Question

Using integration, find the area of the region bounded by the triangle whose vertices are (2, 1), (3, 4) and (5, 2).

Answer 1

$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{points}\mathrm{A}(2,1),\mathrm{B}(3,4)\mathrm{and}\mathrm{C}(5,2) \\ \mathrm{We}\mathrm{need}\mathrm{to}\mathrm{find}\mathrm{area}\mathrm{of}\mathrm{shaded}\mathrm{triangle}\mathrm{ABC} \\ \mathrm{Equation}\mathrm{of}\mathrm{AB}\mathrm{is} \\ y-1=\left(\frac{4-3}{3-2}\right)\left(x-2\right) \\ \Rightarrow 3x-y-5=0...\left(1\right) \\ \mathrm{Equation}\mathrm{of}\mathrm{BC}\mathrm{is} \\ y-4=\left(\frac{2-4}{5-3}\right)\left(x-3\right) \\ \Rightarrow x=y-7=0...\left(2\right) \\ \mathrm{Equation}\mathrm{of}\mathrm{CA}\mathrm{is} \\ y-2=\left(\frac{2-1}{5-2}\right)\left(x-5\right) \\ \Rightarrow x-3y+2=0...\left(3\right) \\ \mathrm{Area}\mathrm{of}\mathrm{\Delta ABC}=\mathrm{Area}\mathrm{of}\mathrm{\Delta ABD}+\mathrm{Area}\mathrm{of}\mathrm{\Delta DBC} \\ \mathrm{In}\mathrm{\Delta ABD}, \\ \mathrm{Consider}\mathrm{point}\mathrm{P}(\mathrm{x},{\mathrm{y}}_2)\mathrm{on}\mathrm{AB}\mathrm{and}\mathrm{Q}(x,y_1)\mathrm{on}\mathrm{AD} \\ \mathrm{Thus},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{approximating}\mathrm{rectangle}\mathrm{with}\mathrm{length}=\left|y_2-y_1\right|\mathrm{and}\mathrm{width}=dx\mathrm{is}\left|y_2-y_1\right|dx \\ \mathrm{The}\mathrm{approximating}\mathrm{rectangle}\mathrm{moves}\mathrm{from}x=2\mathrm{to}x=3 \\ \therefore \mathrm{Area}\mathrm{of}\mathrm{\Delta ABD}={\int }_2^3\left|y_2-y_1\right|dx={\int }_2^3\left(y_2-y_1\right)dx \\ \Rightarrow A={\int }_2^3\left(\left(3x-5\right)-\left(\frac{x+1}{3}\right)\right)dx \\ \Rightarrow A={\int }_2^3\frac{\left(9x-15-x-1\right)}{3}dx \\ \Rightarrow A={\int }_2^3\frac{\left(8x-16\right)}{3}dx \\ \Rightarrow A=\frac{1}{3}{\left[8\frac{x^2}{2}-16x\right]}_2^3 \\ \Rightarrow A=\frac{1}{3}\left[\left(4\times 3^2\right)-\left(16\times 3\right)-\left(4\times 2^2\right)+\left(16\times 2\right)\right] \\ \Rightarrow A=\frac{1}{3}\left(68-64\right) \\ \Rightarrow A=\frac{4}{3}\mathrm{sq}.\mathrm{units} \\ \mathrm{Similarly},\mathrm{for}\mathrm{S}(x,y_4)\mathrm{on}\mathrm{AB}\mathrm{and}\mathrm{R}(x,y_3)\mathrm{on}\mathrm{DC} \\ \mathrm{Area}\mathrm{of}\mathrm{approximating}\mathrm{rectangle}\mathrm{of}\mathrm{length}\left|y_4-y_3\right|\mathrm{and}\mathrm{width}dx=\left|y_4-y_3\right|dx \\ \mathrm{Approximating}\mathrm{rectangle}\mathrm{moves}\mathrm{from}x=3\mathrm{to}x=5 \\ \therefore \mathrm{Area}\mathrm{BDC}={\int }_3^5\left|y_4-y_3\right|dx \\ \Rightarrow A={\int }_3^5\left(\left(7-x\right)-\frac{\left(x+1\right)}{3}\right)dx \\ \Rightarrow A=\frac{1}{3}{\int }_3^5\left(20-4x\right)dx \\ \Rightarrow A=\frac{1}{3}{\left[20x-4\frac{x^2}{2}\right]}_3^5 \\ \Rightarrow A=\frac{1}{3}\left[\left(100-50\right)-\left(60-18\right)\right] \\ \Rightarrow A=\frac{1}{3}\left(50-42\right)=\frac{8}{3}\mathrm{sq}.\mathrm{units} \\ \therefore \mathrm{Area}\mathrm{of}\mathrm{\Delta ABC}=\mathrm{Area}\mathrm{of}\mathrm{\Delta ABD}+\mathrm{Area}\mathrm{of}\mathrm{\Delta DBC}=\frac{4}{3}+\frac{8}{3}=\frac{12}{3}=4\mathrm{sq}.\mathrm{units} \end{gathered}$$