# Ex.9.2 Q4 Areas of Parallelograms and Triangles Solution - NCERT Maths Class 9

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## Question

In the given figure, $$P$$ is a point in the interior of a parallelogram $$ABCD$$. Show that

\begin{align}{\text { (i) ar }(\mathrm{APB})+\text { ar }(\mathrm{PCD})}&={\frac{1}{2} \text { ar }(\mathrm{ABCD})} \\ {\text { (ii) ar }(\mathrm{APD})+\text { ar }(\mathrm{PBC})}&={\text { ar }(\mathrm{APB})+\text { ar }(\mathrm{PCD})}\end{align}

[Hint: Through. $$P$$, draw a line parallel to $$AB$$]

Video Solution
Areas Of Parallelograms And Triangles
Ex 9.2 | Question 4

## Text Solution

What is known?

$$P$$ is a point in the interior of a parallelogram $$ABCD.$$

What is unknown?

How we can show that

(i) $$\text{ar }\left( \text{APB} \right)\text{ + ar }\left( \text{PCD} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ ar }\left( \text{ABCD} \right)$$

(ii) $$\text{ar }\left( \text{APD} \right)\text{ + ar }\left( \text{PBC} \right)\text{ = ar }\left( \text{APB} \right)\text{ + ar }\left( \text{PCD} \right)\text{ }$$

Reasoning:

Draw a line parallel to and $$CD$$ passes through point P. Now we can observe the figure, if a triangle and parallelogram are on same base and between same parallel lines then area of triangle will be half of area of parallelogram so we can find area of triangles $$APB$$ and $$PCD$$ to get the required result. Similarly, we can draw a line parallel to AD and BC to get the required result.

Steps:

(i) Let us draw a line segment $$EF$$, passing through the point $$P$$ and parallel to line segment $$AB$$ in parallelogram $$ABCD$$,

$$AB\; || \;EF$$ (By construction) ... (1)

We know that, $$ABCD$$ is a parallelogram.

$$\therefore AD\; ||\; BC$$ (Opposite sides of a parallelogram are parallel)

$$AE\; || \;BF$$ ... (2)

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

$$AB\; || \;EF$$ and $$AE\; || \;BF$$

Therefore, quadrilateral $$ABFE$$ is a parallelogram. Similarly , it can be deduced that quadrilateral $$EFCD$$ is a parallelogram.

It can be observed that $$\Delta APB$$ and parallelogram $$ABFE$$ are lying on the same base $$AB$$ and between the same set of parallel lines $$AB$$ and $$EF$$.

\begin{align}\therefore \text { Area }(\Delta \mathrm{APB})=\frac{1}{2} \text { Area }(\mathrm{ABFE})\dots(3)\end{align}

Similarly, for $$\Delta PCD$$ and parallelogram $$EFCD$$,

\begin{align}\therefore \rm{Area}\,\,(\Delta \mathrm{PCD})=\frac{1}{2}\text{Area (EFCD) }\dots(4)\end{align}

Adding Equations (3) and (4), we obtain

\begin{align}\text{Area }\left(\Delta \mathrm{APB} \right)\text+\text { Area }(\Delta\mathrm{PCD})=\frac{1}{2} \quad[\text { Area }(\mathrm{ABFE})+\text { Area }(\mathrm{EFCD})]\end{align}

\begin{align}\text{Area }\left(\Delta \mathrm{APB}\right)\text+\text { Area }(\Delta \mathrm{PCD})=\frac{1}{2} \quad \text { Area }(\mathrm{ABCD})\rm{ ...(5)}\end{align}

(ii)

Let us draw a line segment $$MN$$, passing through point $$P$$ and parallel to line segment $$AD$$. In parallelogram $$ABCD$$,

$$MN \;|| \;AD$$ (By construction) ... (6)

We know that, $$ABCD$$ is a parallelogram.

$$\therefore$$  $$AB\; ||\; DC$$ (Opposite sides of a parallelogram are parallel)

$$AM\; ||\; DN$$ … (7)

From Equations (6) and (7), we obtain

$$MN \;|| \;AD$$ and $$AM\; ||\; DN$$

Therefore, quadrilateral $$AMND$$ is a parallelogram.

It can be observed that $$\Delta APD$$ and parallelogram $$AMND$$ are lying on the same base $$AD$$ and between the same parallel lines $$AD$$ and $$MN$$.

\begin{align}\therefore \text { Area }(\Delta \mathrm{APD})=\frac{1}{2} \text { Area (AMND) } \ldots(8)\end{align}

Similarly, for $$\Delta PCB$$ and parallelogram $$MNCB$$,

\begin{align}\therefore \rm{Area}\,(\Delta \mathrm{PCB})=\frac{1}{2} \quad \text { Area }(\mathrm{MNCB})...(9)\end{align}

Adding Equations (8) and (9), we obtain

\begin{align}\text{Area }\left(\Delta \mathrm{APD}\right)+\text { Area }(\Delta \mathrm{PCB})=\frac{1}{2} \quad[\text { Area }(\mathrm{AMND})+\text { Area }(\mathrm{MNCB})]\end{align}

\begin{align}\text{Area }\left(\Delta \mathrm{APD}\right)+\text { Area }(\Delta \mathrm{PCB})=\frac{1}{2} \quad \text { Area }(\mathrm{ABCD})...(10)\end{align}

On comparing Equations (5) and (10), we obtain

\begin{align}\rm{Area}\,\,(\Delta \mathrm{APD})+\text { Area }(\Delta \mathrm{PBC})=\text { Area }(\Delta \mathrm{APB})+\text { Area }(\Delta \mathrm{PCD})\end{align}

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