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

## Question

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

## 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

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

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