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In Fig, AP BQ CR. Prove that ar(AQC) = ar(PBR)

Question

In Fig, $A P ∥ BQ ∥ CR$ . Prove that $a r (A QC) = a r (PBR)$

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In Fig,

A P ∥ BQ ∥ CR

. Prove that

a r (A QC) = a r (PBR)

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

Given:

A P ∥ BQ

We can say that perpendicular distance from

A

BQ

is equal to perpendicular distance from

P

BQ

Let it be

d_{1}

a r (A QB) = \frac{1}{2} \times BQ \times d_{1}

a r (PQB) = \frac{1}{2} \times BQ \times d_{1}

∴

a r (A QB) = a r (PQB)

Given:

CR ∥ BQ

We can say that perpendicular distance from

C

BQ

is equal to perpendicular distance from

R

BQ

Let it be

d_{2}

a r (CQB) = \frac{1}{2} \times BQ \times d_{2}

a r (RQB) = \frac{1}{2} \times BQ \times d_{2}

∴

a r (CQB) = a r (RQB)

a r (A QC) = a r (A QB) + a r (BQC) = a r (PQB) + a r (RQB) = a r (PBR)

∴ a r (A QC) = a r (PBR)

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Question Text	In Fig, $A P ∥ BQ ∥ CR$ . Prove that $a r (A QC) = a r (PBR)$
Answer Type	Text solution:1
Upvotes	150