MathsCBSE Class 9 · Measuring Space: Perimeter and Area

Heron's Formula and Areas of Quadrilaterals: Class 9

Updated 17 October 2026 · Fact-checked

Heron's formula gives the area of a triangle with sides $a, b, c$ and semi-perimeter $s = \frac{a+b+c}{2}$ as $\sqrt{s(s-a)(s-b)(s-c)}$. A quadrilateral is split by a diagonal into two triangles whose areas are added. For a cyclic quadrilateral with sides $a, b, c, d$ and $s = \frac{a+b+c+d}{2}$, Brahmagupta's formula gives the area $\sqrt{(s-a)(s-b)(s-c)(s-d)}$.

Know only the three sides? Find s, multiply s(s−a)(s−b)(s−c), take the root: 210, which matches ½ × 28 × 15.

Illustration drawn for QuestionBuddy.

Understand Heron's Formula and Areas of Quadrilaterals

Area of a triangle from its three sides. When the height is not known but all three sides are, use Heron's formula. Find the semi-perimeter $s = \frac{a + b + c}{2}$ , then the area is
$ $A = s (s - a) (s - b) (s - c) .$ $
For sides $17, 25, 28$ : $s = 35$ , so $A = 35 \times 18 \times 10 \times 7 = 44100 = 210$ . Once the area is known, the altitude on any side follows from $A = \frac{1}{2} \times base \times height$ : on the side $28$ , $h = \frac{2 \times 210}{28} = 15$ . A triangle exists only if each side is shorter than the sum of the other two ( $s - a$ , $s - b$ , $s - c$ must all be positive).

Equilateral triangle. With side $a$ , $s = \frac{3 a}{2}$ and Heron's formula gives $A = \frac{3}{4} a^{2}$ . For $a = 12$ : $A = 363 \approx 62.35$ (with $3 = 1.732$ ). An isosceles triangle with equal sides $13$ and base $10$ has $s = 18$ and $A = 18 \times 5 \times 5 \times 8 = 60$ , the same as $\frac{1}{2} \times 10 \times 12$ using the height $1 3^{2} - 5^{2} = 12$ .

Quadrilaterals by splitting. Draw one diagonal; this cuts the quadrilateral into two triangles. If one is right-angled use $\frac{1}{2} \times base \times height$ , and for the other use Heron's formula. For $A B C D$ with $A B = 9$ , $B C = 40$ , $C D = 28$ , $D A = 15$ and $∠ B = 9 0^{\circ}$ : the diagonal $A C = 81 + 1600 = 41$ ; $△ A B C$ has area $\frac{1}{2} \times 9 \times 40 = 180$ ; $△ A C D$ has sides $41, 28, 15$ with $s = 42$ and area $42 \times 1 \times 14 \times 27 = 126$ ; total $306$ .

Brahmagupta's formula. For a *cyclic* quadrilateral (one whose vertices lie on a circle) with sides $a, b, c, d$ :
$ $A = (s - a) (s - b) (s - c) (s - d), s = \frac{a + b + c + d}{2} .$ $
For sides $7, 24, 20, 15$ (two right triangles sharing the hypotenuse $25$ ): $s = 33$ and $A = 26 \times 9 \times 13 \times 18 = 234$ , which equals $\frac{1}{2} (7) (24) + \frac{1}{2} (15) (20) = 84 + 150$ .

Special quadrilaterals. A parallelogram has area $base \times height$ ; a trapezium with parallel sides $a$ and $b$ and height $h$ has area $\frac{1}{2} (a + b) h$ ; a rhombus or kite with diagonals $d_{1}$ and $d_{2}$ has area $\frac{1}{2} d_{1} d_{2}$ . For a rhombus with diagonals $24$ and $18$ , the diagonals bisect at right angles, so the side is $1 2^{2} + 9^{2} = 15$ , the area is $216$ and the height is $\frac{216}{15} = 14.4$ . For a trapezium with parallel sides $25$ and $10$ and legs $14$ and $13$ , draw a line through one end of the shorter side parallel to a leg: you get a triangle of sides $15, 13, 14$ (area $84$ , height $11.2$ ) and the trapezium has area $\frac{1}{2} (35) (11.2) = 196$ .

Method. Find the perimeter or semi-perimeter, check the triangle inequality, keep the numbers in factors, and simplify the product under the root by pairing factors. State the unit as square units.

Heron's Formula and Areas of Quadrilaterals: formulas to remember

Heron's formula: $A = s (s - a) (s - b) (s - c), s = \frac{a + b + c}{2}$; Needs only the three sides.
Equilateral triangle: $A = \frac{3}{4} a^{2}$; Special case of Heron.
Altitude from area: $h = \frac{2 A}{b}$; Height on side $b$ .
Brahmagupta's formula: $A = (s - a) (s - b) (s - c) (s - d)$; Cyclic quadrilateral only.
Trapezium: $A = \frac{1}{2} (a + b) h$; $a$ , $b$ parallel sides.
Rhombus or kite: $A = \frac{1}{2} d_{1} d_{2}$; Diagonals meet at right angles.

How to solve Heron's Formula and Areas of Quadrilaterals questions

Choose between the base-height formula and Heron's formula depending on what is known.

1List the given sides; if a height is given use $\frac{1}{2} bh$ , otherwise use Heron's formula.
2Compute $s$ and the three differences $s - a$ , $s - b$ , $s - c$ .
3Multiply and look for perfect-square factors to simplify the root.
4For a quadrilateral, draw a diagonal; find the area of each triangle and add.
5For a cyclic quadrilateral, use Brahmagupta's formula and check with the diagonal split if possible.
6Convert the final area into the unit asked and round only at the end.

Heron's Formula and Areas of Quadrilaterals 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.

Heron's formula

Suppose a triangular flower bed has sides 13 m, 14 m and 15 m, and turf costs 120 rupees per square metre.

Find the area of the bed and the cost of turfing it.

Try it first, then show the solutionSolution

$s = \frac{13 + 14 + 15}{2} = 21$ .
Area $= 21 \times 8 \times 7 \times 6 = 7056 = 84$ m $^{2}$ .
Cost $= 84 \times 120 = 10080$ rupees.

Answer: The bed has area 84 m $^{2}$ and turfing costs 10080 rupees.

Memory hook: Semi-perimeter first, then three differences.

Tricks to remember Heron's Formula and Areas of Quadrilaterals

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

s, then s minus each side
Compute $s$ once, then $s - a$ , $s - b$ , $s - c$ and multiply all four.
Use it when: Heron's formula.
Split by a diagonal
Every quadrilateral is two triangles; choose the diagonal that gives a right triangle.
Use it when: Quadrilateral areas.
Cyclic needs a circle
Brahmagupta works only if the four corners lie on one circle.
Use it when: Choosing the formula.

Heron's Formula and Areas of Quadrilaterals: solved examples

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

Example 1: Heron and altitude · 3 marks · Short answerBoard level

The sides of a triangle are 17 cm, 25 cm and 28 cm. Find its area and the length of the altitude on the longest side.

Try it first, then show the solutionSolution

$s = \frac{17 + 25 + 28}{2} = 35$ cm. (½)
Area $= 35 \times 18 \times 10 \times 7 = 44100 = 210$ cm $^{2}$ . (1½)
Altitude on 28 cm: $\frac{1}{2} \times 28 \times h = 210$ , so $h = 15$ cm. (1)

Answer: Area $210$ cm $^{2}$ ; altitude $15$ cm.

Example 2: Equilateral triangle · 3 marks · Short answerBoard level

The perimeter of an equilateral triangle is 36 cm. Find its area by Heron's formula and verify with $\frac{3}{4} a^{2}$ . Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Each side is $12$ cm and $s = 18$ . (½)
Area $= 18 \times 6 \times 6 \times 6 = 6 18 \times 6 = 6108 = 363$ . (1)
$36 \times 1.732 = 62.352$ cm $^{2}$ . (½)
Check: $\frac{3}{4} \times 144 = 363$ . (1)

Answer: Area $= 363 \approx 62.35$ cm $^{2}$ .

Example 3: A four-sided field · 5 marks · Long answerHarder board level

A field $A B C D$ has $A B = 9$ m, $B C = 40$ m, $C D = 28$ m, $D A = 15$ m and $∠ B = 9 0^{\circ}$ . Find its area.

Try it first, then show the solutionSolution

In right $△ A B C$ , $A C = 9^{2} + 4 0^{2} = 1681 = 41$ m. (1)
Area of $△ A B C = \frac{1}{2} \times 9 \times 40 = 180$ m $^{2}$ . (1)
$△ A C D$ has sides $41, 28, 15$ ; $s = 42$ . (1)
Area $= 42 \times 1 \times 14 \times 27 = 15876 = 126$ m $^{2}$ . (1)
Total area $= 180 + 126 = 306$ m $^{2}$ . (1)

Answer: The area of the field is $306$ m $^{2}$ .

Example 4: Cyclic quadrilateral · 3 marks · Short answerHarder board level

A cyclic quadrilateral has sides 7 cm, 24 cm, 20 cm and 15 cm in order. Find its area using Brahmagupta's formula and verify by splitting it into right-angled triangles.

Try it first, then show the solutionSolution

$s = \frac{7 + 24 + 20 + 15}{2} = 33$ . (½)
Area $= (33 - 7) (33 - 24) (33 - 20) (33 - 15) = 26 \times 9 \times 13 \times 18$ . (1)
$26 \times 13 = 338$ and $9 \times 18 = 162$ ; $338 \times 162 = 54756 = 23 4^{2}$ , so the area is $234$ cm $^{2}$ . (1)
Check: the diagonal $25$ makes right triangles $7, 24, 25$ and $15, 20, 25$ with areas $84 + 150 = 234$ . (½)

Answer: Area $= 234$ cm $^{2}$ .

Example 5: Two plots compared · 4 marks · Case-basedHOTS / competency

Case: a triangular plot $P$ has sides 26 m, 28 m and 30 m. A trapezium-shaped plot $Q$ has parallel sides 25 m and 9 m and each non-parallel side 10 m. (i) Find the area of $P$ . (ii) Find the height and area of $Q$ . (iii) By how much is $P$ larger than $Q$ ? (iv) Find the cost of levelling $P$ at 50 rupees per square metre.

Try it first, then show the solutionSolution

(i) $s = 42$ ; area $= 42 \times 16 \times 14 \times 12 = 112896 = 336$ m $^{2}$ . (1)
(ii) The legs are equal, so each overhangs by $\frac{25 - 9}{2} = 8$ m; height $= 1 0^{2} - 8^{2} = 6$ m. Area $= \frac{1}{2} (25 + 9) (6) = 102$ m $^{2}$ . (1½)
(iii) $336 - 102 = 234$ m $^{2}$ . (½)
(iv) $336 \times 50 = 16800$ rupees. (1)

Answer: (i) $336$ m $^{2}$ ; (ii) $6$ m and $102$ m $^{2}$ ; (iii) $234$ m $^{2}$ ; (iv) $16800$ rupees.

Where students lose marks in Heron's Formula and Areas of Quadrilaterals

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

Mistake: Using the perimeter in place of the semi-perimeter
Forgetting the division by 2.
Wrong working
$s = 17 + 25 + 28 = 70$ .
Right working
$s = 35$ .
Do this: Write $s = \frac{a + b + c}{2}$ first.
Usually costs: The whole answer.
Mistake: Applying Brahmagupta to a non-cyclic quadrilateral
Not checking the condition.
Wrong working
Using it for any four sides.
Right working
Use it only if the quadrilateral is cyclic; otherwise split into triangles.
Do this: State "cyclic" in the first line.
Usually costs: The answer.
Mistake: Arithmetic slip in the root
Multiplying large numbers instead of pairing factors.
Wrong working
$35 \times 18 \times 10 \times 7 = 200$ .
Right working
$35 \times 7 = 245$ and $18 \times 10 = 180$ ; $245 \times 180 = 44100 = 21 0^{2}$ .
Do this: Pair factors to create squares, and verify by squaring.
Usually costs: 2 marks.

Exam tips

Write $s$ and the three differences on separate lines; the examiner checks each.
Look for factorisation into squares to avoid large products.
For a quadrilateral, mention the diagonal you draw.
Always give square units.

Practice questions on Heron's Formula and Areas of Quadrilaterals

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
The area of a triangle with sides 8 cm, 15 cm and 17 cm is
(a) 60 cm $^{2}$
(b) 120 cm $^{2}$
(c) 30 cm $^{2}$
(d) 68 cm $^{2}$
Hint
$s = 20$ ; then multiply $20 \times 12 \times 5 \times 3$ .
Answer
(a) 60 cm $^{2}$ . $20 \times 12 \times 5 \times 3 = 3600 = 60$ (the triangle is right-angled: $\frac{1}{2} \times 8 \times 15 = 60$ ).
Q2 · 3 marks · Short answerBoard level
A triangular park has sides 10 m, 24 m and 26 m. Find its area using Heron's formula and check it by showing that the triangle is right-angled.
Hint
Compare $1 0^{2} + 2 4^{2}$ with $2 6^{2}$ .
Answer
$s = 30$ ; area $= 30 \times 20 \times 6 \times 4 = 14400 = 120$ m $^{2}$ . Check: $1 0^{2} + 2 4^{2} = 100 + 576 = 676 = 2 6^{2}$ , so the triangle is right-angled and the area is $\frac{1}{2} \times 10 \times 24 = 120$ m $^{2}$ .
Q3 · 3 marks · Short answerHarder board level
The diagonals of a rhombus are 24 cm and 18 cm. Find its area, side, perimeter and height.
Hint
Diagonals bisect at right angles.
Answer
Area $= \frac{1}{2} \times 24 \times 18 = 216$ cm $^{2}$ . Side $= 1 2^{2} + 9^{2} = 15$ cm. Perimeter $= 60$ cm. Height $= \frac{216}{15} = 14.4$ cm (area = side x height).
Q4 · 5 marks · Long answerHarder board level
A trapezium has parallel sides 25 cm and 10 cm and non-parallel sides 14 cm and 13 cm. Find its height and area.
Hint
Draw a line through one end of the shorter side parallel to the 14 cm leg to form a triangle.
Answer
The line cuts off a parallelogram, leaving a triangle with sides $13$ , $14$ and $25 - 10 = 15$ . Heron: $s = 21$ , area $= 21 \times 8 \times 7 \times 6 = 84$ . The height of the triangle on the side $15$ is $\frac{2 \times 84}{15} = 11.2$ cm, which is the height of the trapezium. Area $= \frac{1}{2} (25 + 10) (11.2) = 196$ cm $^{2}$ .
Q5 · 4 marks · Case-basedHOTS / competency
Case: an isosceles triangular signboard has equal sides of 13 cm and a base of 10 cm. (i) Find its area using Heron's formula. (ii) Find the height on the base. (iii) Find the cost of painting both faces at 0.8 rupees per square centimetre. (iv) Is the signboard equilateral? Give a reason.
Hint
The height divides the base into two parts of 5 cm.
Answer
(i) $s = 18$ , area $= 18 \times 5 \times 5 \times 8 = 3600 = 60$ cm $^{2}$ . (ii) $\frac{1}{2} \times 10 \times h = 60$ gives $h = 12$ cm (check $1 3^{2} - 5^{2} = 12$ ). (iii) $60 \times 2 \times 0.8 = 96$ rupees. (iv) No, the sides are 13, 13 and 10, not all equal.

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.Heron's formula?Show answer
$s (s - a) (s - b) (s - c)$ with $s$ the semi-perimeter.
2.Area of an equilateral triangle of side $a$ ?Show answer
$\frac{3}{4} a^{2}$ .
3.Area of a trapezium?Show answer
$\frac{1}{2} (a + b) h$ .
4.Area of a rhombus from diagonals?Show answer
$\frac{1}{2} d_{1} d_{2}$ .
5.When is Brahmagupta's formula valid?Show answer
For a cyclic quadrilateral.

Heron's Formula and Areas of Quadrilaterals: frequently asked questions

When should I use Heron's formula?

When you know all three sides and no height. If a height is given, use \frac{1}{2} \times \text{base} \times \text{height}.

Is Heron's formula only for acute triangles?

No, it works for every triangle, whatever its angles.

How do I check a Heron answer?

Find an altitude from the area and test it with Pythagoras in a right sub-triangle, or recompute with a different pairing of factors.