Question-137893

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(0,0), D(1,0), A(0,1), C(1,1) B((1/2),1) , E((1/4),0) eq of line AE : x+(1/4)y=(1/4); eq of line FB : y=2x intersect at point G((1/6),(1/3)) eq of line EC : 4x−3y=1 eq of line BD : 2x+y=2 intersect at point H((7/(10)), (3/5)) shadow of area is (1/2)∣ determinant ((((1/4) 0)),(((7/(10)) (3/5) )))+ determinant ((((7/(10)) (3/5))),(((1/2) 1)))+ determinant ((((1/2) 1)),(((1/6) (1/3))))+ determinant ((((1/6) (1/3))),(((1/4) 0)))∣ =(1/2)∣((3/(20))+(4/(10))+0−(1/(12)))∣ = (1/2)(((11)/(20)) −(1/(12))) =(7/(30))

$${by}\:{algebra} \\ $$$${let}\:{F}\left(\mathrm{0},\mathrm{0}\right),\:{D}\left(\mathrm{1},\mathrm{0}\right),\:{A}\left(\mathrm{0},\mathrm{1}\right),\:{C}\left(\mathrm{1},\mathrm{1}\right) \\ $$$${B}\left(\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1}\right)\:,\:{E}\left(\frac{\mathrm{1}}{\mathrm{4}},\mathrm{0}\right) \\ $$$${eq}\:{of}\:{line}\:{AE}\::\:{x}+\frac{\mathrm{1}}{\mathrm{4}}{y}=\frac{\mathrm{1}}{\mathrm{4}};\: \\ $$$${eq}\:{of}\:{line}\:{FB}\::\:\:{y}=\mathrm{2}{x}\: \\ $$$${intersect}\:{at}\:{point}\:{G}\left(\frac{\mathrm{1}}{\mathrm{6}},\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$${eq}\:{of}\:{line}\:{EC}\::\:\mathrm{4}{x}−\mathrm{3}{y}=\mathrm{1} \\ $$$${eq}\:{of}\:{line}\:{BD}\::\:\mathrm{2}{x}+{y}=\mathrm{2} \\ $$$${intersect}\:{at}\:{point}\:{H}\left(\frac{\mathrm{7}}{\mathrm{10}},\:\frac{\mathrm{3}}{\mathrm{5}}\right) \\ $$$${shadow}\:{of}\:{area}\:{is} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}\mid\:\begin{vmatrix}{\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\mathrm{0}}\\{\frac{\mathrm{7}}{\mathrm{10}}\:\:\:\:\frac{\mathrm{3}}{\mathrm{5}}\:}\end{vmatrix}+\begin{vmatrix}{\frac{\mathrm{7}}{\mathrm{10}}\:\:\:\:\frac{\mathrm{3}}{\mathrm{5}}}\\{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\mathrm{1}}\end{vmatrix}+\:\begin{vmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\mathrm{1}}\\{\frac{\mathrm{1}}{\mathrm{6}}\:\:\:\:\frac{\mathrm{1}}{\mathrm{3}}}\end{vmatrix}+\:\begin{vmatrix}{\frac{\mathrm{1}}{\mathrm{6}}\:\:\:\:\frac{\mathrm{1}}{\mathrm{3}}}\\{\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\mathrm{0}}\end{vmatrix}\mid \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mid\left(\frac{\mathrm{3}}{\mathrm{20}}+\frac{\mathrm{4}}{\mathrm{10}}+\mathrm{0}−\frac{\mathrm{1}}{\mathrm{12}}\right)\mid \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{11}}{\mathrm{20}}\:−\frac{\mathrm{1}}{\mathrm{12}}\right)\:=\frac{\mathrm{7}}{\mathrm{30}} \\ $$

Answered by mr W last updated on 08/Apr/21

((EG)/(GA))=((1/4)/(1/2))=(1/2) ⇒EG=(1/2)GA=(1/3)AE ⇒ΔEBG=(1/3)ΔEBA=(1/6)ΔEAC=(([ACDF[)/(12)) ((EH)/(HC))=((3/4)/(1/2))=(3/2) ⇒EH=(3/2)HC=(3/5)EC ⇒ΔEBH=(3/5)ΔEBC=(3/(10))ΔEAC=((3[ACDF])/(20)) shaded area=ΔEBG+ΔEBH =((1/(12))+(3/(20)))[ACDF]=(7/(30))×[ACDF] =(7/(30)) cm^2

$$\frac{{EG}}{{GA}}=\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{EG}=\frac{\mathrm{1}}{\mathrm{2}}{GA}=\frac{\mathrm{1}}{\mathrm{3}}{AE} \\ $$$$\Rightarrow\Delta{EBG}=\frac{\mathrm{1}}{\mathrm{3}}\Delta{EBA}=\frac{\mathrm{1}}{\mathrm{6}}\Delta{EAC}=\frac{\left[{ACDF}\left[\right.\right.}{\mathrm{12}} \\ $$$$\frac{{EH}}{{HC}}=\frac{\frac{\mathrm{3}}{\mathrm{4}}}{\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow{EH}=\frac{\mathrm{3}}{\mathrm{2}}{HC}=\frac{\mathrm{3}}{\mathrm{5}}{EC} \\ $$$$\Rightarrow\Delta{EBH}=\frac{\mathrm{3}}{\mathrm{5}}\Delta{EBC}=\frac{\mathrm{3}}{\mathrm{10}}\Delta{EAC}=\frac{\mathrm{3}\left[{ACDF}\right]}{\mathrm{20}} \\ $$$$ \\ $$$${shaded}\:{area}=\Delta{EBG}+\Delta{EBH} \\ $$$$\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{3}}{\mathrm{20}}\right)\left[{ACDF}\right]=\frac{\mathrm{7}}{\mathrm{30}}×\left[{ACDF}\right] \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{7}}{\mathrm{30}}\:{cm}^{\mathrm{2}} \\ $$

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