Question Number 70592 by behi83417@gmail.com last updated on 05/Oct/19
Commented by behi83417@gmail.com last updated on 05/Oct/19
$$\mathrm{convex}\:\mathrm{ABCD},\mathrm{is}\:\mathrm{given}\:\mathrm{with}: \\ $$$$\mathrm{S}\left(\mathrm{DCE}\right)=\mathrm{9},\mathrm{S}\left(\mathrm{CBE}\right)=\mathrm{10},\mathrm{S}\left(\mathrm{BAE}\right)=\mathrm{13}\:. \\ $$$$\:\:\Rightarrow\:\mathrm{S}\left(\mathrm{ADE}\right)=? \\ $$
Answered by mr W last updated on 06/Oct/19
Commented by mr W last updated on 06/Oct/19
$${S}_{\mathrm{1}} =\frac{{dc}}{\mathrm{2}}=\mathrm{9} \\ $$$${S}_{\mathrm{2}} =\frac{{bc}}{\mathrm{2}}=\mathrm{10} \\ $$$${S}_{\mathrm{3}} =\frac{{ba}}{\mathrm{2}}=\mathrm{13} \\ $$$${S}_{\mathrm{4}} =\frac{{da}}{\mathrm{2}}=? \\ $$$${S}_{\mathrm{1}} {S}_{\mathrm{3}} =\frac{{dcba}}{\mathrm{4}}=\frac{{abcd}}{\mathrm{4}} \\ $$$${S}_{\mathrm{2}} {S}_{\mathrm{4}} =\frac{{bcda}}{\mathrm{4}}=\frac{{abcd}}{\mathrm{4}} \\ $$$$\Rightarrow{S}_{\mathrm{1}} {S}_{\mathrm{3}} ={S}_{\mathrm{2}} {S}_{\mathrm{4}} \\ $$$$\Rightarrow{S}_{\mathrm{4}} =\frac{{S}_{\mathrm{1}} {S}_{\mathrm{3}} }{{S}_{\mathrm{2}} }=\frac{\mathrm{9}×\mathrm{13}}{\mathrm{10}}=\mathrm{11}.\mathrm{7}\: \\ $$
Commented by behi83417@gmail.com last updated on 06/Oct/19
$$\mathrm{nice}\:\mathrm{solution}\:\mathrm{dear}\:\mathrm{master}.\mathrm{thanks}\:\mathrm{in}\: \\ $$$$\mathrm{advance}\:\mathrm{sir}. \\ $$