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Let $ABCD$ be a parallelogram of area $10$ with $AB=3$ and $BC=5$. Locate $E,F$ and $G$ on segments $\overline{AB},\overline{BC}$ and $\overline{AD}$, respectively, with $AE=BF=AG=2$. Let the line through $G$ parallel to $\overline{EF}$ intersect $\overline{CD}$ at $H$. The area of quadrilateral $EFHG$ is
$\text{(A)

ABHISHEK VASHIST , 7 Years ago
Grade 12th pass
anser 1 Answers
VISHNU PREMJI

Last Activity: 7 Years ago

5 its simple draw the triangle marks its sides and try to construct it  so we get GD =3,DH=1,HC=2 This is because opposite sides of parrelogram are equal b join sides EH .,,, we can se a triangle lying on a parrelogram on equal side  THEORITICALLY,the area of this triangle will be half that of parralelogram  area of parrelogram AEHD IS 1/2*10=5    Soareaof triangle  GEH Is 2.5 //ly area of  triangle EFH =2.5    …..   FINALLY  Area of quadrilateral  EFGH IS   5 units
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