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Question
In the given figure, BC ∥ DE.
(a) If area of ΔADC is 20 sq. units, find the area of ΔAEB.
(b) If the area of ΔBFD is 8 square units, find the area of ΔCEF
Solution
(a) Triangles on the same base and between the same parallels are equal in area.
∴ A(ΔDBC) = A9ΔECB) ....(i)
Now,
A(ΔABC) = A(ΔADC) + A(ΔDBC) = A(ΔAEB) + A(ΔECB)
⇒ A(ΔAC) + A(DBC) = A(ΔAEB) + A(ΔECB)
⇒ A(ΔADC) = A(ΔAEB) ....(ii) [from (i)]
Given, A(ΔADC) = 20 sq. units
⇒ A(ΔAEB) = 20 sq. units
(b) A(ΔADC) = A(ΔAEB) ...[From (ii)]
⇒ A(ΔADC) - A(ΔDEF) = A(ΔAEB) - A(ΔDFE)
⇒ A(ΔCEF) = A(ΔBFD)
Given, A(ΔBFD) = 8 sq. units
⇒ A(ΔCEF) = 8 sq. units.
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