Area: Trapezium

Parallel sides of a trapezium are equal to a and b . Find the length of the line segment which is parallel to them and divides the area of the trapezium into two equal parts.
( i.e Find x in terms of a and b .)

Solution

Draw AD and BC produced meet at G .

$\displaystyle \text{Let}\ \alpha (\Delta GDC)=P\ ,\ \alpha (DEFG)=Q\ \ ,\ \alpha (ABFE)=R$ 

$\displaystyle \Delta GDC\sim \Delta GEF\sim \Delta GAB\ \ (\because DC\parallel EF\parallel AB)$ 

$\displaystyle \frac{P}{{P+Q}}=\frac{{{{b}^{2}}}}{{{{x}^{2}}}}$ 

$\displaystyle \frac{P}{Q}=\frac{{{{b}^{2}}}}{{{{x}^{2}}-{{b}^{2}}}}\ ........(1)$

$\displaystyle \frac{P}{{P+Q+R}}=\frac{{{{b}^{2}}}}{{{{a}^{2}}}}$ 

$\displaystyle \frac{P}{{Q+R}}=\frac{{{{b}^{2}}}}{{{{a}^{2}}-{{b}^{2}}}}$

$\displaystyle \frac{P}{{2Q}}=\frac{{{{b}^{2}}}}{{{{a}^{2}}-{{b}^{2}}}}\ \ (\because \ Q=R)$ 

$\displaystyle \frac{P}{Q}=\frac{{2{{b}^{2}}}}{{{{a}^{2}}-{{b}^{2}}}}\ ........(2)$

$\displaystyle \text{From}\ (1)\ \text{and}\ (2)\ ,\ \frac{{{{b}^{2}}}}{{{{x}^{2}}-{{b}^{2}}}}\ =\frac{{2{{b}^{2}}}}{{{{a}^{2}}-{{b}^{2}}}}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2{{x}^{2}}-2{{b}^{2}}={{a}^{2}}-{{b}^{2}}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2{{x}^{2}}={{a}^{2}}+{{b}^{2}}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \ \ \ x=\sqrt{{\frac{{{{a}^{2}}+{{b}^{2}}}}{2}}}$