Question

For each trapezium given below; find its area.

Answer 1

REF.image

Trapezium Area.

$\therefore CD=MN=12cm$

$\therefore AM=NB=\frac{20-12}{2}4cm$

$\to$ As shown in figure

$△AMD$ using

pythagoras theorem

$A{D}^2=A{M}^2+M{D}^2$

$\therefore (10{)}^2=(4{)}^2+M{D}^2$

$\therefore M{D}^2=(10{)}^2-(4{)}^2$

$=100-16$

$\therefore M{D}^2=84$

$\therefore MD=\sqrt{84}$

$\to$ Area of trapezium

$A=\left(\frac{AB+CD}{2}\right)\times$ perpendicular distance

$=\left(\frac{20+12}{2}\right)\times MD$

$=\frac{32}{2}\times \sqrt{84}$

$=16\times \sqrt{84}$

$A=146.64c{m}^2$