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Question Number 159396 by quvonnn last updated on 16/Nov/21

Answered by som(math1967) last updated on 16/Nov/21

$$areaof△BCE=Areaof△ECF$$$$=\frac{1}{2}\times 4\times 3=6sq.unit$$$$CMisbisectorof\mathrm{\angle }C$$$$\therefore \frac{EM}{MB}=\frac{EC}{CB}=\frac{3}{4}$$$$\frac{Ar△EMC}{Ar△CMB}=\frac{3}{4}$$$$\therefore Ar.△EMC=\frac{3}{7}\times 6=\frac{18}{7}squnit$$$$S=Ar.△EMF=6-\frac{18}{7}=\frac{24}{7}squnitans$$

Commented by som(math1967) last updated on 16/Nov/21

Commented by Rasheed.Sindhi last updated on 16/Nov/21

$$areaof△BCE\stackrel{?}{=}Areaof△BCF$$$$Areaof△BCF=4\times 4\times \frac{1}{2}=8{unit}^2$$$$$$

Commented by som(math1967) last updated on 16/Nov/21

$$typosirit△ECF$$

Commented by Avijit007 last updated on 16/Nov/21

$$yes.$$

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