# Ex.9.3 Q11 Areas of Parallelograms and Triangles Solution - NCERT Maths Class 9

## Question

In the given figure, \(ABCDE\) is a pentagon. \(A\) line through \(B\) parallel to \(AC\) meets \(DC\) produced at \(F\). Show that

(i) \(ar (ACB) = ar (ACF) \)

(ii) \(ar (AEDF) = ar (ABCDE) \)

**What is known?**

\(ABCDE\) is a pentagon. \(A\) line through \(B\) parallel to \(AC\) meets \(DC\) produced at \(F.\)

**What is unknown?**

How we can show that

(i) \(ar (ACB) = ar (ACF)\)

(ii) \(ar (AEDF) = ar (ABCDE)\)

**Reasoning:**

We can use theorem for triangles \(ACB\) and \(ACF\), if two triangles are on same base and between same pair of parallel lines then both will have equal area. Now we can add area of quadrilateral \(ACDE\) on both sides to get the second part required result.

## Text Solution

**Steps:**

(i) \(\begin{align}\rm \Delta ACB\, \rm and \,\Delta ACF\end{align}\) lie on the same base \(AC\) and The same parallels \(AC\) and \(BF\).

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

\[\begin{align}\text{Area}\; ( \Delta ACB ) = \text{Area}\; ( \Delta ACF )\end{align}\]

(ii) It can be observed that

\[\begin{align} {\text { Area }(\Delta \mathrm{ACB})}&={\text { Area }(\Delta \mathrm{ACF})} \end{align}\]

\[\begin{align} {\text { Area }(\Delta \mathrm{ACB})+\text { Area }(\mathrm{ACDE})}&={\text { Area }(\Delta \mathrm{ACF})+\text { Area (ACDE) }} \end{align}\]

\[\begin{align} {\text { Adding Equal Areas} \;(ACDE)}\;&{ \text{on both the sides. }}\end{align}\]

\[\begin{align}\text{Area}\, (ABCDE) = \text{Area} \,(AEDF)\end{align}\]