# Ex.9.3 Q13 Areas of Parallelograms and Triangles Solution - NCERT Maths Class 9

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## Question

$$ABCD$$ is a trapezium with $$AB\; || \;DC$$. A line parallel to $$AC$$ intersects $$AB$$ at $$X$$ and $$BC$$ at $$Y$$.

Prove that $$ar (ADX) = ar (ACY)$$.

[Hint: Join $$CX$$.]

Video Solution
Areas Of Parallelograms And Triangles
Ex 9.3 | Question 13

## Text Solution

What is known?

$$ABCD$$ is a trapezium with $$AB || DC$$. A line parallel to $$AC$$ intersects $$AB$$ at $$X$$ and $$BC$$ at $$Y.$$

What is unknown?

How we can prove that $$ar (ADX) = ar (ACY).$$

Reasoning:

We can use theorem for triangles $$ADX$$ and $$ACX$$, 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 $$ACX$$ and $$ACY$$. Now by comparing both the results, we will get the required result.

Steps: It can be observed that \begin{align}\rm \Delta ADX \, \rm and \,\Delta ACX \end{align} lie on the same base $$AX$$ and are between the same parallels $$AB$$ and $$DC$$.

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{ADX})=\text { Area }(\Delta \mathrm{ACX})....\text{(1)}\end{align}

$$\Delta ACY$$ and $$\Delta ACX$$ lie on the same base $$AC$$ and are between the same parallels $$AC$$ and $$XY$$.

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}(\Delta \mathrm{ACY})=\text { Area }(\mathrm{ACX}) \quad \ldots \text (2)\end{align}

From Equations (1) and (2), we obtain

\begin{align}{\rm{Area}}\left( {{\rm{\Delta ADX}}} \right){\rm{ = Area }}\left( {{\rm{\Delta ACY}}} \right)\end{align}