In the given figure B C is parallel to DE- Prove that- area - ABE - area - ACD

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Standard VII

Mathematics

Angle Sum Property

In the given ...

Question

In the given figure $B C$ is parallel to $D E .$ Prove that: area $Δ A B E$ = area $Δ A C D$

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Solution

REF.Image
Given : BC||DE
To prove : area $Δ A B E$ = area $Δ A C D$
$A (Δ B C E) = A (Δ B C D)$ - (i).... (triangle with same base BC and between some parallel lines BC and DE)
Now,
$A (A B E) = A (Δ A C B) + A (B C E) - (i i)$
$A (Δ A B E) = A (Δ A C B) + A (Δ B C D) . . .$ [from (i)]
$A (Δ A B E) = A (Δ A C D)$
Since $A (A C B) + A (Δ B C D) = A (Δ A C D)$

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In the given figure BC is parallel to DE. Prove that: area ΔABE = area ΔACD

REF.ImageGiven : BC||DETo prove : area ΔABE = area ΔACDA(ΔBCE)=A(ΔBCD) - (i).... (triangle with same base BC and between some parallel lines BC and DE)Now, A(ABE)=A(ΔACB)+A(BCE)−(ii)A(ΔABE)=A(ΔACB)+A(ΔBCD)... [from (i)]A(ΔABE)=A(ΔACD)Since A(ACB)+A(ΔBCD)=A(ΔACD)

In the given figure $B C$ is parallel to $D E .$ Prove that: area $Δ A B E$ = area $Δ A C D$