This MCQ module is based on: Heron’s Formula for Triangles
TOPIC 17 OF 25
Heron’s Formula for Triangles
🎓 Class 9
Mathematics
CBSE
Theory
Ch 6 — Measuring Space: Perimeter and Area
⏱ ~40 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Heron’s Formula for Triangles
Targeting Class 9 level in Mensuration, with Intermediate difficulty.
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6.5 Recap — Area of a Triangle
For a triangle with base \(b\) and height \(h\), area \(= \tfrac12 bh\). For a right-angled triangle, the two legs serve as base and height directly. For an equilateral triangle of side \(a\), height = \(\tfrac{\sqrt 3}{2}a\), so area \(=\tfrac{\sqrt 3}{4}a^2\).
But what if the height is unknown — say a scalene triangle is described only by its three side-lengths? Heron's formula handles exactly this case.
6.6 Heron's Formula
Heron's Formula
For a triangle with sides \(a\), \(b\) and \(c\), let \(s = \dfrac{a+b+c}{2}\) be the semi-perimeter?. Then the area is
\[\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}.\]Historical Note
Heron of Alexandria (c. 10–70 CE) was a Greek engineer and mathematician. His proof appears in his book Metrica. The same formula was known earlier in India — Brahmagupta (7th century) generalised it to a four-sided cyclic figure. Heron also designed the aeolipile, often called the world's first steam engine.Triangle ABC with sides \(a\) (opp. A), \(b\) (opp. B), \(c\) (opp. C).
Why It Works (Sketch)
Drop perpendicular from C onto AB at foot D. Let AD = \(x\), so DB = \(c - x\), and the height is \(h = \) CD. Pythagoras in the two right triangles gives \(b^2 = x^2 + h^2\) and \(a^2 = (c-x)^2 + h^2\). Subtracting and solving for \(x\), then for \(h\), and finally substituting in \(\tfrac12 c h\) yields, after careful algebra, exactly \(\sqrt{s(s-a)(s-b)(s-c)}\). The full derivation is a beautiful (if lengthy) algebraic exercise.
Worked Examples
Example 5. Find the area of a triangle with sides 3 cm, 4 cm and 5 cm.
\(s = (3+4+5)/2 = 6\). Area \(= \sqrt{6\cdot3\cdot2\cdot1} = \sqrt{36} = 6\) cm². (Note 3-4-5 is a right triangle, so this also equals \(\tfrac12\cdot3\cdot4=6\) ✓.)
Example 6. A triangular plot has sides 13 m, 14 m and 15 m. What is its area?
\(s = (13+14+15)/2 = 21\). Area \(= \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84\) m².
Example 7. The sides of a triangular signboard are in the ratio 5 : 12 : 13 and its perimeter is 60 cm. Find its area.
Sides: 10, 24, 26 cm (since 5+12+13 = 30 and 60/30 = 2). \(s = 30\). Area \(=\sqrt{30\cdot20\cdot6\cdot4}=\sqrt{14400}=120\) cm².
Example 8. Find the area of an equilateral triangle of side 6 cm using Heron's formula and verify with the standard formula.
\(s = 9\). Heron: \(\sqrt{9\cdot3\cdot3\cdot3} = \sqrt{243} = 9\sqrt{3}\) cm². Standard: \(\tfrac{\sqrt 3}{4}\cdot 36 = 9\sqrt 3\) cm². ✓
Example 9. An isosceles triangle has perimeter 30 cm and equal sides each of length 12 cm. Find its area.
Third side \(= 30 - 24 = 6\) cm. \(s = 15\). Area = \(\sqrt{15\cdot 3\cdot 3\cdot 9} = \sqrt{1215} = 9\sqrt{15}\) ≈ 34.86 cm².
6.7 Median Property
A median of a triangle joins a vertex to the midpoint of the opposite side. Each median divides a triangle into two smaller triangles of equal area. Reason: both smaller triangles share the same height from the vertex and have equal bases (the two halves of the opposite side).
Median AD divides △ABC into two triangles of equal area.
Activity: Verify Heron's Formula
L3 ApplyMaterials: graph paper, pencil, ruler.
Predict: If a triangle has sides 6, 8, 10 (right triangle), Heron's formula should give the same area as ½ × base × height.
- Draw a 6-8-10 right triangle on graph paper.
- Compute area using ½ × base × height = ½ × 6 × 8.
- Now compute s = 12. Apply Heron's formula.
- Compare answers.
Both methods give 24. Heron: \(\sqrt{12\cdot 6\cdot 4\cdot 2}=\sqrt{576}=24\). The formula works for ALL triangles, even when no convenient base or height is given.
6.8 Application to Quadrilaterals (a glimpse)
If a quadrilateral's four side-lengths and one diagonal are known, you can split it into two triangles along the diagonal and compute each area with Heron's formula. The total is the area of the quadrilateral. We will explore this fully in the next part.
Try it: A quadrilateral ABCD has AB = 5, BC = 6, CD = 7, DA = 8 cm and diagonal AC = 9 cm. Find the area. Hint: split into △ABC and △ACD.
Competency-Based Questions
Scenario: A village panchayat is constructing a triangular community garden. The three sides measured along its boundary are 50 m, 70 m and 80 m. The plan also requires laying water pipes from each vertex to the opposite midpoint (the three medians).
Q1. Find the area of the garden using Heron's formula.
L3 Apply(c). \(s = 100\). Area \(= \sqrt{100\cdot 50\cdot 30\cdot 20} = \sqrt{3{,}000{,}000} = 1000\sqrt{6}\) ≈ 2449.5 m².
Q2. The panchayat wants to build a fountain at a point that splits the garden into three equal-area sections. Analyse: where is this point?
L4 AnalyseIt is the centroid — the intersection of the three medians. Each median splits the triangle into two equal-area halves. The centroid divides each median in 2:1 ratio and the three smaller triangles formed by joining centroid to the vertices each have area 1/3 of total ≈ 816.5 m².
Q3. Evaluate: a contractor proposes that "since the perimeter is 200 m, the area is the same as a square of side 50 m, i.e. 2500 m²." Critique.
L5 EvaluateWrong. Perimeter does not determine area. Among all shapes of perimeter 200 m, the circle has the maximum area; among triangles with perimeter 200 m, the equilateral has the maximum. Our scalene triangle has area ≈ 2449.5 m² — close to but different from the contractor's 2500 m². Same perimeter ≠ same area.
Q4. Design: pick three sides (each between 30 m and 100 m) for a triangular plot whose Heron-area is exactly 1200 m². Justify your choice.
L6 CreateSample answer: sides 40, 50, 30 m. \(s = 60\). Area \(=\sqrt{60\cdot20\cdot10\cdot30}=\sqrt{360000}=600\) m². Adjust: scale every side by \(\sqrt{2}\) to double the area: 56.6, 70.7, 42.4 m. Many valid triples exist.
Assertion–Reason Questions
A: Heron's formula always works for a triangle, even when none of its angles is right.
R: The formula uses only the three side lengths and not any angle or height.
R: The formula uses only the three side lengths and not any angle or height.
(a) R is precisely the reason Heron's formula is so general.
A: A triangle with sides 1 cm, 2 cm and 5 cm has area \(\sqrt{4\cdot 3\cdot 2\cdot(-1)}\) — a negative value under the root.
R: The triangle inequality \(a + b > c\) must hold; otherwise the triangle does not exist.
R: The triangle inequality \(a + b > c\) must hold; otherwise the triangle does not exist.
(a) Both are true; R explains A: the negative inside the radical signals that the triangle inequality fails (1 + 2 = 3 < 5), so no real triangle exists.
A: A median of a triangle divides it into two triangles of equal area.
R: The two smaller triangles have equal bases and the same height from the third vertex.
R: The two smaller triangles have equal bases and the same height from the third vertex.
(a) Equal base × same height ⇒ equal area.
Frequently Asked Questions
What is Heron's formula and who discovered it?
Heron's formula gives the area of a triangle in terms of its three sides a, b, c. With s = (a + b + c)/2, the area is Area = sqrt(s(s - a)(s - b)(s - c)). It was published by Heron of Alexandria, a Greek engineer and mathematician, around 60 CE.
How do you find the area of a triangle with sides 13, 14 and 15 cm?
Compute s = (13 + 14 + 15)/2 = 21. Then s - a = 8, s - b = 7, s - c = 6. Area = sqrt(21 x 8 x 7 x 6) = sqrt(7056) = 84 cm^2.
When should you use Heron's formula instead of (1/2) x base x height?
Use Heron's formula when you know all three sides but do not know an altitude. Use (1/2) x base x height when you have the perpendicular height. For right-angled triangles, (1/2) x base x height is fastest because the legs are perpendicular.
How can Heron's formula find the area of a quadrilateral?
Draw a diagonal to split the quadrilateral into two triangles. Compute the sides of each triangle, apply Heron's formula to each, and add the two areas. This works for any quadrilateral, including irregular fields.
What is the semi-perimeter and why is it used in Heron's formula?
The semi-perimeter s = (a + b + c)/2 is half the triangle's perimeter. It appears in Heron's formula because the algebraic identity behind the proof factors neatly through s, s - a, s - b and s - c, giving a symmetric, side-only formula.
Does Heron's formula work for an equilateral triangle of side a?
Yes. For an equilateral triangle s = 3a/2, so s - a = a/2 (three times). Area = sqrt((3a/2)(a/2)^3) = sqrt(3 a^4 / 16) = (sqrt(3)/4) a^2, the standard equilateral-triangle area formula.
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