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Question Number 84892 by Power last updated on 17/Mar/20

Commented by Power last updated on 17/Mar/20

$$\mathbf{please}\mathbf{see}$$

Commented by Power last updated on 17/Mar/20

$$\mathbf{sir}\mathbf{mr}\mathbf{W}\mathbf{please}$$

Commented by Power last updated on 17/Mar/20

$$\mathrm{sir}\mathrm{mind}\mathrm{is}\mathrm{power}\mathrm{pls}$$

Commented by Power last updated on 18/Mar/20

$$\mathrm{solution}\mathrm{sir}\mathrm{plese}$$

Answered by mr W last updated on 18/Mar/20

Commented by Power last updated on 18/Mar/20

$$\mathrm{great}\mathrm{sir}$$$$$$

Commented by mr W last updated on 18/Mar/20

$$makeNG//AP$$$$PR//AB,\frac{DP}{AD}=\frac{PR}{AB}=\frac{2}{3}$$$$\Rightarrow \frac{DP}{AP}=\frac{2}{3+2}=\frac{2}{5},\frac{AD}{AP}=\frac{3}{3+2}=\frac{3}{5}$$$$\frac{{\mathrm{\Delta }}_{DPR}}{{\mathrm{\Delta }}_{DAB}}={\left(\frac{PR}{AB}\right)}^2={\left(\frac{2}{3}\right)}^2=\frac{4}{9}$$$$\frac{{\mathrm{\Delta }}_{DAB}}{{\mathrm{\Delta }}_{PAB}}=\frac{AD}{AP}=\frac{3}{5}$$$$\frac{{\mathrm{\Delta }}_{PAB}}{{\mathrm{\Delta }}_{ABC}}=\frac{BP}{BC}=\frac{1}{3}$$$$\frac{{\mathrm{\Delta }}_{DPR}}{{\mathrm{\Delta }}_{ABC}}=\frac{{\mathrm{\Delta }}_{DPR}}{{\mathrm{\Delta }}_{DAB}}\times \frac{{\mathrm{\Delta }}_{DAB}}{{\mathrm{\Delta }}_{PAB}}\times \frac{{\mathrm{\Delta }}_{PAB}}{{\mathrm{\Delta }}_{ABC}}$$$$\Rightarrow \frac{{\mathrm{\Delta }}_{DPR}}{{\mathrm{\Delta }}_{ABC}}=\frac{4}{9}\times \frac{3}{5}\times \frac{1}{3}=\frac{4}{45}$$$$$$$$\frac{NG}{AP}=\frac{BG}{BP}=\frac{BN}{AB}=\frac{1}{3}$$$$\Rightarrow NG=\frac{AP}{3}$$$$\Rightarrow BG=\frac{BP}{3}=\frac{1}{3}\times \frac{BC}{3}=\frac{BC}{9}$$$$\Rightarrow GC=BC-BG=\frac{8}{9}\times BC$$$$\frac{EP}{NG}=\frac{PC}{GC}=\frac{2}{3}\times \frac{BC}{GC}=\frac{2}{3}\times \frac{9}{8}=\frac{3}{4}$$$$\frac{EP}{DP}=\frac{EP}{NG}\times \frac{NG}{AP}\times \frac{AP}{DP}$$$$\Rightarrow \frac{EP}{DP}=\frac{3}{4}\times \frac{1}{3}\times \frac{5}{2}=\frac{5}{8}$$$$\frac{DE}{DP}=\frac{8-5}{8}=\frac{3}{8}$$$$similarly$$$$\frac{DF}{DR}=\frac{3}{8}$$$$\Rightarrow \mathrm{\Delta }DEF\sim \mathrm{\Delta }DPR$$$$\frac{{\mathrm{\Delta }}_{DEF}}{{\mathrm{\Delta }}_{DPR}}={\left(\frac{DF}{DR}\right)}^2={\left(\frac{3}{8}\right)}^2=\frac{9}{64}$$$$\frac{{\mathrm{\Delta }}_{DEF}}{{\mathrm{\Delta }}_{ABC}}=\frac{{\mathrm{\Delta }}_{DEF}}{{\mathrm{\Delta }}_{DPR}}\times \frac{{\mathrm{\Delta }}_{DPR}}{{\mathrm{\Delta }}_{ABC}}$$$$\Rightarrow \frac{{\mathrm{\Delta }}_{DEF}}{{\mathrm{\Delta }}_{ABC}}=\frac{9}{64}\times \frac{4}{45}=\frac{1}{80}$$$$i.e.theareaofeachshadedsmall$$$$triangleis\frac{1}{80}oftheareaofthebig$$$$triangleABC,thereforethesumof$$$$areasofallshadedregionsis\frac{3}{80}of$$$$theareaofthetriangleABC.$$$$A_{shaded}=\frac{3}{80}{\mathrm{\Delta }}_{ABC}=\frac{3}{80}\times 35=\frac{21}{16}$$

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