Question

In a scalene triangle, one side exceeds the other two sides by 4 cm and 5 cm respectively and the perimeter of the triangle is 36 cm. The area of the triangle is _________.

Answer 1

Given: The perimeter of the triangle is 36 cm.

Let the sides of the triangle be a, b and c.

According to the question,
b = a − 4 and c = ​a − 5


Perimeter of the triangle = a + b + c
$$\begin{gathered} \Rightarrow 36=a+a-4+a-5 \\ \Rightarrow 36=3a-9 \\ \Rightarrow 3a=36+9 \\ \Rightarrow 3a=45 \\ \Rightarrow a=15\mathrm{cm} \\ \Rightarrow b=a-4=15-4=11\mathrm{cm} \\ \Rightarrow c=a-5=15-5=10\mathrm{cm} \\ \mathrm{Therefore},\mathrm{the}\mathrm{sides}\mathrm{are}15\mathrm{cm},11\mathrm{cm}\mathrm{and}10\mathrm{cm}. \end{gathered}$$

Now, using Heron's formula:
If a, b and c are three sides of a triangle, then
area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.


Here, $s=\frac{15+11+10}{2}=\frac{36}{2}=18$

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{18\left(18-15\right)\left(18-11\right)\left(18-10\right)} \\ =\sqrt{18\left(3\right)\left(7\right)\left(8\right)} \\ =\sqrt{\left(3\times 3\times 2\right)\left(3\right)\left(7\right)\left(2\times 2\times 2\right)} \\ =\left(3\times 2\times 2\right)\sqrt{21} \\ =12\sqrt{21}{\mathrm{cm}}^2 \end{gathered}$$


Hence, the area of the triangle is $\underline{12\sqrt{21}{\mathrm{cm}}^2}$.