Question

the perimeter of a triangle is 50cm. one sid of thetriangle is 4cm longer than the smaller side and the third side is 6cm less than twice the smaller side. Find the area of thetriangle.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{smallest}\mathrm{side}\mathrm{of}\mathrm{a}△\mathrm{is}x\mathrm{cm} \\ \mathrm{then},\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{second}\mathrm{side}=\left(x+4\right)\mathrm{cm} \\ \mathrm{and}\mathrm{length}\mathrm{of}\mathrm{third}\mathrm{side}=\left(2x-6\right)\mathrm{cm} \\ \mathrm{Then}\mathrm{perimeter}\mathrm{of}\mathrm{a}\mathrm{given}△=\left(x+x+4+2x-6\right)\mathrm{cm} \\ =\left(4x-2\right)\mathrm{cm} \\ \mathrm{But}\mathrm{perimeter}\mathrm{of}\mathrm{a}\mathrm{given}△=50\mathrm{cm} \\ \mathrm{Hence},4x-2=50 \\ \Rightarrow 4x=50+2 \\ \Rightarrow 4x=52 \\ \Rightarrow x=\frac{52}{4}=13\mathrm{cm} \\ \mathrm{Hence},\mathrm{the}\mathrm{smallest}\mathrm{side}=13\mathrm{cm} \\ \mathrm{second}\mathrm{side}=13+4=17\mathrm{cm} \\ \&\mathrm{third}\mathrm{side}=2\times 13-6=20\mathrm{cm} \\ \mathrm{Hence},a=13\mathrm{cm},b=17\mathrm{cm}\mathrm{and}c=20\mathrm{cm} \\ \therefore s=\frac{a+b+c}{2} \\ \Rightarrow s=\frac{13+17+20}{2}=\frac{50}{2}=25\mathrm{cm} \end{gathered}$$

$$\begin{gathered} \mathrm{Area}\mathrm{of}△=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{25\left(25-13\right)\left(25-17\right)\left(25-20\right)} \\ =\sqrt{25\times 12\times 8\times 5} \\ =\sqrt{5\times 5\times 2\times 2\times 3\times 2\times 2\times 2\times 5} \\ =2\times 2\times 5\sqrt{2\times 3\times 5} \\ =20\sqrt{30}\mathrm{sq}.\mathrm{cm} \\ =20\times 5.477\mathrm{sq}.\mathrm{cm} \\ =109.54\mathrm{sq}.\mathrm{cm} \end{gathered}$$