ABCD is a parallelogram and EFCD is a rectangle- AL - DC- then arABCD is equal toA- ar - ADL B- ar- FBC C- arEFCDD- arEALD

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Theorem 4: Triangle and Parallelogram

ABCD is a par...

Question

ABCD is a parallelogram and EFCD is a rectangle. AL ⊥ DC, then ar(ABCD) is equal to

ar(∆ADL)

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ar(∆FBC)

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ar(EALD)

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ar(EFCD)

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Solution

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⊥

D C

A L

A B C D

E F C D

A L ⊥ D C

a r (A B C D) = a r (E F C D)

a r (A B C D) = D C \times A L

a r (A B C D) = a r (A B E F)

a r (A B C D) = A B \times D G

ABCD and ABFE are parallelograms as shown in the figure. If ar (ABCD) = 36 $c m^{2}$ and ar(ABFE) = 18 $c m^{2}$ , then ar (EFCD) [in $c m^{2}$ ] is

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