Question Number 137893 by bemath last updated on 07/Apr/21 | ||
Answered by benjo_mathlover last updated on 08/Apr/21 | ||
$${by}\:{algebra} \\ $$$${let}\:{F}\left(\mathrm{0},\mathrm{0}\right),\:{D}\left(\mathrm{1},\mathrm{0}\right),\:{A}\left(\mathrm{0},\mathrm{1}\right),\:{C}\left(\mathrm{1},\mathrm{1}\right) \\ $$$${B}\left(\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1}\right)\:,\:{E}\left(\frac{\mathrm{1}}{\mathrm{4}},\mathrm{0}\right) \\ $$$${eq}\:{of}\:{line}\:{AE}\::\:{x}+\frac{\mathrm{1}}{\mathrm{4}}{y}=\frac{\mathrm{1}}{\mathrm{4}};\: \\ $$$${eq}\:{of}\:{line}\:{FB}\::\:\:{y}=\mathrm{2}{x}\: \\ $$$${intersect}\:{at}\:{point}\:{G}\left(\frac{\mathrm{1}}{\mathrm{6}},\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$${eq}\:{of}\:{line}\:{EC}\::\:\mathrm{4}{x}−\mathrm{3}{y}=\mathrm{1} \\ $$$${eq}\:{of}\:{line}\:{BD}\::\:\mathrm{2}{x}+{y}=\mathrm{2} \\ $$$${intersect}\:{at}\:{point}\:{H}\left(\frac{\mathrm{7}}{\mathrm{10}},\:\frac{\mathrm{3}}{\mathrm{5}}\right) \\ $$$${shadow}\:{of}\:{area}\:{is} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}\mid\:\begin{vmatrix}{\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\mathrm{0}}\\{\frac{\mathrm{7}}{\mathrm{10}}\:\:\:\:\frac{\mathrm{3}}{\mathrm{5}}\:}\end{vmatrix}+\begin{vmatrix}{\frac{\mathrm{7}}{\mathrm{10}}\:\:\:\:\frac{\mathrm{3}}{\mathrm{5}}}\\{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\mathrm{1}}\end{vmatrix}+\:\begin{vmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\mathrm{1}}\\{\frac{\mathrm{1}}{\mathrm{6}}\:\:\:\:\frac{\mathrm{1}}{\mathrm{3}}}\end{vmatrix}+\:\begin{vmatrix}{\frac{\mathrm{1}}{\mathrm{6}}\:\:\:\:\frac{\mathrm{1}}{\mathrm{3}}}\\{\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\mathrm{0}}\end{vmatrix}\mid \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mid\left(\frac{\mathrm{3}}{\mathrm{20}}+\frac{\mathrm{4}}{\mathrm{10}}+\mathrm{0}−\frac{\mathrm{1}}{\mathrm{12}}\right)\mid \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{11}}{\mathrm{20}}\:−\frac{\mathrm{1}}{\mathrm{12}}\right)\:=\frac{\mathrm{7}}{\mathrm{30}} \\ $$ | ||
Answered by mr W last updated on 08/Apr/21 | ||
$$\frac{{EG}}{{GA}}=\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{EG}=\frac{\mathrm{1}}{\mathrm{2}}{GA}=\frac{\mathrm{1}}{\mathrm{3}}{AE} \\ $$$$\Rightarrow\Delta{EBG}=\frac{\mathrm{1}}{\mathrm{3}}\Delta{EBA}=\frac{\mathrm{1}}{\mathrm{6}}\Delta{EAC}=\frac{\left[{ACDF}\left[\right.\right.}{\mathrm{12}} \\ $$$$\frac{{EH}}{{HC}}=\frac{\frac{\mathrm{3}}{\mathrm{4}}}{\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow{EH}=\frac{\mathrm{3}}{\mathrm{2}}{HC}=\frac{\mathrm{3}}{\mathrm{5}}{EC} \\ $$$$\Rightarrow\Delta{EBH}=\frac{\mathrm{3}}{\mathrm{5}}\Delta{EBC}=\frac{\mathrm{3}}{\mathrm{10}}\Delta{EAC}=\frac{\mathrm{3}\left[{ACDF}\right]}{\mathrm{20}} \\ $$$$ \\ $$$${shaded}\:{area}=\Delta{EBG}+\Delta{EBH} \\ $$$$\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{3}}{\mathrm{20}}\right)\left[{ACDF}\right]=\frac{\mathrm{7}}{\mathrm{30}}×\left[{ACDF}\right] \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{7}}{\mathrm{30}}\:{cm}^{\mathrm{2}} \\ $$ | ||