Question

ABCD IS A TREPIZIUM IN WHICH AB||CD . O IS THE MIDPOINT OF BC. THROUGH TH POINT O LINE PQ||AD HAS DRAWN WHICH INTERSECT AB AT Q AND DC PRODUCED AT P .PROVE THAT ar(ABCD)=ar(AQPD)

Answer 1

$\mathbb{ELLO \ THERE!}$

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Question: ABCD is a trapezium in which AB║CD. O is the midpoint of BC. Through the point O, a line PQ is drawn parallel to AD, to intersect AB at Q. DC is produced at P. Prove that ar(ABCD) = ar(AQPD)

[Figure attached for reference]

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$\underline\mathsf{ANSWER}}$

Given

ABCD is a trapezium.

AB║CD

BO = CO

PQ║AD

To Prove

ar(ABCD) = ar(AQPD)

Proof

In Δ BOQ and ΔCOP

∠BOQ = ∠COP [V.O.A]

OB = OC  [O is the midpoint]

AB║DP

∠BQO = ∠CPO [Alternate Interior Angles]

∴ ΔBOQ ≅ ΔCOP by AAS Congruency.

⇒ ar(BOQ) = ar(COP)

[Congruent figures have equal areas]

Now,

→ ar(ABCD) = ar(AQOCD) + ar(BOQ)

→ But, we have proven that ar(BOQ) = ar(COP)

→ ar(ABCD) = ar(AQOCD) + ar(COP)

→ ar(ABCD) = ar(AQPD)

Hence Proved!

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Regards,

Tomboyish44.