Question Number 172453 by mr W last updated on 27/Jun/22
Commented by infinityaction last updated on 27/Jun/22
$$\mathrm{22}\:???? \\ $$
Commented by mr W last updated on 27/Jun/22
$${yes}.\:{please}\:{show}\:{your}\:{working}. \\ $$
Commented by infinityaction last updated on 27/Jun/22
Commented by infinityaction last updated on 27/Jun/22
$$\:\:\:\bigtriangleup{OQR}\sim\bigtriangleup{RTQ} \\ $$$$\:\:\:\:\frac{\mathrm{6}}{\mathrm{8}}\:\:=\:\:\left(\frac{{QR}}{{RT}}\right)^{\mathrm{2}} \:\:\Rightarrow\:\:\frac{{QR}}{{RT}}\:\:\:=\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:{let}\:\:\:\:\:{QR}\:\:=\:\:\sqrt{\mathrm{3}}{k}\:\:{and}\:\:{RT}\:\:=\:\:\mathrm{2}{k} \\ $$$$\:\:\:\:{then}\:\:{QT}\:\:=\:\:\:\sqrt{\mathrm{4}{k}^{\mathrm{2}} −\mathrm{3}{k}^{\mathrm{2}} \:}\:\:=\:\:{k} \\ $$$$\:\:\:{in}\:\bigtriangleup{PSQ} \\ $$$$\:\:\:\:\mathrm{tan30}°\:\:=\:\:\frac{\sqrt{\mathrm{3}}{k}}{{PQ}} \\ $$$$\:\:\:\:{PQ}\:\:=\:\:\mathrm{3}{k} \\ $$$$\:\:\:\:\:{PT}\:\:=\:\:{PQ}−{TQ}\:=\:\mathrm{3}{k}−{k}\:=\:\mathrm{2}{k} \\ $$$$\:\:\:\:{now}\:\:\:\bigtriangleup{PQS}\:\sim\bigtriangleup{TQR} \\ $$$$\:\:\:\frac{{A}+\mathrm{2}}{\mathrm{8}}\:\:\:=\:\:\left(\frac{\mathrm{3}{k}}{\:\sqrt{\mathrm{3}}{k}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:{A}+\mathrm{2}\:\:=\:\:\mathrm{24} \\ $$$$\:\:\:\:{A}\:\:=\:\mathrm{22} \\ $$
Commented by mr W last updated on 27/Jun/22
$${thanks}\:{sir}! \\ $$
Commented by infinityaction last updated on 27/Jun/22
$$\:\:\:{METHOD}\:−\mathrm{2} \\ $$$$\:\:\:\bigtriangleup{OQR}\sim\bigtriangleup{RTQ} \\ $$$$\:\:\:\:\frac{\mathrm{6}}{\mathrm{8}}\:\:=\:\:\left(\frac{{QR}}{{RT}}\right)^{\mathrm{2}} \:\:\Rightarrow\:\:\frac{{QR}}{{RT}}\:\:\:=\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:{let}\:\:\:\:\:{QR}\:\:=\:\:\sqrt{\mathrm{3}}{k}\:\:{and}\:\:{RT}\:\:=\:\:\mathrm{2}{k} \\ $$$$\:\:\:\:{then}\:\:{QT}\:\:=\:\:\:\sqrt{\mathrm{4}{k}^{\mathrm{2}} −\mathrm{3}{k}^{\mathrm{2}} \:}\:\:=\:\:{k} \\ $$$$\:\:\:{in}\:\bigtriangleup{PSQ} \\ $$$$\:\:\:\:\mathrm{tan30}°\:\:=\:\:\frac{\sqrt{\mathrm{3}}{k}}{{PQ}} \\ $$$$\:\:\:\:{PQ}\:\:=\:\:\mathrm{3}{k} \\ $$$$\:\:\:\:\:{PT}\:\:=\:\:{PQ}−{TQ}\:=\:\mathrm{3}{k}−{k}\:=\:\mathrm{2}{k} \\ $$$$\:\:\:\:{A}\:=\:\:\bigtriangleup{PQS}\:−\:\bigtriangleup{OTQ} \\ $$$$\:\:\:\:{A}\:\:=\:\:\:\frac{\mathrm{3}\sqrt{\mathrm{3}}{k}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2} \\ $$$$\:\:\:\:{A}\:\:=\:\:\frac{\mathrm{3}\sqrt{\mathrm{3}}{k}^{\mathrm{2}} −\mathrm{4}}{\mathrm{2}} \\ $$$${in}\bigtriangleup{RTQ} \\ $$$$\:\:\mathrm{8}\:\:=\:\:\frac{\sqrt{\mathrm{3}}{k}^{\mathrm{2}} }{\mathrm{2}}\:\:\:\Rightarrow\:\:{k}^{\mathrm{2}} \:\:=\:\:\frac{\mathrm{16}}{\:\sqrt{\mathrm{3}}} \\ $$$${A}\:\:\:\:=\:\:\:\frac{\mathrm{3}\sqrt{\mathrm{3}}×\frac{\mathrm{16}}{\:\sqrt{\mathrm{3}}}\:−\:\mathrm{4}}{\mathrm{2}} \\ $$$${A}\:\:=\:\:\:\frac{\mathrm{48}−\mathrm{4}}{\mathrm{2}}\:\:\:\:\Rightarrow\:\:{A}\:\:=\:\:\mathrm{22} \\ $$
Answered by mr W last updated on 27/Jun/22
Commented by mr W last updated on 27/Jun/22
$$\Delta{EFB}\sim\Delta{BFC}\sim\Delta{CFD} \\ $$$$\left(\frac{{BF}}{{EF}}\right)^{\mathrm{2}} =\frac{\left[\Delta{BFC}\right]}{\left[\Delta{EFB}\right]}=\frac{\mathrm{6}}{\mathrm{2}}=\mathrm{3}\:\Rightarrow\frac{{BF}}{{EF}}=\sqrt{\mathrm{3}} \\ $$$$\frac{{CF}}{{BF}}=\frac{{BF}}{{EF}}=\sqrt{\mathrm{3}} \\ $$$$\frac{{CF}}{{EF}}=\frac{{CF}}{{BF}}×\frac{{BF}}{{EF}}=\sqrt{\mathrm{3}}×\sqrt{\mathrm{3}}=\mathrm{3} \\ $$$$\frac{\left[\Delta{CFD}\right]}{\left[\Delta{EFB}\right]}=\left(\frac{{CF}}{{EF}}\right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} =\mathrm{9} \\ $$$${S}=\left[\Delta{CFD}\right]=\mathrm{9}×\left[\Delta{EFB}\right]=\mathrm{9}×\mathrm{2}=\mathrm{18} \\ $$$${T}+\mathrm{2}={S}+\mathrm{6}=\mathrm{18}+\mathrm{6}=\mathrm{24} \\ $$$$?={T}=\mathrm{24}−\mathrm{2}=\mathrm{22}\:\checkmark \\ $$
Commented by infinityaction last updated on 27/Jun/22
$${nice}\:{solution} \\ $$
Commented by mr W last updated on 27/Jun/22
$${in}\:{principle}\:{we}\:{had}\:{similar}\:{way}\:{for} \\ $$$${solution}. \\ $$
Commented by infinityaction last updated on 27/Jun/22
$${yes}\:{sir} \\ $$
Commented by Tawa11 last updated on 28/Jun/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$