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Question

In given figure DE||BC and AD:DB=5:4 Find Area(DEF)Area(CFB).
1009451_2d82594a7d834efab8539e4486dade98.png

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Solution

In ABC, we have
DE||BC

ADE=ABC and AED=ACB [Corresponding angles]

Thus, in triangles ADE and ABC, we have

A=A [Common]

ADE=ABC

and, AED=ACB

AEDABC [By AAA similarity]

ADAB=DEBC

We have,
ADDB=54

DBAD=45

DBAD+1=45+1

DB+ADAD=95

ABAD=95ADAB=59

DEBC=59

In DFE and CFB, we have

1=3 [Alternate interior angles]

2=4 [Vertically opposite angles]

Therefore, by AA-similarity criterion, we have

DFECFB

Area(DFE)Area(CFB)=DE2BC2

Area(DFE)Area(CFB)=(59)2=2581.....[Using (i)]

1032069_1009451_ans_3a5df9b4ba314f97b40a55eb2e91784c.png

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