How do you find the area of a triangle with 3 sides without the height

Let us learn how to find the area of triangle with 3 sides given; that is when the length of three sides of a triangle is given. We find the area of a triangle whose three sides are given by using Heron’s formula. The area of a two-dimensional, closed geometric figure is the amount of region enclosed by that figure. It is expressed in terms of square units. Generally, we find the area of a triangle by halving the product of base and height. But in certain cases of triangles, it is impossible to find the height of the given triangle, and then we use Heron’s formula to determine the area of triangle with 3 sides.

The area of triangle with 3 sides was first derived by the Greek mathematician Heron of Alexandria. He gave a formula which could calculate the area of a triangle without any requirement of measure of angles or any other distances, rather than the length of the sides of the triangle.

Let ABC be a triangle such that the length of the 3 sides of the triangle is AB = c, BC = a and CA = b.

The semi-perimeter of triangle ABC = s = (a + b + c)/2

Then, the area of triangle ABC = √[s × (s – a) × (s – b) × (s – c)].

Learn how to find the area of different types of triangles using Heron’s formula.

Derivation of the Formula for Area of Triangle with 3 Sides

We shall derive the formula for the area of triangles with 3 given sides using the,

• Cosine rule
• Pythagoras formula

Area of Triangle with 3 Sides Formula Using Cosine Rule

Let us recall the law of cosines for any given triangle. According to the rule, the square of any side is equal to twice the product of the other two sides with the cosine of the angle in between them, subtracted from the sum of the square of the other two sides.

Mathematically,

• a2 = b2 + c2 – 2bc cos 𝛼
• b2 = a2 + c2 – 2ac cos 𝛽
• c2 = a2 + b2 – 2ab cos 𝛾

Consider a triangle ABC with angles 𝛼, 𝛽 and 𝛾, and the length of the sides opposite to these angles be a, b, and c, respectively.

From the figure, it is clear that the altitude of the triangle ABC is b sin 𝛾.

Now, from cosine law,

cos 𝛾 = (a2 + b2 – c2)/2ab ….(i)

Then, sin 𝛾 = √(1 – cos2 𝛾)

\(\begin{array}{l}\Rightarrow sin \gamma = \sqrt{1-\frac{(a^{2}+b^{2}-c^{2})^{2}}{4a^{2}b^{2}}}\end{array} \)

\(\begin{array}{l}\Rightarrow sin \gamma = \frac{\sqrt{4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}{2ab}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{1}{2}ab\:sin \gamma = \frac{1}{4}\sqrt{4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}\:\:\:…(ii)\end{array} \)

Now, a is the base of the triangle, and b sin 𝛾 is the altitude.

∴ the area of the triangle ABC = ½ × base × altitude = ½ × a × b sin 𝛾

From (ii), we get,

\(\begin{array}{l}Area\:of\:triangle\:ABC = \frac{1}{4}\sqrt{4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}\end{array} \)

\(\begin{array}{l}= \frac{1}{4}\sqrt{[2ab+(a^{2}+b^{2}-c^{2})][2ab-(a^{2}+b^{2}-c^{2})]}\end{array} \)

\(\begin{array}{l}= \frac{1}{4}\sqrt{[c^{2}-(a-b)^{2}][(a+b)^{2}-c^{2}]}\end{array} \)

\(\begin{array}{l}= \sqrt{\frac{[c^{2}-(a-b)^{2}][(a+b)^{2}-c^{2}]}{16}}\end{array} \)

\(\begin{array}{l}= \sqrt{\frac{(c-a+b)(c+a-b)(a+b-c)(a+b+c)}{16}}\end{array} \)

\(\begin{array}{l}= \sqrt{\frac{(a+b+c)}{2}\frac{(b+c-a)}{2}\frac{(a+c-b)}{2}\frac{(a+b-c)}{2}}\:\:\:…(iii)\end{array} \)

Now, semi-perimeter of triangle ABC = s = (a + b + c)/2

Then, s – a = (b + c – a)/2, s – b = (a + c – b)/2 and s – c = (a + b – c)/2

Substituting all the values in (iii), we get,

\(\begin{array}{l}Area\:of\:triangle\:ABC=\sqrt{s(s-a)(s-b)(s-c)}\end{array} \)

Area of Triangle with 3 Sides Formula Using Pythagoras Theorem

To prove Heron’s formula to calculate the area of triangle with 3 sides, we shall use Pythagoras’ theorem, according to which, in a right triangle, the square of the longest side is equal to the sum of the square of the other two sides.

Consider a triangle ABC, whose length of the three sides are a, b and c.

BD is perpendicular to AC, so ΔADB and ΔCDB are two right-angled triangles.

In ΔADB, applying Pythagoras’ theorem, we get,

AB2 = BD2 + AD2

⇒ c2 = h2 + x2

⇒ x = √(c2 – h2) ….(i)

In ΔCDB, again applying Pythagoras’ theorem, we get,

BC2 = CD2 + BD2

⇒ a2 = (b – x)2 + h2

⇒ a2 = b2 – 2bx + x2 + h2

Substituting the value of x and x2, we get,

⇒ a2 = b2 – 2b√(c2 – h2) + (c2 – h2) + h2

⇒ b2 + c2 – a2 = 2b√(c2 – h2)

On squaring both the sides, we get,

(b2 + c2 – a2)2 = 4b2(c2 – h2)

⇒ (b2 + c2 – a2)2 ÷ 4b2 = c2 – h2

\(\begin{array}{l}\Rightarrow h^{2}=c^{2}-\frac{(b^{2}+c^{2}-a^{2})^{2}}{4b^{2}}\end{array} \)

\(\begin{array}{l}\Rightarrow h^{2}=\frac{4b^{2}c^{2}-(b^{2}+c^{2}-a^{2})^{2}}{4b^{2}}\end{array} \)

\(\begin{array}{l}\Rightarrow h^{2}=\frac{[2bc-(b^{2}+c^{2}-a^{2})][2bc+(b^{2}+c^{2}-a^{2})]}{4b^{2}}\end{array} \)

\(\begin{array}{l}\Rightarrow h^{2}=\frac{[a^{2}-(b-c)^{2}][(b+c)^{2}-a^{2}]}{4b^{2}}\end{array} \)

\(\begin{array}{l}\Rightarrow h^{2}=\frac{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}{4b^{2}}\:\:\:\:…..(ii)\end{array} \)

Now, perimeter of triangle ABC = P = a + b + c

Then, P – 2a = (a + b + c) – 2a = b + c – a

Similarly,

P – 2b = a + c – b and P – 2c = a + b – c

Substituting all these values in (ii), we get,

h2 = [P(P – 2a)(P – 2b)(P – 2c)]/4b2

Taking square root on both sides, we get

\(\begin{array}{l}\Rightarrow h=\frac{\sqrt{P(P-2a)(P-2b)(P-2c)}}{2b}\end{array} \)

Now, the area of triangle ABC = ½ × AC × BD = ½ × b × h

\(\begin{array}{l}=\frac{1}{2}b \times \frac{\sqrt{P(P-2a)(P-2b)(P-2c)}}{2b}\end{array} \)

\(\begin{array}{l}=\frac{1}{4} \times \sqrt{P(P-2a)(P-2b)(P-2c)}\end{array} \)

\(\begin{array}{l}= \sqrt{\frac{P(P-2a)(P-2b)(P-2c)}{16}}\end{array} \)

\(\begin{array}{l}= \sqrt{\frac{P}{2}\frac{(P-2a)}{2}\frac{(P-2b)}{2}\frac{(P-2c)}{2}}\:\:\:\:…(iii)\end{array} \)

Now, semi-perimeter of triangle ABC = P/2 = s (let), then from (iii), we have

\(\begin{array}{l}Area\:of\:triangle\:ABC= \sqrt{s(s-a)(s-b)(s-c)}\end{array} \)

Hence, we derived the formula to determine the area of triangle with 3 sides.

How to Find the Area of Triangle with 3 Sides

To find the area of a triangle whose length of three sides is given:

• First, find the semi-perimeter of the triangle, s = (a+b+c)/2, where a, b and c are the length of the three sides of the triangle.
• Then, find the value of (s – a), (s – b) and (s – c).
• Lastly, find the area of the given triangle using Heron’s formula.

Note: If the area of a triangle using Heron’s formula results in zero or a negative quantity, the triangle with the given sides is not possible.

Try out: Heron’s formula calculator

Solved Examples on Area of Triangle with 3 Sides

Example 1:

Find the area of triangle with 3 sides, 7 cm, 6 cm and 5 cm, respectively.

Solution:

Let a = 7 cm, b = 6 cm and c = 5 cm.

Semi-perimeter of the triangle = s = (a+b+c)/2 = (7 + 6 + 5)/2 = 18/2 = 9 cm.

Area of the given triangle = √[s(s – a)(s – b)(s – c)] = √[9(9 – 7)(9 – 6)(9 – 5)]

= √[9 × 2 × 3 × 4] = 3 × 2 × √[2 × 3] = 6√6 cm2

∴ The area of the given triangle is 6√6 cm2.

Example 2:

Find the area of an isosceles triangle whose equal sides are of length 9 cm, and the third side is 12 cm.

Solution:

Let a = b = 9 cm and c = 12 cm.

Semi-perimeter of the triangle = s = (a+b+c)/2 = (9 + 9 + 12)/2 = 30/2 = 15 cm.

Area of the given triangle = √[s(s – a)(s – b)(s – c)] = √[15(15 – 9)(15 – 9)(15 – 12)]

= √[15 × 6 × 6 × 3] = 6 × 3 × √5 = 18√5 cm2

∴ the area of the given triangle is 18√5 cm2.

Related Articles

• Area of Scalene Triangle
• Area of Isosceles Triangle
• Area of Equilateral Triangle
• Heron’s Formula Class 9
• Important Questions Class 9 Maths Chapter 12 Herons Formula

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Practice Questions on Area of Triangle with 3 Sides

1. Find the area of triangle whose length of the three sides are:

(i) 3 cm, 4 cm and 5 cm

(ii) 8 cm, 7 cm and 11 cm

(iii) 2 m, 4 m and 3.5 m

2. The semi-perimeter of an isosceles triangle is 15 cm, and the length of the unequal side is 12 cm. Find the length of the other two sides and the area of the triangle.

Frequently Asked Questions on Area of Triangle with 3 Sides