Question

If a quadrilateral $ABCD$ is such that $AB=b,AD=d$ and $AC=mb+pd$ $(m+p\ge 1),$ then the area of the quadrilateral is $k(p+m)|b\times d|,$ where $k$ is equal to

• A. $\frac{1}{4}$
• B. $\frac{1}{8}$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{2}$

We have $BC=BA+AC=$ $-b+mb+pd=b(m-1)+pd$

and $CD=CA+AD=$ $-mb-pd+d=mb+d(1-p)$

Area of the triangle $ABD$ $=\frac{1}{2}|b\times d|$

Area of triangle $BCD$ $=\frac{1}{2}|BC\times CD|=\frac{1}{2}\left|(b(m-1)+pd)(-mb+d)(1-p)\right|$

$=\frac{1}{2}\left||m-1+p||b\times d|\right|$

Hence area of the quadrilateral $ABCD=$ $=\frac{1}{2}\left|(m+p)|b\times d|\right|\Rightarrow k=\frac{1}{2}$