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Question

In the figure, DE is parallel to BC and ADDB=32. Area of ΔDEF is 81cm2. Area of ΔCFB is .

A
180 cm2
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B
250 cm2
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C
225 cm2
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D
195 cm2
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Solution

The correct option is C 225 cm2

Here, DE || BC (given)

In ΔADE and ΔABC,

A=A (common)

ADE=ABC (corresponding angles)

ADE ABC (AA similarity criterion)

ADAB=DEBC ---- (1)

Given, ADDB=32

DBAD=23

Adding 1 on both sides, we get

DBAD+1=23+1DB+ADAD=2+33 ABAD=53

ADAB=35 --- (2)

From (1) and (2)

DEBC=35

In ΔDFE and ΔBFC,

DFE+BFC (vertically opposite angle)

FCB=EDF (alternate angle)

ΔDFE CFB (by AA similarity criterion)

ar(ΔDFE)ar(ΔCFB)=DE2BC2

(The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides)

81ar(ΔCFB)=(DEBC)281ar(ΔCFB)=(35)2ar(ΔCFB)=81×259=225 cm2


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