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

## Question

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

(i)

\(\begin{align}ar (APB)\!+\!ar (PCD)\!=\!\frac{1}{2} ar(ABCD)\end{align}\)

(ii)

\(ar(APD)\!+\!ar(PBC)\!=\!ar(APB)\!+\!ar(PCD)\)

[Hint: Through. \(P\), draw a line parallel to \(AB\)]

## 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)\(ar(APB)\!+\!ar(PCD)\!=\!\frac{1}{2}ar(ABCD)\)

(ii) \(ar (APD)\!+\!ar(PBC)\!=ar(APB)\!+\!ar(PCD)\)

**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 \(EF\).

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

Similarly, for \(\Delta PCD\) and parallelogram \(EFCD\),

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

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

\(\begin{align}\begin{bmatrix}\text{Area }(\Delta {APB})+\\ \text { Area }(\Delta{PCD})\end{bmatrix}=\frac{1}{2} \begin{bmatrix}\text { Area }({ABFE})+\\\text { Area }({EFCD})]\end{bmatrix}\end{align}\)

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

(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 are lying on the same base and between the same parallel lines and .

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

Similarly, for \(\Delta PCB\) and parallelogram ,

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

Adding Equations () and (), we obtain

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

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

On comparing Equations () and (), we obtain

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