Question

Find area of an isosceles triangle which has perimeter 50m and whose unequal side is 10m (use root 15= 3.87) .

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{measure}\mathrm{of}\mathrm{equal}\mathrm{sides}\mathrm{of}\mathrm{an}\mathrm{isosceles}△\mathrm{be}x\mathrm{m} \\ \mathrm{Now},\mathrm{sides}\mathrm{of}\mathrm{the}\mathrm{isosceles}△\mathrm{are}: \\ a=x\mathrm{m} \\ b=x\mathrm{m} \\ c=10\mathrm{m} \\ \mathrm{Now},\mathrm{perimeter}=50\mathrm{m} \\ \Rightarrow a+b+c=50 \\ \Rightarrow x+x+10=50 \\ \Rightarrow 2x=40 \\ \Rightarrow x=20 \\ \mathrm{So},a=b=20\mathrm{m} \\ \mathrm{Now},\mathrm{semiperimeter},s=\frac{50}{2}=25\mathrm{m} \\ \mathrm{Now},\mathrm{area}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{25\left(25-20\right)\left(25-20\right)\left(25-10\right)} \\ =\sqrt{25\times 5\times 5\times 15} \\ =\sqrt{5\times 5\times 5\times 5\times 15} \\ =25\sqrt{15}{\mathrm{m}}^2 \\ =25\times 3.87 \\ =96.75{\mathrm{m}}^2 \end{gathered}$$