MathsICSE Class 9 · Mensuration

Heron's Formula and Areas of Triangles, Parallelograms, Rhombuses and Trapeziums (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

Heron's formula gives the area of a triangle with sides a, b, c as √(s(s − a)(s − b)(s − c)), where s is half the perimeter. A parallelogram has area base × height, a rhombus ½ × d₁ × d₂ and a trapezium ½ × (sum of parallel sides) × height. Split other quadrilaterals into triangles.

Understand Heron's Formula and Areas of Triangles, Parallelograms, Rhombuses and Trapeziums

Heron's formula. For a triangle with sides $a, b, c$ and semi-perimeter $s = \frac{a + b + c}{2}$ :
$ $Area = s (s - a) (s - b) (s - c) .$ $
*Example.* Sides $13, 14, 15$ : $s = 21$ , area $= 21 \times 8 \times 7 \times 6 = 7056 = 84$ cm².

From area to height. Since area $= \frac{1}{2} \times base \times height$ , the altitude on a side is $\frac{2 \times area}{side}$ .

Quadrilaterals.

Parallelogram: $base \times height$ . A parallelogram and a rectangle on the same base and between the same parallels have equal areas.
Rhombus: $\frac{1}{2} d_{1} d_{2}$ . Its diagonals bisect each other at right angles, so side $= (\frac{d _{1}}{2})^{2} + (\frac{d _{2}}{2})^{2}$ .
Trapezium: $\frac{1}{2} (a + b) h$ with $a, b$ the parallel sides. For an isosceles trapezium with parallel sides $a > b$ and equal legs $l$ , drop perpendiculars; each cuts off $\frac{a - b}{2}$ and $h = l^{2} - (\frac{a - b}{2})^{2}$ .
Any quadrilateral: draw a diagonal and add the areas of the two triangles (use Heron when only sides are known).

Perimeter. Add the sides; do not mix perimeter with area (cm against cm²).

Cost problems. Cost $=$ area $\times$ rate per unit area.

Level. Class IX is a school-exam class. Heron with large numbers and the isosceles-trapezium height need careful arithmetic and are at the hardest school-exam level.

Heron's Formula and Areas of Triangles, Parallelograms, Rhombuses and Trapeziums: formulas to remember

Heron's formula: $A = s (s - a) (s - b) (s - c), s = \frac{a + b + c}{2}$; Triangle from three sides.
Altitude: $h = \frac{2 A}{base}$; From the area.
Rhombus: $A = \frac{1}{2} d_{1} d_{2}$; Side = √((d₁/2)² + (d₂/2)²).
Trapezium: $A = \frac{1}{2} (a + b) h$; a, b parallel sides.
Isosceles trapezium height: $h = l^{2} - (\frac{a - b}{2})^{2}$; Legs l, parallel sides a and b.

How to solve Heron's Formula and Areas of Triangles, Parallelograms, Rhombuses and Trapeziums questions

Choose the formula by the data given, compute s or the diagonals first, then the area.

1List what is given: three sides, diagonals, base and height.
2For Heron: compute $s$ , then each of $s - a$ , $s - b$ , $s - c$ ; check they are positive.
3Multiply under the root and simplify (look for perfect squares).
4For a rhombus or trapezium use the diagonal or height formula; find missing lengths with Pythagoras.
5For a general quadrilateral, split by a diagonal.
6State units (cm², m²) and cost if asked.

Heron's Formula and Areas of Triangles, Parallelograms, Rhombuses and Trapeziums in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A triangular field

Suppose a field is a triangle with sides 51 m, 37 m and 20 m, to be levelled at ₹12 per m².

Find its area, the cost, and the altitude on the 51 m side.

Try it first, then show the solutionSolution

$s = 54$ ; area $= 54 \times 3 \times 17 \times 34 = 93636 = 306$ m².
Cost $= 306 \times 12 = ₹3672$ .
Altitude $= \frac{2 \times 306}{51} = 12$ m.

Answer: 306 m²; ₹3,672; 12 m.

Memory hook: Heron first, then the area gives everything else.

Paving a trapezium court

Imagine an isosceles-trapezium court with parallel sides 25 m and 9 m and equal legs 17 m, paved at ₹40 per m².

Find the height, the area and the cost.

Try it first, then show the solutionSolution

Each leg cuts off $\frac{25 - 9}{2} = 8$ m, so $h = 1 7^{2} - 8^{2} = 15$ m.
Area $= \frac{1}{2} (25 + 9) \times 15 = 255$ m².
Cost $= 255 \times 40 = ₹10200$ .

Answer: Height 15 m; area 255 m²; ₹10,200.

Memory hook: A 8-15-17 triangle sits in each end.

Tricks to remember Heron's Formula and Areas of Triangles, Parallelograms, Rhombuses and Trapeziums

Short memory aids. Say each one aloud once, then use it on a question the same day.

s then three differences
Compute s, s − a, s − b, s − c in a column before multiplying.
Use it when: Heron without slips.
Half the diagonal product
A rhombus is half the rectangle on its diagonals.
Use it when: Rhombus area.
Cut off the ends
In an isosceles trapezium, the ends are two equal right triangles.
Use it when: Finding height.

Heron's Formula and Areas of Triangles, Parallelograms, Rhombuses and Trapeziums: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Heron's formula · 1 mark · MCQBoard level

The area of a triangle with sides $13$ cm, $14$ cm and $15$ cm is:
(a) 42 cm²
(b) 84 cm²
(c) 91 cm²
(d) 105 cm²

Try it first, then show the solutionSolution

$s = 21$ ; $21 \times 8 \times 7 \times 6 = 7056 = 84$ .

Answer: (b) 84 cm²

Example 2: Rhombus · 3 marks · Short answerBoard level

The diagonals of a rhombus are $24$ cm and $10$ cm. Find its area, side and perimeter.

Try it first, then show the solutionSolution

Area $= \frac{1}{2} \times 24 \times 10 = 120$ cm².
Side $= 1 2^{2} + 5^{2} = 13$ cm.
Perimeter $= 4 \times 13 = 52$ cm.

Answer: 120 cm²; 13 cm; 52 cm

Example 3: Area and altitude · 4 marks · Short answerHarder board level

A triangle has sides $17$ cm, $25$ cm and $26$ cm. Find its area and the altitude on the $17$ cm side.

Try it first, then show the solutionSolution

$s = 34$ ; $s - a = 17, 9, 8$ .
Area $= 34 \times 17 \times 9 \times 8 = 41616 = 204$ cm².
Altitude $= \frac{2 \times 204}{17} = 24$ cm.

Answer: 204 cm²; 24 cm

Example 4: Quadrilateral by diagonal · 4 marks · Short answerHarder board level

In quadrilateral $A B C D$ , $A B = 9$ cm, $B C = 12$ cm, $C D = 13$ cm, $D A = 14$ cm and the diagonal $A C = 15$ cm. Find its area.

Try it first, then show the solutionSolution

$△ A B C$ : $9^{2} + 1 2^{2} = 225 = 1 5^{2}$ , a right angle at $B$ ; area $= \frac{1}{2} \times 9 \times 12 = 54$ cm².
$△ A C D$ (15, 13, 14): $s = 21$ , area $= 21 \times 6 \times 8 \times 7 = 84$ cm².
Total $= 54 + 84 = 138$ cm².

Answer: 138 cm²

Example 5: Isosceles trapezium · 5 marks · Long answerHOTS / competency

An isosceles trapezium has parallel sides $25$ m and $9$ m and equal legs $17$ m. Find its height, area, and the cost of paving at ₹40 per m².

Try it first, then show the solutionSolution

Each end cuts off $\frac{25 - 9}{2} = 8$ m.
$h = 1 7^{2} - 8^{2} = 225 = 15$ m.
Area $= \frac{1}{2} (25 + 9) (15) = 255$ m².
Cost $= 255 \times 40 = ₹10, 200$ .

Answer: 15 m; 255 m²; ₹10,200

Where students lose marks in Heron's Formula and Areas of Triangles, Parallelograms, Rhombuses and Trapeziums

The slips examiners see most often, and what to write instead.

Mistake: Using the perimeter for s
s is half the perimeter.
Wrong working
$s = 108$ for sides 51, 37, 20.
Right working
$s = 54$ .
Do this: Write s = (a + b + c)/2 first.
Usually costs: Whole answer.
Mistake: Taking a slant side as the trapezium height
The height is perpendicular to the parallel sides.
Wrong working
Area $= \frac{1}{2} (25 + 9) \times 17$ .
Right working
Area $= \frac{1}{2} (25 + 9) \times 15$ .
Do this: Find h with Pythagoras.
Usually costs: Whole answer.
Mistake: Wrong unit
Area is in square units.
Wrong working
255 m.
Right working
255 m².
Do this: Check the unit.
Usually costs: One mark.

Exam tips

Class IX is examined in school papers; Heron questions are 3 to 4 marks with perfect-square products.
Factor the product under the root into primes to simplify the square root.
Always mention units and, for cost problems, the rate.
Check Heron results by finding an altitude and seeing if it is a sensible number.

Practice questions on Heron's Formula and Areas of Triangles, Parallelograms, Rhombuses and Trapeziums

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A parallelogram has area $90$ cm² and base $12$ cm. Its height is:
(a) 7.5 cm
(b) 8 cm
(c) 10.8 cm
(d) 78 cm
Hint
Area ÷ base.
Answer
(a) 7.5 cm
Q2 · 4 marks · Short answerHarder board level
A triangle has perimeter $36$ cm and sides in the ratio $3 : 4 : 5$ . Find its sides, area by Heron's formula, and show it is right-angled.
Hint
One part = 36 ÷ 12.
Answer
Sides $9, 12, 15$ ; $s = 18$ ; area $= 18 \times 9 \times 6 \times 3 = 54$ cm². $81 + 144 = 225$ , so right-angled.
Q3 · 3 marks · Short answerHarder board level
A rhombus has side $10$ cm and one diagonal $12$ cm. Find the other diagonal and the area.
Hint
Half-diagonals and side form a right triangle.
Answer
Half of the other diagonal $= 100 - 36 = 8$ , so it is $16$ cm. Area $= \frac{1}{2} \times 12 \times 16 = 96$ cm².
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A parallelogram and a rectangle on the same base and between the same parallels have equal areas.
Reason (R): Area of a parallelogram $=$ base $\times$ height, with the same base and height here.
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
Same base, same height.
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: A triangular plot has sides $51$ m, $37$ m and $20$ m and is to be levelled at ₹12 per m².
(a) Find the semi-perimeter.
(b) Find the area.
(c) Find the cost of levelling.
(d) Find the altitude on the $51$ m side.
Hint
$s - a = 3, 17, 34$ .
Answer
(a) 54 m (b) 306 m² (c) ₹3,672 (d) 12 m.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State Heron's formula.Show answer
√(s(s − a)(s − b)(s − c)) with s = (a + b + c)/2.
2.Area of a rhombus?Show answer
Half the product of the diagonals.
3.Area of a trapezium?Show answer
Half the sum of the parallel sides times the height.
4.How do you find the area of a general quadrilateral?Show answer
Split by a diagonal into two triangles.

Heron's Formula and Areas of Triangles, Parallelograms, Rhombuses and Trapeziums: frequently asked questions

When should I use Heron's formula?

When all three sides are known and no height is given.

Is the trapezium in the syllabus?

The Class IX mensuration chapter covers the area and perimeter of quadrilaterals, including trapeziums.

Can the triangle sides be any three numbers?

No; each side must be less than the sum of the other two.