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