# Ex.9.3 Q14 Areas of Parallelograms and Triangles Solution - NCERT Maths Class 9

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## Question

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

Video Solution
Areas Of Parallelograms And Triangles
Ex 9.3 | Question 14

## Text Solution

What is known?

$$AP || BQ || CR.$$

What is unknown?

How we can prove that $$ar (AQC) = ar (PBR).$$

Reasoning:

We can use theorem for triangles $$ABQ$$ and $$PBQ$$, if two triangles are on same base and between same pair of parallel lines then both will have equal area. Similarly for triangles we can use for triangles $$BCQ$$ and $$BRQ$$. Now by adding both the results, we will get the required result.

Steps: Since \begin{align}\rm \Delta ABQ\, \rm and \,\Delta PBQ\end{align} lie on the same base $$BQ$$ and are between the same parallels $$AP$$ and $$BQ$$,

According to Theorem 9.2 : Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

\begin{align}\therefore \text { Area }(\Delta \mathrm{ABQ})=\text { Area }(\triangle \mathrm{PBQ}) . . \text (1)\end{align}

Again, \begin{align}\rm (\Delta BCQ)\, \rm and \,(\Delta BRQ)\end{align} lie on the same base $$BQ$$ and are between the same parallels $$BQ$$ and $$CR$$.

\begin{align}\therefore \text { Area }(\Delta B C Q)=\text { Area }(\Delta B R Q) \ldots \text (2)\end{align}

On adding Equations (1) and (2), we obtain

\begin{align} {\text { Area }(\Delta \mathrm{ABQ})+\text { Area }(\Delta \mathrm{BCQ})}&={\text { Area }(\Delta \mathrm{PBQ})}{+\text { Area }(\Delta \mathrm{BRQ})} \end{align}

\begin{align} {\therefore \text { Area }(\Delta \mathrm{AQC})}&={\text { Area }(\Delta \mathrm{PBR})}\end{align}