Heron's Formula

1. Area of a Triangle — Base and Height Recap

From Class 6 onwards you used the formula:

This works perfectly when the base and the corresponding height (altitude) are both known. A few quick examples:

• Right triangle with legs $5$ cm and $12$ cm: $A = \\frac{1}{2} \\times 5 \\times 12 = 30$ cm² — here the two legs are perpendicular, so one leg is the base and the other is the height, given for free.
• Any triangle with base $10$ cm and altitude $8$ cm: $A = \\frac{1}{2} \\times 10 \\times 8 = 40$ cm².
• Equilateral triangle with side $a$: the altitude $= \\frac{\\sqrt{3}}{2}a$, so $A = \\frac{1}{2} \\times a \\times \\frac{\\sqrt{3}}{2}a = \\frac{\\sqrt{3}}{4}a^{2}$.

The problem: in many real-life triangles — especially scalene ones (all sides different) — only the three side lengths are known from measurements. Computing the altitude requires an extra step (Pythagoras or trigonometry). Heron's formula eliminates that extra step entirely.

Derivation sketch (for the curious): starting from $A = \\frac{1}{2}bh$, express $h$ in terms of $a, b, c$ using the Pythagorean theorem, then substitute and simplify. After considerable algebra, you arrive at the elegant form below. The derivation is not required in board exams but understanding the idea helps you trust and remember the formula.

2. Heron's Formula

Named after the Greek mathematician Heron of Alexandria (c. 10–70 CE), who described the result in his work Metrica. The formula is also written as Hero's formula. For a triangle with sides $a$, $b$, $c$:

where $s$ is the semi-perimeter (defined in the next section). What makes this extraordinary:

• No height, no angles, no trigonometry — only the three side lengths.
• Works for any triangle: scalene, isosceles, equilateral, right-angled.
• Fully symmetric in $a, b, c$: labelling the sides differently does not change the answer.

Memory trick — "s minus each side": subtract each side from $s$ in turn, multiply all four quantities ($s$ and the three remainders) together, and take the square root. A right-triangle check always confirms your arithmetic.

3. Semi-perimeter $s$

The semi-perimeter is simply half the perimeter:

Key observations:

• $(s - a) + (s - b) + (s - c) = 3s - (a+b+c) = 3s - 2s = s$. The three "remainders" add back to $s$.
• Each of $(s-a), (s-b), (s-c)$ must be strictly positive — guaranteed by the triangle inequality. If any remainder is zero or negative, the three lengths do not form a valid triangle.
• For an equilateral triangle with side $a$: $s = \\frac{3a}{2}$, so $(s-a) = \\frac{a}{2}$ for all three sides.
• Always compute $s$ first, then the three brackets — do not try to shortcut straight to the formula without $s$.

4. Step-by-Step Method

Follow these four steps every time to avoid errors:

1. Label the sides. Identify $a$, $b$, $c$ clearly. The labelling is arbitrary — any side can be called $a$.
2. Compute the semi-perimeter. $s = \\dfrac{a+b+c}{2}$. Double-check: $2s$ must equal $a+b+c$.
3. Compute the three brackets. Find $(s-a)$, $(s-b)$, $(s-c)$ individually and verify that all three are positive.
4. Apply the formula and simplify. $A = \\sqrt{s(s-a)(s-b)(s-c)}$. Prime-factorise the product inside the root to extract perfect squares, then attach the correct unit squared.

5. Application to Quadrilaterals

Heron's formula applies to triangles only — but every quadrilateral can be split into two triangles by drawing one diagonal. Then:

Procedure:

1. Choose a diagonal that divides the quadrilateral into two triangles whose three sides are all known (or computable via Pythagoras).
2. Apply Heron's formula to each triangle separately.
3. Add the two areas for the total.

Common shapes tested in board exams:

• Rhombus (all sides equal): a diagonal splits it into two congruent triangles. Compute Heron's for one and double it. If both diagonals $d_1, d_2$ are given, use $A = \\frac{1}{2}d_1 d_2$ instead (faster). When only one diagonal and the side are given, use Heron's.
• Parallelogram (opposite sides equal): same idea — two congruent triangles from a diagonal. Heron's gives the area without needing the height.
• Trapezium (one pair of parallel sides): diagonal gives two different triangles — compute Heron's for each and add.
• Irregular quadrilateral: must know all four sides plus the diagonal. Apply Heron's to each of the two resulting triangles.

Important: choose the diagonal that gives you enough information. When a right angle is present in the quadrilateral, always use Pythagoras to find the diagonal first — it simplifies the subsequent Heron's calculation.

6. Special Case — Equilateral Triangle

Derive the area formula for an equilateral triangle with side $a$ directly from Heron's formula:

• $s = \\dfrac{a+a+a}{2} = \\dfrac{3a}{2}$.
• $(s-a) = \\dfrac{3a}{2} - a = \\dfrac{a}{2}$ (same for all three sides).
• $A = \\sqrt{\\dfrac{3a}{2} \\cdot \\dfrac{a}{2} \\cdot \\dfrac{a}{2} \\cdot \\dfrac{a}{2}} = \\sqrt{\\dfrac{3a^4}{16}} = \\dfrac{a^2\\sqrt{3}}{4}$.

Memorise this — it saves considerable time. Quick applications:

• Equilateral triangle of side $10$ cm: $A = \\frac{\\sqrt{3}}{4} \\times 100 = 25\\sqrt{3} \\approx 43.3$ cm².
• Equilateral triangle of side $60$ cm (traffic sign): $A = \\frac{\\sqrt{3}}{4} \\times 3600 = 900\\sqrt{3} \\approx 1558.8$ cm².

Other useful quick formulas (for reference):

• Right triangle with legs $p$ and $q$: $A = \\frac{1}{2}pq$.
• Isosceles triangle with equal sides $a$ and base $b$: $A = \\frac{b}{4}\\sqrt{4a^{2}-b^{2}}$ (derived by Heron's).
• Rhombus with diagonals $d_1, d_2$: $A = \\frac{1}{2}d_1 d_2$.

Exercise 10.1 (All 6 questions)

Q1. A traffic signal board indicating "SCHOOL AHEAD" is an equilateral triangle with side $a$. Find the area of the signal board using Heron's formula. If its perimeter is 180 cm, what will be the area?

Solution: $s = \\frac{3a}{2}$. Then $(s-a) = \\frac{a}{2}$ for all three sides.

$A = \\sqrt{\\frac{3a}{2} \\cdot \\frac{a}{2} \\cdot \\frac{a}{2} \\cdot \\frac{a}{2}} = \\sqrt{\\frac{3a^{4}}{16}} = \\frac{\\sqrt{3}}{4}a^{2}$.

For perimeter $= 180$ cm: $a = 60$ cm. $A = \\frac{\\sqrt{3}}{4} \\times 3600 = 900\\sqrt{3}$ cm².

Q2. The triangular side walls of a flyover have sides 122 m, 22 m and 120 m. The advertisements yield Rs 5000 per m² per year. A company hired one wall for 3 months. How much rent did it pay?

Solution: Same as Example 2. Area $= 1320$ m². Rent for 3 months $=$ Rs $16{,}50{,}000$.

Q3. There is a slide in a park. A triangular cross-section has sides 1.7 m, 1.8 m and 1 m. Find the area of the cross-section.

Solution: $s = \\frac{1.7+1.8+1}{2} = \\frac{4.5}{2} = 2.25$ m.

$(s-1.7)=0.55$,  $(s-1.8)=0.45$,  $(s-1)=1.25$.

$A = \\sqrt{2.25 \\times 0.55 \\times 0.45 \\times 1.25}$.

Product $= 2.25 \\times 0.55 \\times 0.45 \\times 1.25 = 0.6992...$

$A \\approx \\sqrt{0.6992} \\approx 0.836$ m² $\\approx 0.84$ m².

Q4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Solution: Third side $= 42 - 18 - 10 = 14$ cm.

$s = \\frac{42}{2} = 21$ cm.  $(s-18)=3$,  $(s-10)=11$,  $(s-14)=7$.

$A = \\sqrt{21 \\times 3 \\times 11 \\times 7} = \\sqrt{4851} = \\sqrt{9 \\times 539} = 3\\sqrt{539}$ cm².

$= 3\\sqrt{7 \\times 77} = 3 \\times 7\\sqrt{11} = 21\\sqrt{11}$ cm² $\\approx 69.65$ cm².

Q5. Sides of a triangle are in the ratio 12:17:25 and its perimeter is 540 cm. Find the area.

Solution: $k(12+17+25) = 540 \\Rightarrow 54k = 540 \\Rightarrow k = 10$. Sides: $120$ cm, $170$ cm, $250$ cm.

$s = \\frac{540}{2} = 270$ cm.  $(s-120)=150$,  $(s-170)=100$,  $(s-250)=20$.

$A = \\sqrt{270 \\times 150 \\times 100 \\times 20} = \\sqrt{81{,}000{,}000} = 9000$ cm².

Q6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Solution: Base $= 30 - 12 - 12 = 6$ cm.

$s = \\frac{30}{2} = 15$ cm.  $(s-12)=3$,  $(s-12)=3$,  $(s-6)=9$.

$A = \\sqrt{15 \\times 3 \\times 3 \\times 9} = \\sqrt{1215} = \\sqrt{81 \\times 15} = 9\\sqrt{15}$ cm² $\\approx 34.85$ cm².

Common Mistakes and Exam Tips

• Using full perimeter instead of semi-perimeter: $s = \\frac{a+b+c}{2}$, NOT $a+b+c$. Divide by 2 before substituting.
• Arithmetic inside the root: compute the product $s(s-a)(s-b)(s-c)$ as a single number. Use prime factorisation to pull out perfect squares before taking the root — never root each factor separately (e.g. $\\sqrt{21}\\times\\sqrt{8}$ is harder than $\\sqrt{21\\times8}=\\sqrt{168}$).
• Wrong units: area is always in square units. Sides in m give area in m², not m.
• Quadrilateral — finding the diagonal: you must know (or compute) the diagonal to split the quadrilateral. If a right angle is present, use Pythagoras. Never assume the diagonal is given if it is not stated.
• Equilateral formula misapplied: $\\frac{\\sqrt{3}}{4}a^{2}$ is valid only when all three sides are equal. Do not use it for isosceles or scalene triangles.
• Cross-check with right-triangle shortcut: whenever the sides satisfy $a^{2}+b^{2}=c^{2}$, verify your Heron's answer with $A=\\frac{1}{2}\\times a\\times b$. Any discrepancy signals an arithmetic error.