Question
Find the area of pentagon $ABCDE$ in which $\text{BL}\perp\text{AC},\text{DM}\perp\text{AC}$ and $\text{EN}\perp\text{AC}$ such that $AC = 18\ cm, AM = 14\ cm, AN = 6\ cm, BL = 4\ cm, DM = 12\ cm$ and $EN = 9\ cm.$
✓
Answer
In the given pentagon $ABCDE,$
$\text{BL}\perp\text{AC},\text{DM}\perp\text{AC},\text{EN}\perp\text{AC}$
$AC = 18cm, AM = 14cm, AN = 6cm,$
$BL = 4cm, DM = 12cm$ and $EN = 9cm$
$MN = AM - AN = 14 - 6 = 8cm$
$MC = AC - AM = 18 - 14 = 4cm$
Now area of $\triangle\text{ABC}=\frac{\text{AC}\times\text{BL}}{2}$
$=\frac{18\times4}{2}\text{cm}^2=36\text{cm}^2$
Area $\triangle\text{AEN}=\frac{\text{AN}\times\text{EN}}{2}$
$=\frac{6\times9}{2}=27\text{cm}^2$
Area $\triangle\text{DML}=\frac{\text{MC}\times\text{DM}}{2}$
$=\frac{4\times12}{2}=24\text{cm}^2$
And area of trap. $NEDM,$
$=\frac{1}{2}(\text{EN}+\text{DM})\times\text{NM}$
$=\frac{1}{2}(9+12)\times8\text{cm}^2$
$=\frac{1}{2}\times21\times8=84\text{cm}^2$
Area of figure $ABCDE =$ Area of $\triangle\text{ABC}\ +$ Area $\triangle\text{AEN}\ +$ Area $\triangle\text{DML}\ +$ Area trap. $NEDM,$
$= (36 + 27 + 24 + 84)cm^2 = 171cm^2.$
$\text{BL}\perp\text{AC},\text{DM}\perp\text{AC},\text{EN}\perp\text{AC}$
$AC = 18cm, AM = 14cm, AN = 6cm,$
$BL = 4cm, DM = 12cm$ and $EN = 9cm$
$MN = AM - AN = 14 - 6 = 8cm$
$MC = AC - AM = 18 - 14 = 4cm$
Now area of $\triangle\text{ABC}=\frac{\text{AC}\times\text{BL}}{2}$
$=\frac{18\times4}{2}\text{cm}^2=36\text{cm}^2$
Area $\triangle\text{AEN}=\frac{\text{AN}\times\text{EN}}{2}$
$=\frac{6\times9}{2}=27\text{cm}^2$
Area $\triangle\text{DML}=\frac{\text{MC}\times\text{DM}}{2}$
$=\frac{4\times12}{2}=24\text{cm}^2$
And area of trap. $NEDM,$
$=\frac{1}{2}(\text{EN}+\text{DM})\times\text{NM}$
$=\frac{1}{2}(9+12)\times8\text{cm}^2$
$=\frac{1}{2}\times21\times8=84\text{cm}^2$
Area of figure $ABCDE =$ Area of $\triangle\text{ABC}\ +$ Area $\triangle\text{AEN}\ +$ Area $\triangle\text{DML}\ +$ Area trap. $NEDM,$
$= (36 + 27 + 24 + 84)cm^2 = 171cm^2.$
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