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 = 18cm, AM = 14cm, AN = 6cm, BL = 4cm, DM = 12cm and EN = 9cm.
✓
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
$MN=AM-AN=14-6=8 cm$
$MC=AC-AM=18-14=4 cm$
Now area of $\triangle ABC =\frac{ AC \times BL }{2}$
$=\frac{18 \times 4}{2} cm^2=36 cm^2$
Area $\triangle AEN =\frac{ AN \times EN }{2}$
$=\frac{6 \times 9}{2}=27 cm^2$
Area $\triangle DML =\frac{ MC \times DM }{2}$
$=\frac{4 \times 12}{2}=24 cm^2$
And area of trap. NEDM,
$=\frac{1}{2}(EN+DM) \times NM$
$=\frac{1}{2}(9+12) \times 8 cm^2$
$=\frac{1}{2} \times 21 \times 8=84 cm^2$
Area of figure $ABCDE =$ Area of $\triangle ABC +$ Area $\triangle AEN +$ Area $\triangle DML +$ Area trap. NEDM,
$=(36+27+24+84) cm ^2=171 cm^2$.
$\text{BL}\perp\text{AC},\text{DM}\perp\text{AC},\text{EN}\perp\text{AC}$
AC = 18cm, AM = 14cm, AN = 6cm,
BL = 4cm, DM = 12cm and EN = 9cm
$MN=AM-AN=14-6=8 cm$
$MC=AC-AM=18-14=4 cm$
Now area of $\triangle ABC =\frac{ AC \times BL }{2}$
$=\frac{18 \times 4}{2} cm^2=36 cm^2$
Area $\triangle AEN =\frac{ AN \times EN }{2}$
$=\frac{6 \times 9}{2}=27 cm^2$
Area $\triangle DML =\frac{ MC \times DM }{2}$
$=\frac{4 \times 12}{2}=24 cm^2$
And area of trap. NEDM,
$=\frac{1}{2}(EN+DM) \times NM$
$=\frac{1}{2}(9+12) \times 8 cm^2$
$=\frac{1}{2} \times 21 \times 8=84 cm^2$
Area of figure $ABCDE =$ Area of $\triangle ABC +$ Area $\triangle AEN +$ Area $\triangle DML +$ Area trap. NEDM,
$=(36+27+24+84) cm ^2=171 cm^2$.
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