Question

In a triangle, the sum of any two sides exceeds the third by 6 cm. The area of the triangle is __________.

Answer 1

Given:
In a triangle, the sum of any two sides exceeds the third by 6 cm.

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

According to the question,
$$\begin{gathered} a+b-c=6...\left(1\right) \\ b+c-a=6...\left(2\right) \\ a+c-b=6...\left(3\right) \\ \mathrm{Adding}\left(1\right),\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ 2a+2b+2c-a-b-c=18 \\ \Rightarrow a+b+c=18...\left(4\right) \\ \mathrm{Subtracting}\left(1\right)\mathrm{from}\left(4\right),\mathrm{we}\mathrm{get} \\ a+b+c-a-b+c=18-6 \\ \Rightarrow 2c=12 \\ \Rightarrow c=6...\left(5\right) \\ \mathrm{Subtracting}\left(2\right)\mathrm{from}\left(4\right),\mathrm{we}\mathrm{get} \\ a+b+c-b-c+a=18-6 \\ \Rightarrow 2a=12 \\ \Rightarrow a=6...\left(6\right) \\ \mathrm{Subtracting}\left(3\right)\mathrm{from}\left(4\right),\mathrm{we}\mathrm{get} \\ a+b+c-a-c+b=18-6 \\ \Rightarrow 2b=12 \\ \Rightarrow b=6...\left(7\right) \end{gathered}$$

From (5), (6) and (7),
a = b = c = 6 cm
⇒ triangle is equilateral

$$\begin{gathered} \mathrm{Area}=\frac{\sqrt{3}}{4}\times {\left(\mathrm{side}\right)}^2 \\ =\frac{\sqrt{3}}{4}\times {\left(6\right)}^2 \\ =\frac{\sqrt{3}}{4}\times 36 \\ =9\sqrt{3}{\mathrm{cm}}^2 \end{gathered}$$


Hence, the area of the triangle is $\underline{9\sqrt{3}{\mathrm{cm}}^2}.$