Question Number 144888 by imjagoll last updated on 30/Jun/21
Commented by imjagoll last updated on 30/Jun/21
$$\mathrm{find}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{area}. \\ $$
Answered by nimnim last updated on 30/Jun/21
Commented by mr W last updated on 30/Jun/21
$${how}\:{can}\:{you}\:{say}\:{BC}//{AD}? \\ $$
Commented by nimnim last updated on 30/Jun/21
$$\mathrm{Ok}.\:\mathrm{I}'\mathrm{m}\:\mathrm{done}.\:\mathrm{Sir},\:\mathrm{Please}\:\mathrm{can}\:\mathrm{you}\:\mathrm{show}\:\mathrm{me}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{way} \\ $$
Answered by mr W last updated on 30/Jun/21
Commented by mr W last updated on 30/Jun/21
$${BC}^{\mathrm{2}} =\left(\mathrm{4}+\mathrm{5}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} =\mathrm{90} \\ $$$${AC}^{\mathrm{2}} ={BC}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} =\mathrm{81} \\ $$$${AC}=\mathrm{9} \\ $$$${DE}=\frac{\mathrm{4}}{\mathrm{4}+\mathrm{5}}×\mathrm{3}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${EF}=\frac{\mathrm{5}}{\mathrm{4}+\mathrm{5}}×\mathrm{3}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$${A}_{{shade}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{9}×\mathrm{3}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}×\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{5}×\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\:\:\:\:=\mathrm{15} \\ $$
Commented by nimnim last updated on 01/Jul/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}.\:\mathrm{May}\:\mathrm{the}\:\mathrm{good}\:\mathrm{Lord}\:\mathrm{bless}\:\mathrm{you}. \\ $$