MyQuestionIcon

9

You visited us 9 times! Enjoying our articles? Unlock Full Access!

Triangles on the Same Base and between the Same Parallels

In the given ...

Question

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

Open in App

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,
$A r e a (Δ A B Q) + A r e a (Δ B C Q) = A r e a (Δ P B Q) + A r e a (Δ B R Q)$
$∴ A r e a (Δ A Q C) = A r e a (Δ P B R)$

Suggest Corrections

thumbs-up

1

Similar questions

Q. In the figure, AP || BQ || CR. Prove that

a r (△ A Q C) = a r (△ P B R)

A P ∥ B Q ∥ C R

a r (A Q C) = a r (P B R)

In the given figure, AP $∥$ BQ $∥$ CR. Prove that ar ( $△$ AQC) = ar ( $△$ PBR).
[2 MARKS]

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).

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Triangles on the Same Base and between the Same Parallels

Standard IX Mathematics

Join BYJU'S Learning Program

CrossIcon