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Heron's Formula - Area of a Triangle by Heron's Formula

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Basic Area of a Triangle: For any triangle where the base bb and the perpendicular height hh are known, the area is calculated as 12×b×h\frac{1}{2} \times b \times h. Visually, this is equivalent to half the area of a rectangle or parallelogram with the same base and height, as seen when a rectangle is cut diagonally into two equal right-angled triangles.

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Need for Heron's Formula: When only the lengths of the three sides (aa, bb, and cc) are known and the height is not easily measurable, Heron's formula is used. Visually, imagine a scalene triangle where no side is perfectly horizontal or vertical; this formula allows us to find the interior space using only the boundary measurements.

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Semi-perimeter (ss): The semi-perimeter is defined as half of the triangle's total perimeter, calculated as s=a+b+c2s = \frac{a+b+c}{2}. Visually, if you were to unfold the three sides of the triangle into a single straight line, the semi-perimeter ss would be the point exactly halfway along that line.

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Heron's Formula: The area is calculated using the formula s(s−a)(s−b)(s−c)\sqrt{s(s-a)(s-b)(s-c)}. Visually, the terms under the square root represent the geometric weight of each side relative to the semi-perimeter, effectively measuring the 2D space enclosed by the three segments.

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Simplification Technique: To calculate the square root easily, factorize ss and the three differences (s−a,s−b,s−c)(s-a, s-b, s-c) into their prime factors rather than multiplying them into one large number. Visually, this is like 'pairing up' identical numbers under the root sign; for every two identical factors inside the square root, one factor can be moved outside.

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Application to Equilateral Triangles: In a triangle where all sides are equal (a=b=ca=b=c), the formula simplifies significantly to 34a2\frac{\sqrt{3}}{4} a^2. Visually, this represents the area of a perfectly symmetrical shape with three equal 60∘60^{\circ} angles and equal side lengths.

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Application to Quadrilaterals: Heron's formula can be used to find the area of a quadrilateral if the length of a diagonal is known. Visually, you divide the quadrilateral into two distinct triangles using the diagonal, calculate the area of each triangle separately using their sides, and sum the results.

📐Formulae

s=a+b+c2s = \frac{a+b+c}{2}

Area of Triangle=s(s−a)(s−b)(s−c)\text{Area of Triangle} = \sqrt{s(s-a)(s-b)(s-c)}

Area of Equilateral Triangle=34a2\text{Area of Equilateral Triangle} = \frac{\sqrt{3}}{4} a^2

Perimeter=a+b+c=2s\text{Perimeter} = a + b + c = 2s

💡Examples

Problem 1:

Find the area of a triangle whose sides are 13 cm13\text{ cm}, 14 cm14\text{ cm}, and 15 cm15\text{ cm}.

Solution:

  1. Calculate semi-perimeter ss: s=13+14+152=422=21 cms = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21\text{ cm}
  2. Apply Heron's Formula: Area=s(s−a)(s−b)(s−c)\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} Area=21(21−13)(21−14)(21−15)\text{Area} = \sqrt{21(21-13)(21-14)(21-15)} Area=21×8×7×6\text{Area} = \sqrt{21 \times 8 \times 7 \times 6}
  3. Factorize to simplify: Area=(3×7)×(2×2×2)×7×(2×3)\text{Area} = \sqrt{(3 \times 7) \times (2 \times 2 \times 2) \times 7 \times (2 \times 3)} Grouping pairs: Area=72×32×24\text{Area} = \sqrt{7^2 \times 3^2 \times 2^4} Area=7×3×4=84\text{Area} = 7 \times 3 \times 4 = 84 Result: 84 cm284\text{ cm}^2

Explanation:

We first identify the three sides a,b,ca, b, c. We calculate the semi-perimeter ss, then find the differences between the semi-perimeter and each side. By substituting these into Heron's formula and using prime factorization, we find the area without needing the height.

Problem 2:

An isosceles triangle has a perimeter of 32 cm32\text{ cm} and the ratio of its equal side to its base is 3:23:2. Find the area of the triangle.

Solution:

  1. Find side lengths: Let the sides be 3x,3x,3x, 3x, and 2x2x. 3x+3x+2x=32  ⟹  8x=32  ⟹  x=43x + 3x + 2x = 32 \implies 8x = 32 \implies x = 4 Sides are 12 cm,12 cm,12\text{ cm}, 12\text{ cm}, and 8 cm8\text{ cm}.
  2. Find semi-perimeter ss: s=322=16 cms = \frac{32}{2} = 16\text{ cm}
  3. Apply Heron's Formula: Area=16(16−12)(16−12)(16−8)\text{Area} = \sqrt{16(16-12)(16-12)(16-8)} Area=16×4×4×8\text{Area} = \sqrt{16 \times 4 \times 4 \times 8}
  4. Simplify: Area=16×16×4×2=16×22=322\text{Area} = \sqrt{16 \times 16 \times 4 \times 2} = 16 \times 2\sqrt{2} = 32\sqrt{2} Result: 322 cm232\sqrt{2}\text{ cm}^2

Explanation:

We use the given ratio and total perimeter to solve for the individual side lengths. After finding the sides, we calculate the semi-perimeter and apply Heron's formula, simplifying the final radical for the result.