Question

The base of an isosceles triangle is 10 cm and one of its equal sides is 13 cm. Find its area using Heron’s formula.

Answer 1

Hint:
We will find the semi-perimeter of the triangle and the difference between each side and the semi-perimeter. We will substitute these values in Heron’s formula. Then we will solve the expression and find the area.

Formulas used:
We will use the following formulas:
1) Semi-perimeter of a triangle is half times the sum of the length of its sides:
\[s = \dfrac{{a + b + c}}{2}\]
2) According to the Heron’s formula, the area of a triangle with sides \[a\], \[b\] and \[c\] and semi-perimeter \[s\] is:
\[A = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \]

Complete step by step solution:
We know that an isosceles triangle is a triangle with 2 equal sides. We have the length of one of the equal sides as 13 cm, so the length of the other equal side will also be 13 cm.
We will first find the semi-perimeter of the triangle. We will substitute 10 for \[a\] and 13 for \[b\] and \[c\] in the formula for semi-perimeter. Therefore, we get
\[\begin{array}{l} \Rightarrow s = \dfrac{{10 + 13 + 13}}{2}\\ \Rightarrow s = \dfrac{{36}}{2}\\ \Rightarrow s = 18\end{array}\]
Now, we will find the value of \[s - a\]. Substituting 10 for \[a\] and 18 for \[s\], we get
\[ \Rightarrow 18 - 10 = 8\]
We will find the value of \[s - b\]. Substituting 13 for \[b\] and 18 for \[s\], we get
\[ \Rightarrow 18 - 13 = 5\]
We will find the value of \[s - c\]. Substituting 13 for \[c\] and 18 for \[s\], we get
\[ \Rightarrow 18 - 13 = 5\]
Now we will find the area of the triangle.
Substituting 18 for \[s\], 8 for \[s - a\] and 5 for \[s - b\] and \[s - c\] in the Heron’ formula, \[A = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \], we get
\[ \Rightarrow A = \sqrt {18 \times 8 \times 5 \times 5} \]
We will write the numbers as the product of their prime factors. Therefore, we get
\[ \Rightarrow A = \sqrt {2 \times 3 \times 3 \times 2 \times 2 \times 2 \times 5 \times 5} \]
We will rearrange the terms and group the same numbers; then we will take the terms that occur in pairs out of the square root:
\[\begin{array}{l} \Rightarrow A = \sqrt {2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5} \\ \Rightarrow A = 2 \times 2 \times 3 \times 5\end{array}\]
Multiplying the terms, weget
\[ \Rightarrow A = 60\]

$\therefore $ The area of the triangle is 60 square centimetres.

Note:
The Heron’s formula was given by a Greek mathematician, Hero Alexandria. It can be derived using the Pythagoras Theorem and the Law of Cosines of a triangle. With the help of Heron’s formula we can find the area of any type of triangle, whether it is an isosceles triangle or equilateral triangle. Heron’s formula is usually used in the application of trigonometry to prove different laws of trigonometry.