NCERT Solutions for Class 9 Maths Chapter 9: Areas of Parallelograms and Triangles - Exercise 9.2

Exercise (9.2)

1. In the given figure, $\text{ABCD}$ is a parallelogram, $\text{AE}\bot \text{DC}$ and $\text{CF}\bot \text{AD}$. If $\text{AB = 16 cm}$, $\text{AE = 8 cm}$ and $\text{CF = 10 cm}$, find $\text{AD}$.

Ans: We are given the following: $\text{AB = 16 cm}$, $\text{AE = 8 cm}$ and $\text{CF = 10 cm}$.

We know that in a parallelogram the opposite sides are parallel and equal in length.

In parallelogram $\text{ABCD}$, the sides $\text{AB}$ and $\text{CD}$ are parallel and equal.

So, $\text{CD = AB = 16 cm}$

We know the area of parallelogram $\text{= base  }\!\!\times\!\!\text{  corresponding height}$

So, the area of parallelogram $\text{ABCD}$ will be:

$\text{Area of parallelogram = CD }\!\!\times\!\!\text{ AE = AD }\!\!\times\!\!\text{ CF}$

$\Rightarrow \text{16 cm  }\!\!\times\!\!\text{  8 cm = AD  }\!\!\times\!\!\text{  10 cm}$

$\Rightarrow \text{AD = }\frac{\text{16  }\!\!\times\!\!\text{  8}}{\text{10}}\text{ cm}$

$\Rightarrow \text{AD = 12}\text{.8 cm}$

Therefore, the length of $\text{AD}$ is $\text{12}\text{.8 cm}$.

2. If $\text{E}$, $\text{F}$, $\text{G}$ and $\text{H}$ are respectively the mid-points of the sides of a parallelogram $\text{ABCD}$, show that $\text{ar}\left( \text{EFGH} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ABCD} \right)$.

Ans:

Let us join the points $\text{H}$ and $\text{F}$ in the figure.

We know that in a parallelogram the opposite sides are parallel and equal in length.

So, in parallelogram $\text{ABCD}$:

$\text{AB = CD}$, $\text{AD = BC}$, $\text{AB }\parallel \text{ CD}$ and $\text{AD }\parallel \text{ BC}$

It is given that $\text{H}$ and $\text{F}$ are mid-points of the sides $\text{AD}$ and $\text{BC}$ and $\text{AD = BC}$ and $\text{AD }\parallel \text{ BC}$.

$\therefore \frac{\text{1}}{\text{2}}\text{AD = }\frac{\text{1}}{\text{2}}\text{BC}$

$\Rightarrow \text{AH = BF}$ and $\text{AH }\parallel \text{ BF}$

Hence, we can now consider $\text{ABFH}$ as a parallelogram.

Now in the figure, we can see that the parallelogram $\text{ABFH}$ and $\text{ }\!\!\Delta\!\!\text{ EHF}$ have a common base $\text{HF}$ and both of them lie between the parallel lines $\text{AB}$ and $\text{HF}$.

\[\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ HEF} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ABFH} \right)\]

Similarly, both the parallelogram $\text{HFCD}$ and $\text{ }\!\!\Delta\!\!\text{ HGF}$ have a common base $\text{HF}$ and both of them lie between the parallel lines $\text{DC}$ and $\text{HF}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ HGF} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{DCFH} \right)$

Now we find the area of $\text{EFGH}$.

$\therefore \text{ar}\left( \text{EFGH} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ HEF} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ HGF} \right)$

$\Rightarrow \text{ar}\left( \text{EFGH} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ABFH} \right)\text{ + }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{DCFH} \right)$

$\Rightarrow \text{ar}\left( \text{EFGH} \right)\text{ = }\frac{\text{1}}{\text{2}}\left[ \text{ar}\left( \text{ABFH} \right)\text{ + ar}\left( \text{DCFH} \right) \right]$

$\Rightarrow \text{ar}\left( \text{EFGH} \right)\text{ = }\frac{\text{1}}{\text{2}}\left[ \text{ar}\left( \text{ABCD} \right) \right]$

Therefore, we have proved that $\text{ar}\left( \text{EFGH} \right)\text{ = }\frac{\text{1}}{\text{2}}\left[ \text{ar}\left( \text{ABCD} \right) \right]$.

3. $\text{P}$ and $\text{Q}$ are any two points lying on the sides $\text{DC}$ and $\text{AD}$ respectively of a parallelogram $\text{ABCD}$. Show that $\text{ar}\left( \text{APB} \right)\text{ = ar}\left( \text{BQC} \right)$.

Ans: It is given that $\text{P}$ and $\text{Q}$ are any two points lying on the sides $\text{DC}$ and $\text{AD}$ respectively of a parallelogram $\text{ABCD}$.

Now in the figure, we can see that the parallelogram $\text{ABCD}$ and $\text{ }\!\!\Delta\!\!\text{ APB}$ have a common base $\text{AB}$ and both of them lie between the parallel lines $\text{AB}$ and $\text{DC}$.

\[\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APB} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ABCD} \right)\]               $\left( \text{1} \right)$

Similarly, we can see that the parallelogram $\text{ABCD}$ and $\text{ }\!\!\Delta\!\!\text{ BQC}$ have a common base $\text{BC}$ and both of them lie between the parallel lines $\text{AD}$ and $\text{BC}$.

\[\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BQC} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ABCD} \right)\]              $\left( \text{2} \right)$ 

Therefore, from equations $\left( \text{1} \right)$ and $\left( \text{2} \right)$, we can prove that \[\text{ar}\left( \text{APB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BQC} \right)\].

4. In the given figure, $\text{P}$ is a point in the interior of a parallelogram $\text{ABCD}$. Show that

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

(Hint: Through $\text{P}$, draw a line parallel to $\text{AB}$)

Ans: Let us draw a line segment $\text{EF}$ which is parallel to the line segment $\text{AB}$ and passes through the point $\text{P}$.

We know that in a parallelogram the opposite sides are parallel and equal in length.

So, in parallelogram $\text{ABCD}$:

$\text{AB = CD}$, $\text{AD = BC}$, $\text{AB }\parallel \text{ CD}$ and $\text{AD }\parallel \text{ BC}$

In the parallelogram $\text{ABCD}$, we drew $\text{AB}\parallel \text{EF}$.

But we can also write $\text{AD }\parallel \text{ BC}$.

$\therefore \text{AE }\parallel \text{ BF}$

So, we can say that $\text{ABFE}$ as a parallelogram.

Now in the figure, we can see that the parallelogram $\text{ABFE}$ and $\text{ }\!\!\Delta\!\!\text{ APB}$ have a common base $\text{AB}$ and both of them lie between the parallel lines $\text{AB}$ and $\text{EF}$.

\[\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APB} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ABFE} \right)\]

Similarly, both the parallelogram $\text{EFCD}$ and $\text{ }\!\!\Delta\!\!\text{ PCD}$ have a common base $\text{CD}$ and both of them lie between the parallel lines $\text{DC}$ and $\text{EF}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ PCD} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{EFCD} \right)$

Now we find the sum of \[\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APB} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ PCD} \right)\].

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APB} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ PCD} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ABFE} \right)\text{ + }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{EFCD} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APB} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ PCD} \right)\text{ = }\frac{\text{1}}{\text{2}}\left[ \text{ar}\left( \text{ABFE} \right)\text{ + ar}\left( \text{EFCD} \right) \right]$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APB} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ PCD} \right)\text{ = }\frac{\text{1}}{\text{2}}\left[ \text{ar}\left( \text{ABCD} \right) \right]$

Hence, we have shown that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APB} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ PCD} \right)\text{ = }\frac{\text{1}}{\text{2}}\left[ \text{ar}\left( \text{ABCD} \right) \right]$.

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

Ans: Let us draw a line segment $\text{GH}$ which is parallel to the line segment $\text{AD}$ and passes through the point $\text{P}$.

We know that in a parallelogram the opposite sides are parallel and equal in length.

So, in parallelogram $\text{ABCD}$:

$\text{AB = CD}$, $\text{AD = BC}$, $\text{AB }\parallel \text{ CD}$ and $\text{AD }\parallel \text{ BC}$

In the parallelogram $\text{ABCD}$, we drew $\text{AD}\parallel \text{GH}$.

But we can also write $\text{AB }\parallel \text{ DC}$.

$\therefore \text{AG }\parallel \text{ DH}$

So, we can say that $\text{AGHD}$ as a parallelogram.

Now in the figure, we can see that the parallelogram $\text{AGHD}$ and $\text{ }\!\!\Delta\!\!\text{ APD}$ have a common base $\text{AD}$ and both of them lie between the parallel lines $\text{AD}$ and $\text{GF}$.

\[\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APD} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{AGHD} \right)\]

Similarly, both the parallelogram $\text{GBCH}$ and $\text{ }\!\!\Delta\!\!\text{ PBC}$ have a common base $\text{BC}$ and both of them lie between the parallel lines $\text{BC}$ and $\text{GH}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ PBC} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{GBCH} \right)$

Now we find the sum of \[\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APD} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ PBC} \right)\].

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APD} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ PBC} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{AGHD} \right)\text{ + }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{GBCH} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APD} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ PBC} \right)\text{ = }\frac{\text{1}}{\text{2}}\left[ \text{ar}\left( \text{AGHD} \right)\text{ + ar}\left( \text{GBCH} \right) \right]$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APD} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ PBC} \right)\text{ = }\frac{\text{1}}{\text{2}}\left[ \text{ar}\left( \text{ABCD} \right) \right]$

But from previous part $\left( \text{i} \right)$, we found out that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APB} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ PCD} \right)\text{ = }\frac{\text{1}}{\text{2}}\left[ \text{ar}\left( \text{ABCD} \right) \right]$

So, by comparing them both, we obtain the following:

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

Therefore, we have shown that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APD} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ PBC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ APB} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ PCD} \right)$.

5. In the given figure, $\text{PQRS}$ and $\text{ABRS}$ are parallelograms and $\text{X}$ is any point on side $\text{BR}$. Show

i. $\text{ar}\left( \text{PQRS} \right)\text{ = ar}\left( \text{ABRS} \right)$

Ans: From the figure, we can see that the parallelogram $\text{PQRS}$ and $\text{ABRS}$ have a common base $\text{SR}$ and both of them lie between the parallel lines $\text{PB}$ and $\text{SR}$.

\[\therefore \text{ar}\left( \text{PQRS} \right)\text{ = ar}\left( \text{ABRS} \right)\]

Hence, we have shown that \[\text{ar}\left( \text{PQRS} \right)\text{ = ar}\left( \text{ABRS} \right)\].

ii. $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AXS} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{PQRS} \right)$

Ans: From the figure, we can see that the parallelogram $\text{ABRS}$ and $\text{ }\!\!\Delta\!\!\text{ AXS}$ have a common base $\text{AS}$ and both of them lie between the parallel lines $\text{AS}$ and $\text{BR}$.

\[\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AXS} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ABRS} \right)\]

But form previous part $\left( \text{i} \right)$, we found out that \[\text{ar}\left( \text{PQRS} \right)\text{ = ar}\left( \text{ABRS} \right)\].

So, we can write the following:

\[\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AXS} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{PQRS} \right)\]

Hence, we have shown that \[\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AXS} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{PQRS} \right)\].

6. A farmer was having a field in the form of a parallelogram $\text{PQRS}$. She took any point $\text{A}$ on $\text{RS}$ and joined it to points $\text{P}$ and $\text{Q}$. In how many parts the field is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

Ans:

It is given that there is a point $\text{A}$ on the side $\text{RS}$ of the parallelogram $\text{PQRS}$. Then the point $\text{A}$ is joined by points $\text{P}$ and $\text{Q}$.

So, we can see from the figure that the field is divided into three parts.

These parts are all in the shape of triangles.

These parts are $\text{ }\!\!\Delta\!\!\text{ APQ}$, $\text{ }\!\!\Delta\!\!\text{ APS}$ and $\text{ }\!\!\Delta\!\!\text{ AQR}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APQ} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ APS} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ AQR} \right)\text{ = ar}\left( \text{PQRS} \right)$               …$\left( \text{i} \right)$

If a parallelogram and a triangle are situated on the same base and both of them lie between two parallel lines, then the area of a triangle is half of that of the area of the area of the parallelogram.

From the figure, we can see that the parallelogram $\text{PQRS}$ and $\text{ }\!\!\Delta\!\!\text{ APQ}$ have a common base $\text{PQ}$ and both of them lie between the parallel lines $\text{PQ}$ and $\text{SR}$.

\[\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APQ} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{PQRS} \right)\]

We will put this value in equation $\left( \text{i} \right)$.

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APQ} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ APS} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ AQR} \right)\text{ = ar}\left( \text{PQRS} \right)$

$\Rightarrow \frac{\text{1}}{\text{2}}\text{ar}\left( \text{PQRS} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ APS} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ AQR} \right)\text{ = ar}\left( \text{PQRS} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APS} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ AQR} \right)\text{ = ar}\left( \text{PQRS} \right)\text{ - }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{PQRS} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APS} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ AQR} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{PQRS} \right)$

Hence, we can see that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ APS} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ AQR} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ APQ} \right)$

So, the farmer can sow wheat in \[\text{ }\!\!\Delta\!\!\text{ APQ}\] and pulses in the other triangles $\text{ }\!\!\Delta\!\!\text{ APS}$ and $\text{ }\!\!\Delta\!\!\text{ AQR}$ or vice versa.

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.2

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