Question

the edges of a triangular board are 6 cm , 8 cm , 10 cm . calculate the cost of painting in at the rate of 9 paise per cm²

Answer 1

Dear Student,
Here is the solution of your asked query:
$$\begin{gathered} \mathrm{Edges}\mathrm{of}\mathrm{triangular}\mathrm{board}\mathrm{are}6\mathrm{cm},8\mathrm{cm}\mathrm{and}10\mathrm{cm}. \\ \mathrm{a}=6,\mathrm{b}=8\mathrm{and}\mathrm{c}=10 \\ \mathrm{So},\mathrm{semi}-\mathrm{perimeter}\mathrm{of}\mathrm{triangular}\mathrm{board};\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\frac{6+8+10}{2}=12\mathrm{cm} \\ \mathrm{Then},\mathrm{area}\mathrm{of}\mathrm{triangular}\mathrm{board}\mathrm{is}\mathrm{given}\mathrm{by}; \\ \mathrm{A}=\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)} \\ =\sqrt{12\left(12-6\right)\left(12-8\right)\left(12-10\right)} \\ =\sqrt{12\times 6\times 4\times 2} \\ =\sqrt{2\times 2\times 3\times 2\times 3\times 2\times 2\times 2} \\ =2\times 2\times 2\times 3 \\ =24 \\ \mathrm{So}\mathrm{area}\mathrm{of}\mathrm{triangular}\mathrm{board}=24{\mathrm{cm}}^2 \\ \mathrm{Rate}\mathrm{of}\mathrm{painting}=9\mathrm{paise}/{\mathrm{cm}}^2 \\ \mathrm{So}\mathrm{total}\mathrm{cost}\mathrm{of}\mathrm{painting}=24\times 9=216\mathrm{paise} \end{gathered}$$
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