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Triangles on the Same Base and between the Same Parallels

In the given ...

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Question 14
In the given figure, AP || BQ || CR. Prove that ar(AQC) = ar(PBR).

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Solution

$Δ A B Q$ and $Δ P B Q$ lie on the same base BQ and are between the same parallels AP and BQ,
$\Rightarrow A r e a (Δ A B Q) = A r e a (Δ P B Q) . . . (1)$
Again, $Δ B C Q$ and $Δ B R Q$ lie on the same base BQ and are between the same parallels BQ and CR.
$\Rightarrow A r e a (Δ B C Q) = A r e a (Δ B R Q) . . . (2)$
On adding equations (1) and (2), we obtain,
$A r e a (Δ A B Q) + A r e a (Δ B C Q) = A r e a (Δ P B Q) + A r e a (Δ B R Q)$
$∴ A r e a (Δ A Q C) = A r e a (Δ P B R)$

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Q. Question 14
In the given figure, AP || BQ || CR. Prove that ar(AQC) = ar(PBR).

In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

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Triangles on the Same Base and between the Same Parallels

Standard IX Mathematics

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