Question

The difference between the semi-perimeter and the sides of a ∆ABC are 8 cm, 7 cm and 5 cm respectively. The area of the triangle is
(a) $20\sqrt{7}{\mathrm{cm}}^2$
(b) $10\sqrt{14}{\mathrm{cm}}^2$
(c) $20\sqrt{14}{\mathrm{cm}}^2$
(d) 140 cm²

Answer 1

(c) $20\sqrt{14}{\mathrm{cm}}^2$
Let the sides of the triangle be a cm, b cm and c cm and the semi-perimeter be s cm.
Given:
$s-a=8\mathrm{cm},s-b=7\mathrm{cm}\mathrm{and}s-c=5\mathrm{cm}$
By adding all these, we get:
$$\begin{gathered} s-a+s-b+s-c=8+7+5 \\ \Rightarrow 3s-(a+b+c)=20\left[2s=\left(a+b+c\right)\right] \\ \Rightarrow 3s-2s=20 \\ \Rightarrow s=20\mathrm{cm} \end{gathered}$$

$$\begin{gathered} \mathrm{By}\mathrm{applying}\mathrm{Heron}\text{'}\mathrm{s}\mathrm{formula},\mathrm{we}\mathrm{get}: \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{20\times 8\times 7\times 5} \\ =\sqrt{5\times 4\times 7\times 5\times 4\times 2} \\ =5\times 4\times \sqrt{14} \\ =20\sqrt{14}{\mathrm{cm}}^2 \end{gathered}$$