ .
Inradius: the circle that touches all three sides has radius $r$ , and area $= r \times s$ . For a right triangle, $r = \frac{a + b - c}{2}$ with $c$ the hypotenuse.

Heron's formula and the inradius

s = \frac{a + b + c}{2} \qquad A = \sqrt{s(s - a)(s - b)(s - c)} \qquad A = rs \qquad A_{\text{isosceles}} = \frac{b}{4}\sqrt{4a^2 - b^2}

ssemi-perimeter
rradius of the inscribed circle

Worked example

Find the area of a triangle whose sides are 13 cm, 14 cm and 15 cm.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 26 Sep 2025, Shift 2 · Q73Easy

Example 2 · Mensuration 2D · Areas of Triangles

A triangle with sides 5 cm, 12 cm, and 13 cm has an inradius of 2 cm. What is its area?

Putting the perimeter into Heron's formula

Heron uses the SEMI-perimeter

s

. With the full perimeter every bracket is wrong and the answer is far too large.

Running Heron on a right triangle

Sides such as 7, 24, 25 or 15, 20, 25 form a right triangle. Half the product of the two shorter sides gives the area in one step.

Drill 3 more on three sides: semi-perimeter, heron's formula and the inradius

Concept 3 of 4: Area of an equilateral triangle and a regular hexagon

An equilateral triangle has one size number, its side, so its height and area follow from that alone. Six equilateral triangles meeting at a point make a regular hexagon, so the hexagon needs nothing new.

Definition

Equilateral triangle of side $a$ : perimeter $3a$ , height $\frac{\sqrt{3}}{2}a$ , area $\frac{\sqrt{3}}{4}a^2$ .
Regular hexagon of side $a$ : six equilateral triangles of side $a$ , so area $= 6 \times \frac{\sqrt{3}}{4}a^2 = \frac{3\sqrt{3}}{2}a^2$ . Its perimeter is $6a$ .
A hexagon is therefore 6 times the equilateral triangle on the same side.
Fast route: $\frac{\sqrt{3}}{4}a^2 = \left(\frac{a}{2}\right)^2\sqrt{3}$ . Halve the side, square it, attach $\sqrt{3}$ : side 16 gives $64\sqrt{3}$ . Always safe for an equilateral triangle.
When a decimal answer is wanted, use $\sqrt{3} \approx 1.732$ .

Equilateral triangle and regular hexagon

A = \frac{\sqrt{3}}{4}a^2 \qquad h = \frac{\sqrt{3}}{2}a \qquad A_{\text{hexagon}} = \frac{3\sqrt{3}}{2}a^2

aside of the triangle or hexagon

Worked example

An equilateral triangle has a perimeter of 30 cm. Find its area and its height.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 23 Sep 2025, Shift 1 · Q59Easy

Example 3 · Mensuration 2D · Areas of Triangles

A regular hexagon has a perimeter of 72 cm. What is its area?

Mixing up the height and the area

The height of an equilateral triangle is

\frac{\sqrt{3}}{2}a

, a length. The area is

\frac{\sqrt{3}}{4}a^2

. Writing

\frac{\sqrt{3}}{2}a^2

for the area doubles it.

Taking a hexagon as three triangles

Lines from the centre to the six corners cut a regular hexagon into SIX equilateral triangles, each with the hexagon's side.

Drill 6 more on area of an equilateral triangle and a regular hexagon

Concept 4 of 4: Similar figures: scale the sides, square the area

Make every length of a figure twice as long and it holds four copies of the old one, two along and two across. Area always grows as the square of the length scale, whatever the shape.

Definition

Two figures of the same shape (two squares, two equilateral triangles, two regular hexagons, two regular polygons with the same number of sides) are similar.
If every side is multiplied by $k$ , the area is multiplied by $k^2$ . So area ratio $= (\text{side ratio})^2$ , and side ratio $= \sqrt{\text{area ratio}}$ .
Percentages: a side up by $x\%$ makes the area change by $2x + \frac{x^2}{100}$ percent. Up 50% gives $1.5^2 = 2.25$ , that is 125% more. Down 10% gives $0.9^2 = 0.81$ , that is 19% less.
When it breaks: the rule needs every length scaled by the same factor. Doubling only the length of a rectangle doubles its area; it does not multiply it by 4.

Scaling of area

\frac{A_2}{A_1} = \left(\frac{s_2}{s_1}\right)^2 \qquad \text{area change} = 2x + \frac{x^2}{100}\ \%

$s_1, s_2$ matching sides of the two figures
xpercentage change in every side

Worked example

The side of a square is increased by 30%. By what percentage does its area increase?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 22 Sep 2025, Shift 1 · Q59Easy

Example 4 · Mensuration 2D · Areas of Triangles

The area of a regular triangle increases by

44\%

. By what percentage did the side length increase?

Doubling the side doubles the area

Area scales as the square of the side. Doubling every side of a figure makes its area 4 times as large; tripling makes it 9 times.

A percentage change in area is not the side's change

A side up by 10% raises the area by 21%, not 10%. Square the factor

1.1

to get

1.21

Drill 2 more on similar figures: scale the sides, square the area

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 (4)

Half base times height, and two sides with the angle between them
Area of a triangle
$A = \frac{1}{2}bh \qquad A = \frac{1}{2}ab\sin C$
Three sides: semi-perimeter, Heron's formula and the inradius
Heron's formula and the inradius
$s = \frac{a + b + c}{2} \qquad A = \sqrt{s(s - a)(s - b)(s - c)} \qquad A = rs \qquad A_{\text{isosceles}} = \frac{b}{4}\sqrt{4a^2 - b^2}$
Area of an equilateral triangle and a regular hexagon
Equilateral triangle and regular hexagon
$A = \frac{\sqrt{3}}{4}a^2 \qquad h = \frac{\sqrt{3}}{2}a \qquad A_{\text{hexagon}} = \frac{3\sqrt{3}}{2}a^2$
Similar figures: scale the sides, square the area
Scaling of area
$\frac{A_2}{A_1} = \left(\frac{s_2}{s_1}\right)^2 \qquad \text{area change} = 2x + \frac{x^2}{100}\ \%$

Watch out for (8)

Pairing a base with the wrong height
→ Half base times height, and two sides with the angle between them
Reading an area ratio as a base ratio
→ Half base times height, and two sides with the angle between them
Putting the perimeter into Heron's formula
→ Three sides: semi-perimeter, Heron's formula and the inradius
Running Heron on a right triangle
→ Three sides: semi-perimeter, Heron's formula and the inradius
Mixing up the height and the area
→ Area of an equilateral triangle and a regular hexagon
Taking a hexagon as three triangles
→ Area of an equilateral triangle and a regular hexagon
Doubling the side doubles the area
→ Similar figures: scale the sides, square the area
A percentage change in area is not the side's change
→ Similar figures: scale the sides, square the area

Test yourself on Mensuration 2D

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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