Let `A(z_(1)), B(z_(2)), C(z_(3)` and `D(z_(4))` be the vertices of a trepezium in an Argand plane such that `AB||CD` Let `|z_(1)-z_(2)|=4, |z_(3),z_(4)|=10` and the diagonals `AC` and `BD` intersects at `P`. It is given that `Arg((z_(4)-z_(2))/(z_(3)-z_(1)))=(pi)/2` and `Arg((z_(3)-z_(2))/(z_(4)-z_(1)))=(pi)/4`
Which of the following option(s) is/are correct?

Options

A. Area of trepezium `ABCD` is equla to `140/3` sq. units
B. Area of the trapezium `ABCD` is equal to `70/3` sq. units
C. Area of the triangle `BCP` is equal, to `100/21` sq. units
D. Area of the triangle `BCP` is equal to `200/42` sq. units

Answer

A

Explanation

`DeltaABP` and `DeltaCDP` are similar if `AP=2x`
`BP=2y` then `CP=5x, DP=5y`
Area of trapezium `ABCD=49/2xy`
`tan alpha=(2x)/(5y), tan beta =(2y)/(2x)` also `alpha+beta=45^(@)`
`impliesxy=10/21(x^(2)+y^(2))`
Also `AB^(2)=AP^(2)+BP^(2)impliesx^(2)+y^(2)=4impliesxy=40/21`
`Ar(DeltaPCD)=5xy`