Question
Find the area of pentagon ABCDE in which $\text{BL}\perp\text{AC},\text{CM}\perp\text{AD}$ and $\text{EN}\perp\text{AD}$ such that AC = 10cm, AD = 12cm, BL = 3cm, CM = 7cm and EN = 5cm.
✓
Answer
In the given pentagon $A B C D E$,
$BL \perp AC , CM \perp AD , EN \perp AD$
$AC=10 cm, D=12 cm, BL=3 cm,$
$CM=7 cm \text { and } EN=5 cm$
Area $\triangle ABC =\frac{1}{2} AC \times BL$
$=\frac{1}{2} \times 10 \times 3=15 cm^2$
$\text { Area } \triangle ACD=\frac{1}{2} AD \times CM$
$=\frac{1}{2} \times 12 \times 7=42 cm^2$
$\text { Area } \triangle AED=\frac{1}{2} AD \times EN$
$=\frac{1}{2} \times 12 \times 5=30 cm^2$
$\therefore \text { Area } ABCDE=\text { Area } \triangle ABC+\text { Area } \triangle ACD+\text { Area } \triangle AED$
$=(15+42+30) cm^2$
$BL \perp AC , CM \perp AD , EN \perp AD$
$AC=10 cm, D=12 cm, BL=3 cm,$
$CM=7 cm \text { and } EN=5 cm$
Area $\triangle ABC =\frac{1}{2} AC \times BL$
$=\frac{1}{2} \times 10 \times 3=15 cm^2$
$\text { Area } \triangle ACD=\frac{1}{2} AD \times CM$
$=\frac{1}{2} \times 12 \times 7=42 cm^2$
$\text { Area } \triangle AED=\frac{1}{2} AD \times EN$
$=\frac{1}{2} \times 12 \times 5=30 cm^2$
$\therefore \text { Area } ABCDE=\text { Area } \triangle ABC+\text { Area } \triangle ACD+\text { Area } \triangle AED$
$=(15+42+30) cm^2$
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