Question

Diagonal AC and BD of a trapezium ABCD with AB||DC intersect each other at O. Prove That ar(AOD)=ar(BOC).​

Answer 1

Dear user:

Given:

AB||DC

TO PROVE:

ar(AOD) =ar(BOC)

Proof:

ΔADC and ΔBDC are on the same base DC and between same parallel AB and DC.

∴ ar(ΔADC) = ar(ΔBDC) [triangles on the same base and between same parallel are equal in area]

Subtract Area (ΔDOC) from both side.

ar(ΔADC) – ar(ΔDOC) = ar(ΔBDC) – ar(ΔDOC)

ar(ΔAOD) = ar(ΔBOC)

Hence proved☺☺

#AN

Answer 2

$\huge{\underline{\bf\orange{Answer :-}}}$

Given:

ABCD is Trapezium with AB||DC.

To Find:

ar(AOD) =ar((BOC)

Solution:

it can be observed that ∆DAC and ∆DBC lie on the same base DC and between same parallels AB and CD.

∴ ar(∆DAC) = ar(∆DBC)

Subtract Area (ΔDOC) from both side.

=ar(ΔDAC) – ar(ΔDOC) = ar(ΔDBC)–ar(ΔDOC)

= ar (ΔAOD) = ar (ΔBOC)

$Hence \: proved$