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

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## 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.

Video Solution
Areas Of Parallelograms And Triangles
Ex 9.3 | Question 11

## 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}