Question

Find the area of pentagon ABCDE in which $BL\perp AC,$ $DM\perp AC$ and $EN\perp AC$ such that $AC=18cm,$ $AM=14cm,$ $AN=6cm,$ $BL=4cm,$ $DM=12cm$ and $EN=9cm.$

Answer 1

Given: $BL\perp AC,$ $DM\perp AC$, and $EN\perp AC$ such that $AC=18cm,$ $AM=14cm,$ $AN=6cm,$ $BL=4cm,$ $DM=12cm$ and $EN=9cm.$

Area of triangle $=\frac{1}{2}\times$ base $\times$ height

Area of trapezium $=\frac{1}{2}\times$ (Sum of parallel sides) $\times$ height

Area of pentagon ABCDE = Area of $△AEN$ $+$ Area of trapezium EDMN $+$ Area of $△DMC+$Area of $△ACB$

$=[\frac{1}{2}\times AN\times EN]+[\frac{1}{2}\times (EN+DM)\times NM]+[\frac{1}{2}\times MC\times DM]+[\frac{1}{2}\times AC\times BL]$

$=[\frac{1}{2}\times AN\times EN]+[\frac{1}{2}\times (EN+DM)\times (AM-AN\left)\right]+[\frac{1}{2}\times (AC-AM)\times DM]+[\frac{1}{2}\times AC\times BL]$

$=\frac{1}{2}\times 6\times 9+\frac{1}{2}\times (9+12)\times (14-6)+\frac{1}{2}\times (18-14)\times 12+\frac{1}{2}\times 18\times 4c{m}^2$

$=27+84+24+36c{m}^2$

$=171c{m}^2$
Hence, the area of the given pentagon is $171c{m}^2.$