Question

In the figure quadrilateral ABCD is a square of side 1 unit, E and F are midpoints of AD and AB respectively. I is the point of intersection of BE and CF, then area of quadrilateral IEDC is​

Answer 1

$\text { Let, 'a' be the length of the side of the square. }$

$\text { Area of } E I C D=A B C D-A B E-B C \mid$

$\text { Now, Equation of line CF }$

$x / a=(y-a / 2) /(-a / 2)$

$\text { or, } x+2 y=a \ldots$

$\text { Equation of line BE }$

$x /(a / 2)=y / a$

$\text { or, } 2 \mathrm{x}=\mathrm{y} \ldots . . .(2)$

$\text { Solving equations }(1) \text { and (2) we get, } x=a / 5, y=2 a / 5$

$\text { Co-ordinate of } I \text { is }(a / 5,2 a / 5)$

$\text { Area of } \mathrm{BIC} \text { is }\{0 \times(0-2 \mathrm{a} / 5)+\mathrm{a} \times(2 \mathrm{a} / 5-0)+\mathrm{a} / 5 \times(0-0)\}=2 \mathrm{a}^{2} / 5$

$\text { Area of } \mathrm{ABE}=1 / 2 \times \mathrm{a} / 2 \times \mathrm{a}=\mathrm{a}^{2} / 4$

$\text { Area of } A B C D=a^{2}$

$\text { Area of } E I C D=a^{2}-a^{2} / 4-2 a^{2} / 5=7 a^{2} / 20$

Hence the area of quadrilateral IEDC is $7 a^{2} / 20$.

Answer 2

Answer:

11/20 unit²

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