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# NCERT Solutions for Class 9 Maths Chapter 9: Areas of Parallelograms and Triangles - Exercise 9.3

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## NCERT Solutions for Class 9 Maths Chapter 9 (Ex 9.3)

Free PDF download of NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.3 (Ex 9.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise CBSE Solutions in your emails. Download free NCERT Solutions Class 9 Maths to amp up your preparations and to score well in your examinations. Students can also avail of NCERT Solutions for Class 9 Science from our website.

 Class: NCERT Solutions for Class 9 Subject: Class 9 Maths Chapter Name: Chapter 9 - Areas of Parallelograms and Triangles Exercise: Exercise - 9.3 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

### Important Topics under NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles (Ex 9.3) Exercise 9.3

The NCERT Chapter 9 Areas of Parallelograms and Triangles is an important chapter of Class 9 Maths and is included under the ‘Menstruation’ unit.

Class 9 students are introduced to various basic topics under Maths Chapter 9 of NCERT.  We have mentioned these topics for you so that you may know what all you will learn in the NCERT Chapter 9 and how will you set your revision timetable to practice the NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles (Ex 9.3) Exercise 9.3.

Since these Class 9 Maths Chapter 9 topics are quite complex, our expert teachers have curated NCERT Solutions for Maths for you in detail. We want you to make the most of them and do not miss any topic. This way you can score well in your examinations.

Here are the important topics from Class 9 NCERT Maths Chapter 9 in the tabular form to provide a clear picture of what's in the chapter Areas of Parallelograms and Triangles.

 Sl. No. Important Topics 1. Figures on the Same Base and Between the Same Parallels 2. Parallelograms on the same Base and Between the same Parallels 3. Triangles on the same Base and between the same Parallels
Competitive Exams after 12th Science

## Access NCERT solutions for Class 9 Maths Chapter 9 - Areas of Parallelograms and Triangles

Exercise (9.3)

1. In the given figure, $\text{E}$ is any point on median $\text{AD}$ of a $\text{ }\!\!\Delta\!\!\text{ ABC}$. Show that $\text{ar}\left( \text{ABE} \right)\text{ = ar}\left( \text{ACE} \right)$

Ans: It is given that there is a point $\text{E}$ on a median $\text{AD}$ of the $\text{ }\!\!\Delta\!\!\text{ ABC}$.

We know the median divides the triangle into two equal area triangles.

So, median $\text{AD}$ will divide $\text{ }\!\!\Delta\!\!\text{ ABC}$ into two triangles of equal area.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABD} \right)\text{ = ar}\left( \text{ACD} \right)$               …$\left( \text{1} \right)$

The median $\text{ED}$ will divide $\text{ }\!\!\Delta\!\!\text{ EBC}$ into two triangles of equal area.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ EBD} \right)\text{ = ar}\left( \text{ECD} \right)$                …$\left( \text{2} \right)$

Now we will subtract the equation $\left( \text{2} \right)$ from $\left( \text{1} \right)$.

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABD} \right)\text{ - ar}\left( \text{ }\!\!\Delta\!\!\text{ EBD} \right)\text{ = ar}\left( \text{ACD} \right)\text{ - ar}\left( \text{ECD} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABE} \right)\text{ = ar}\left( \text{ACE} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABE} \right)\text{ = ar}\left( \text{ACE} \right)$.

2. In a triangle $\text{ABC}$, $\text{E}$ is the mid-point of median $\text{AD}$. Show that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BED} \right)\text{ = }\frac{\text{1}}{\text{4}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)$.

Ans: It is given that point $\text{E}$ is a mid-point on a median $\text{AD}$ of the $\text{ }\!\!\Delta\!\!\text{ ABC}$.

We know the median divides the triangle into two equal area triangles.

So, median $\text{AD}$ will divide $\text{ }\!\!\Delta\!\!\text{ ABC}$ into two triangles of equal area.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABD} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACD} \right)$

Since the two triangles are of equal area, we can now write $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABD} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)$

Similarly, median $\text{BE}$ will divide $\text{ }\!\!\Delta\!\!\text{ ABD}$ into two triangles of equal area.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BED} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BEA} \right)$

Since the two triangles are of equal area, we can now write $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BED} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \Delta \text{ABD} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BED} \right)\text{ = }\frac{\text{1}}{\text{2}}\left[ \frac{\text{1}}{\text{2}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right) \right]$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BED} \right)\text{ = }\frac{\text{1}}{\text{4}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BED} \right)\text{ = }\frac{\text{1}}{\text{4}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)$

3. Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Ans: Let us assume that $\text{ABCD}$ be a parallelogram.

If we join the diagonals $\text{AC}$ and $\text{BD}$, let $\text{E}$ be their intersection.

We know that the diagonals of a parallelogram bisect each other.

So, we can say that $\text{E}$ is the mid-point of both the diagonals.

We know the median divides the triangle into two equal area triangles.

So, median $\text{DE}$ will divide $\text{ }\!\!\Delta\!\!\text{ ADC}$ into two triangles of equal area.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ CED} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ DEA} \right)$                                       …$\left( \text{1} \right)$

The median $\text{CE}$ will divide $\text{ }\!\!\Delta\!\!\text{ DCB}$ into two triangles of equal area.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BEC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ CED} \right)$                                        …$\left( \text{2} \right)$

Similarly we can write $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AEB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BEC} \right)$      …$\left( \text{3} \right)$

From equations $\left( \text{1} \right)$, $\left( \text{2} \right)$ and $\left( \text{3} \right)$, we can write

$\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEA} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ CED} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BEC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AEB} \right)$

Therefore, we have shown that the diagonals of a parallelogram divide it into four triangles of equal area.

4. In the given figure, $\text{ABC}$ and $\text{ABD}$ are two triangles on the same base $\text{AB}$. If line-segment $\text{CD}$ is bisected by $\text{AB}$ at $\text{O}$, show that $\text{ar}\left( \text{ABC} \right)\text{ = ar}\left( \text{ABD} \right)$

Ans: It is given that $\text{CD}$ is bisected by $\text{AB}$ at $\text{O}$.

Let us first consider $\text{ }\!\!\Delta\!\!\text{ ACD}$.

We are given that $\text{CO = DO}$.

Since a median divides the triangle into two equal area triangles, we can say that $\text{AO}$ is the median of $\text{ }\!\!\Delta\!\!\text{ ACD}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AOC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AOD} \right)$                             …$\left( \text{1} \right)$

Similarly $\text{BO}$ is the median of $\text{ }\!\!\Delta\!\!\text{ BCD}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BOC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BOD} \right)$                              …$\left( \text{2} \right)$

We will now add equations $\left( \text{1} \right)$ and $\left( \text{2} \right)$:

$\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AOC} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ BOC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BOD} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ AOD} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ABD} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ABC} \right)\text{ = ar}\left( \text{ABD} \right)$.

5. $\text{D}$, $\text{E}$ and $\text{F}$ are respectively the midpoints of the sides $\text{BC}$, $\text{CA}$ and $\text{AB}$ of a $\text{ }\!\!\Delta\!\!\text{ ABC}$. Show that:

i. $\text{BDEF}$ is a parallelogram

Ans: It is given that $\text{F}$ and $\text{E}$ are midpoints of $\text{AB}$ and $\text{AC}$ respectively.

We know that the line joining the mid-points of two sides of a triangle is parallel to the third side and also half of its length.

$\therefore \text{FE}\parallel \text{BC}$ and $\text{FE = }\frac{\text{1}}{\text{2}}\text{BC}$

Since, $\text{D}$ is the mid-point of $\text{BC}$ we know $\text{BD = }\frac{\text{1}}{\text{2}}\text{BC}$.

$\therefore \text{FE = BD}$ and $\text{FE}\parallel \text{BD}$                                          …$\left( \text{1} \right)$

It is given that $\text{D}$ and $\text{E}$ are midpoints of $\text{BC}$ and $\text{AC}$ respectively.

We know that the line joining the mid-points of two sides of a triangle is parallel to the third side and also half of its length.

$\therefore \text{DE}\parallel A\text{B}$ and $\text{DE = }\frac{\text{1}}{\text{2}}\text{AB}$

Since, $\text{F}$ is the mid-point of $\text{AB}$ we know $\text{BF = }\frac{\text{1}}{\text{2}}\text{AB}$.

$\therefore \text{DE = BF}$ and $\text{DE}\parallel \text{BF}$                                           …$\left( \text{2} \right)$

So, from equations $\left( \text{1} \right)$ and $\left( \text{2} \right)$, we obtain:

$\text{FE = BD}$ and $\text{FE}\parallel \text{BD}$

$\text{DE = BF}$ and $\text{DE}\parallel \text{BF}$

Therefore, we have shown that $\text{BDEF}$ is a parallelogram.

ii. $\text{ar}\left( \text{DEF} \right)\text{ = }\frac{\text{1}}{\text{4}}\text{ar}\left( \text{ABC} \right)$

Ans: We have found out that $\text{BDEF}$ is a parallelogram.

A diagonal divides the parallelogram into two equal area triangles.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BDF} \right)\text{ = ar}\left( \text{DEF} \right)$                              …$\left( \text{1} \right)$

Similarly, we can say that $\text{CDFE}$ is also a parallelogram.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ CDE} \right)\text{ = ar}\left( \text{DEF} \right)$                              …$\left( \text{2} \right)$

Similarly, we can say that $\text{AEDF}$ is also a parallelogram.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AEF} \right)\text{ = ar}\left( \text{DEF} \right)$                               …$\left( \text{3} \right)$

From equations $\left( \text{1} \right)$, $\left( \text{2} \right)$ and $\left( \text{3} \right)$, we obtain

$\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BDF} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ CDE} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AEF} \right)$

But we know that $ar\left( \Delta ABC \right)=\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ BDF} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ CDE} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ AEF} \right)$

$\Rightarrow ar\left( \Delta ABC \right)=\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)\text{ = 4 }\!\!\times\!\!\text{ ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ }$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ = }\frac{\text{1}}{\text{4}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ = }\frac{\text{1}}{\text{4}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)$.

iii. $\text{ar}\left( \text{BDEF} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ABC} \right)$

Ans: We know $\text{ar}\left( \text{BDEF} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BDF} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)$

But $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BDF} \right)$.

So, $\text{ar}\left( \text{BDEF} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)$

$\Rightarrow \text{ar}\left( \text{BDEF} \right)\text{ = 2 }\!\!\times\!\!\text{ ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)$

$\Rightarrow \text{ar}\left( \text{BDEF} \right)\text{ = 2 }\!\!\times\!\!\text{ }\left[ \frac{\text{1}}{\text{4}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right) \right]$

$\Rightarrow \text{ar}\left( \text{BDEF} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)$

Therefore, we have shown that $\text{ar}\left( \text{BDEF} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)$.

6. In the given figure, diagonals $\text{AC}$ and $\text{BD}$ of quadrilateral $\text{ABCD}$ intersect at $\text{O}$ such that $\text{OB = OD}$. If $\text{AB = CD}$, then show that:

i. $\text{ar}\left( \text{DOC} \right)\text{ = ar}\left( \text{AOB} \right)$

(Hint: From $\text{D}$ and $\text{B}$, draw perpendiculars to $\text{AC}$)

Ans: In the figure, let us draw $\text{BP}\bot \text{AC}$ and $\text{DQ}\bot \text{AC}$.

Let us consider in $\text{ }\!\!\Delta\!\!\text{ DOQ}$ and $\text{ }\!\!\Delta\!\!\text{ BOP}$:

$\angle \text{DQO = }\angle \text{BPO = 9}{{\text{0}}^{\text{o}}}$ (By drawing perpendicular lines)

$\angle \text{DOQ = }\angle \text{BOP}$ (They are vertically opposite angles)

$\text{OB = OD}$ (This is given)

Therefore, by $\text{AAS}$ congruence rule, $\text{ }\!\!\Delta\!\!\text{ DOQ }\cong \text{ }\!\!\Delta\!\!\text{ BOP}$

So, we can write $\text{DQ = BP}$                                             …$\left( \text{1} \right)$

The congruent triangles also have equal areas.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DOQ} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BOP} \right)$                                             …$\left( \text{2} \right)$

Let us consider in $\text{ }\!\!\Delta\!\!\text{ DQC}$ and $\text{ }\!\!\Delta\!\!\text{ BPA}$:

$\angle \text{DQC = }\angle \text{BPA = 9}{{\text{0}}^{\text{o}}}$ (By drawing perpendicular lines)

$\text{CD = AB}$ (This is given)

$\text{DQ = BP}$ [From equation $\left( \text{1} \right)$]

Therefore, by $\text{RHS}$ congruence rule, $\text{ }\!\!\Delta\!\!\text{ DQC }\cong \text{ }\!\!\Delta\!\!\text{ BPA}$

The congruent triangles also have equal areas.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DQC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BPA} \right)$                                             …$\left( \text{3} \right)$

We will now add equations $\left( \text{2} \right)$ and $\left( \text{3} \right)$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DOQ} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ DQC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BOP} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ BPA} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DOC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AOB} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DOC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AOB} \right)$.

ii. $\text{ar}\left( \text{DCB} \right)\text{ = ar}\left( \text{ACB} \right)$

Ans: From the above part, we obtained that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DOC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AOB} \right)$.

We will add $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ OCB} \right)$ on both sides.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DOC} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ OCB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AOB} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ OCB} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DCB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACB} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DCB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACB} \right)$.

iii. $\text{DA}\parallel \text{CB}$ or $\text{ABCD}$ is a parallelogram.

Ans: From the previous part, we obtained that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DCB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACB} \right)$.

We know that if two triangles have equal areas and the same base, then they will both lie between two parallel lines.

$\therefore \text{CB }\parallel \text{ DA}$

It is given that $\text{AB = CD}$.

So, In quadrilateral $\text{ABCD}$, a pair of opposite sides is equal $\left( \text{AB = CD} \right)$ and the other pair of opposite sides is parallel $\left( \text{CB }\parallel \text{ DA} \right)$.

Hence, we say that $\text{ABCD}$ is a parallelogram.

7. $\text{D}$ and $\text{E}$ are points on sides $\text{AB}$ and $\text{AC}$ respectively of $\text{ }\!\!\Delta\!\!\text{ ABC}$ such that $\text{ar}\left( \text{DBC} \right)\text{ = ar}\left( \text{EBC} \right)$. Prove that $\text{DE}\parallel \text{BC}$.

Ans: It is given that $\text{ar}\left( \text{DBC} \right)\text{ = ar}\left( \text{EBC} \right)$.

We know that if two triangles have equal areas and same base, then they will both lie between two parallel lines.

$\therefore \text{DE}\parallel \text{BC}$

Hence, we have proved that $\text{DE}\parallel \text{BC}$.

8. $\text{XY}$ is a line parallel to side $\text{BC}$ of a triangle $\text{ABC}$. If $\text{BE}\parallel \text{AC}$ and $\text{CF}\parallel \text{AB}$ meet $\text{XY}$ at $\text{E}$ and $\text{F}$ respectively, show that $\text{ar}\left( \text{ABE} \right)\text{ = ar}\left( \text{ACF} \right)$.

Ans: It is given that $\text{XY}$ is a line parallel to side $\text{BC}$ of a triangle $\text{ABC}$.

$\therefore \text{XY}\parallel \text{BC}$

Since $\text{E}$ is an extension of line $\text{XY}$, we can also write $\text{EY}\parallel \text{BC}$

It is given that $\text{BE}\parallel \text{AC}$.

Since $\text{Y}$ is a point of line $\text{AC}$, we can also write $\text{BE}\parallel CY$.

So, in quadrilateral $\text{EBCY}$, we have found out that the pair of opposite sides are parallel.

Hence, $\text{EBCY}$ is a parallelogram.

Similarly, we know that $\text{XY}\parallel \text{BC}$.

Since $\text{F}$ is an extension of line $\text{XY}$, we can also write $\text{XF}\parallel \text{BC}$.

It is given that $\text{CF}\parallel \text{AB}$.

Since $\text{X}$ is a point of line $\text{AB}$, we can also write $\text{CF}\parallel BX$.

So, in quadrilateral $\text{FCBX}$, we have found out that the pair of opposite sides are parallel.

Hence, $\text{FCBX}$ is a parallelogram.

We know that if two parallelograms have same base and they lie between two parallel lines, then they have the same areas.

The parallelograms $\text{EBCY}$ and $\text{FCBX}$ lie on the same base $\text{BC}$ and between the parallel lines $\text{BC}$ and $\text{EF}$.

$\therefore \text{ar}\left( \text{EBCY} \right)\text{ = ar}\left( \text{FCBX} \right)$                                              …$\left( \text{1} \right)$

Now let us consider parallelogram $\text{EBCY}$ and $\text{ }\!\!\Delta\!\!\text{ ABE}$.

They both lie on the same base $\text{BE}$ and between two parallel lines $\text{AC}$ and $\text{EB}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABE} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{EBCY} \right)$                                           …$\left( \text{2} \right)$

Similarly, consider parallelogram $\text{FCBX}$ and $\text{ }\!\!\Delta\!\!\text{ ACF}$.

They both lie on the same base $\text{FC}$ and between two parallel lines $\text{AB}$ and $\text{FC}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ACF} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{FCBX} \right)$                                            …$\left( \text{3} \right)$

From equations  $\left( \text{1} \right)$, $\left( \text{2} \right)$ and $\left( \text{3} \right)$, we can write:

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABE} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACF} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABE} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACF} \right)$.

9. The side $\text{AB}$ of a parallelogram $\text{ABCD}$ is produced to any point$\text{P}$. A line through $\text{A}$ and parallel to $\text{CP}$ meets $\text{CB}$ produced at $\text{Q}$ and then parallelogram $\text{PBQR}$ is completed (see the following figure). Show that $\text{ar}\left( \text{ABCD} \right)\text{ = ar}\left( \text{PBQR} \right)$.

Ans: Let us join the lines $\text{AC}$ and $\text{PQ}$.

We know that if two triangles have the same base and they lie between two parallel lines, then they have the same areas.

$\text{ }\!\!\Delta\!\!\text{ ACQ}$ and $\text{ }\!\!\Delta\!\!\text{ QPA}$ lie on the same base $\text{AQ}$ and between two parallel lines $\text{AQ}$ and $\text{CP}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ACQ} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ QPA} \right)$

Subtracting $\text{ar}\left( \text{ABQ} \right)$ on both sides:

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ACQ} \right)\text{ - ar}\left( \text{ABQ} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ QPA} \right)\text{ - ar}\left( \text{ABQ} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ QBP} \right)$                                    …$\left( \text{1} \right)$

It is given that $\text{AC}$ and $\text{PQ}$ are diagonals of parallelograms $\text{ABCD}$ and $\text{QBPR}$ respectively. The diagonal divide the parallelogram into two triangles of equal areas.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ABCD} \right)$                                 …$\left( \text{2} \right)$

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ QBP} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{QBPR} \right)$                                  …$\left( \text{3} \right)$

From equations $\left( \text{1} \right)$, $\left( \text{2} \right)$ and $\left( \text{3} \right)$, we will obtain $\text{ar}\left( \text{ABCD} \right)\text{ = ar}\left( \text{QBPR} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ABCD} \right)\text{ = ar}\left( \text{QBPR} \right)$.

10. Diagonals $\text{AC}$ and $\text{BD}$ of a trapezium $\text{ABCD}$ with $\text{AB}\parallel \text{DC}$ intersect each other at $\text{O}$. Prove that $\text{ar}\left( \text{AOD} \right)\text{ = ar}\left( \text{BOC} \right)$.

Ans: It is given that $\text{AB}\parallel \text{DC}$.

We know that if two triangles have the same base and they lie between two parallel lines, then they have the same areas.

Here, we can see that $\text{ }\!\!\Delta\!\!\text{ ADC}$ and $\text{ }\!\!\Delta\!\!\text{ BCD}$ lie on the same base $\text{DC}$ and between two parallel lines $\text{DC}$ and $\text{AB}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ADC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BCD} \right)$

Subtracting $\text{ar}\left( \text{DOC} \right)$ on both sides:

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ADC} \right)\text{ - ar}\left( \text{ }\!\!\Delta\!\!\text{ DOC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BCD} \right)\text{ - ar}\left( \text{ }\!\!\Delta\!\!\text{ DOC} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AOD} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BOC} \right)$

Therefore, we have proved that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AOD} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BOC} \right)$.

11. In the given figure, $\text{ABCDE}$ is a pentagon. A line through $\text{B}$ parallel to $\text{AC}$ meets $\text{DC}$ produced at $\text{F}$. Show that

i. $\text{ar}\left( \text{ACB} \right)\text{ = ar}\left( \text{ACF} \right)$

Ans: We know that if two triangles have the same base and they lie between two parallel lines, then they have the same areas.

Here, we can see that $\text{ }\!\!\Delta\!\!\text{ ACB}$ and $\text{ }\!\!\Delta\!\!\text{ ACF}$ lie on the same base $\text{AC}$ and between two parallel lines $\text{AC}$ and $\text{BF}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ACB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACF} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ACB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACF} \right)$

ii. $\text{ar}\left( \text{AEDF} \right)\text{ = ar}\left( \text{ABCDE} \right)$

Ans: From the previous part, we obtained that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ACB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACF} \right)$.

We add $ar\left( \text{ACDE} \right)$ on both sides.

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ACB} \right)\text{ + ar}\left( \text{ACDE} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACF} \right)\text{ + ar}\left( \text{ACDE} \right)$

$\Rightarrow \text{ar}\left( \text{ABCDE} \right)\text{ = ar}\left( \text{AEDF} \right)$

Therefore, we have proved that $\text{ar}\left( \text{AEDF} \right)\text{ = ar}\left( \text{ABCDE} \right)$.

12. A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.

Ans: Let us assume quadrilateral $\text{ABCD}$ be the original shape of the land.

Let us join the points $\text{B}$ and $\text{D}$ to form the diagonal of $\text{ABCD}$.

Now we draw a line parallel to $\text{BD}$ through point $\text{A}$. Let it meet the extending of side $\text{CD}$ at new point $\text{E}$.

Now we join $\text{BE}$ and $\text{AD}$ and let their intersection be $\text{F}$.

So, we assume that the portion $\text{ }\!\!\Delta\!\!\text{ AFB}$ can be cut from the original field so that the villager gets the new shape of his field as $\text{ }\!\!\Delta\!\!\text{ DEF}$.

So, we will prove this assumption of ours that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AFB} \right)$.

We know that if two triangles have the same base and they lie between two parallel lines, then they have the same areas.

Here, we can see that $\text{ }\!\!\Delta\!\!\text{ DEB}$ and $\text{ }\!\!\Delta\!\!\text{ BAD}$ lie on the same base $\text{DB}$ and between two parallel lines $\text{DB}$ and $\text{EA}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BAD} \right)$

We subtract $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DFB} \right)$ on both sides.

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEB} \right)\text{ - ar}\left( \text{ }\!\!\Delta\!\!\text{ DFB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BAD} \right)\text{ - ar}\left( \text{ }\!\!\Delta\!\!\text{ DFB} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AFB} \right)$

So, our assumption is correct.

Therefore, the land must be cut according to the figure drawn.

13. $\text{ABCD}$ is a trapezium with $\text{AB}\parallel \text{DC}$. A line parallel to $\text{AC}$ intersects $\text{AB}$ at $\text{X}$ and $\text{BC}$ at $\text{Y}$. Prove that $\text{ar}\left( \text{ADX} \right)\text{ = ar}\left( \text{ACY} \right)$. (Hint: Join $\text{CX}$)

Ans:

It is given that $\text{AB}\parallel \text{DC}$ and $\text{XY}\parallel A\text{C}$.

We know that if two triangles have the same base and they lie between two parallel lines, then they have the same areas.

Here, we can see that $\text{ }\!\!\Delta\!\!\text{ ADX}$ and $\text{ }\!\!\Delta\!\!\text{ XCA}$ lie on the same base $\text{AX}$ and between two parallel lines $\text{AX}$ and $\text{DC}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ADX} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ XCA} \right)$                       …$\left( \text{1} \right)$

Similarly, $\text{ }\!\!\Delta\!\!\text{ ACY}$ and $\text{ }\!\!\Delta\!\!\text{ ACX}$ lie on the same base $\text{AC}$ and between two parallel lines $\text{AC}$ and $\text{XY}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ACY} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ XCA} \right)$                       …$\left( \text{2} \right)$

Hence, from equations $\left( \text{1} \right)$ and $\left( \text{2} \right)$, we obtain:

$\text{ar}\left( \text{ADX} \right)\text{ = ar}\left( \text{ACY} \right)$

Therefore, we proved that $\text{ar}\left( \text{ADX} \right)\text{ = ar}\left( \text{ACY} \right)$.

14. In the given figure, $\text{AP}\parallel \text{BQ}\parallel \text{CR}$. Prove that $\text{ar}\left( \text{AQC} \right)\text{ = ar}\left( \text{PBR} \right)$.

Ans: We know that if two triangles have the same base and they lie between two parallel lines, then they have the same areas.

Here, we can see that $\text{ }\!\!\Delta\!\!\text{ ABQ}$ and $\text{ }\!\!\Delta\!\!\text{ PBQ}$ lie on the same base $\text{BQ}$ and between two parallel lines $\text{BQ}$ and $\text{AP}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABQ} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ PBQ} \right)$                            …$\left( \text{1} \right)$

Similarly, $\text{ }\!\!\Delta\!\!\text{ BQC}$ and $\text{ }\!\!\Delta\!\!\text{ BQR}$ lie on the same base $\text{BQ}$ and between two parallel lines $\text{BQ}$ and $\text{CR}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BQC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BQR} \right)$                            …$\left( \text{2} \right)$

We will add equations $\left( \text{1} \right)$ and $\left( \text{2} \right)$.

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABQ} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ BQC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BQR} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ PBQ} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AQC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ PBR} \right)$

Therefore, we have proved that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AQC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ PBR} \right)$.

15. Diagonals $\text{AC}$ and $\text{BD}$ of a quadrilateral $\text{ABCD}$ intersect at $\text{O}$ in such a way that $\text{ar}\left( \text{AOD} \right)\text{ = ar}\left( \text{BOC} \right)$. Prove that $\text{ABCD}$ is a trapezium.

Ans: It is given that $\text{ar}\left( \text{AOD} \right)\text{ = ar}\left( \text{BOC} \right)$.

We add $\text{ar}\left( \text{AOB} \right)$ on both sides.

$\Rightarrow \text{ar}\left( \text{AOD} \right)\text{ + ar}\left( \text{AOB} \right)\text{ = ar}\left( \text{BOC} \right)\text{ + ar}\left( \text{AOB} \right)$

$\Rightarrow \text{ar}\left( \text{ABD} \right)\text{ = ar}\left( \text{ABC} \right)$

We know that if two triangles have the same base and they lie between two parallel lines, then they have the same areas.

So, these two triangles lie between two parallel lines $\text{AB}$ and $\text{DC}$.

Therefore, we can say that $\text{ABCD}$ is a trapezium.

16. In the given figure, $\text{ar}\left( \text{DRC} \right)\text{ = ar}\left( \text{DPC} \right)$ and $\text{ar}\left( \text{BDP} \right)\text{ = ar}\left( \text{ARC} \right)$. Show that both the quadrilaterals $\text{ABCD}$ and $\text{DCPR}$ are trapeziums.

Ans: It is given that $\text{ar}\left( \text{DRC} \right)\text{ = ar}\left( \text{DPC} \right)$.

We know that if two triangles have the same base and they have the same areas, then they must lie between two parallel lines.

Here, we can see that $\text{ }\!\!\Delta\!\!\text{ DRC}$ and $\text{ }\!\!\Delta\!\!\text{ DPC}$ lie on the same base $\text{DC}$. So, they must lie between two parallel lines.

$\therefore \text{DC}\parallel \text{RP}$

Hence, we can say that $\text{DCPR}$ is a trapezium.

It is also given that $\text{ar}\left( \text{ARC} \right)\text{ = ar}\left( \text{BDP} \right)$

So, we will subtract this equation from the previous equation$\text{ar}\left( \text{DRC} \right)\text{ = ar}\left( \text{DPC} \right)$

$\Rightarrow \text{ar}\left( \text{ARC} \right)\text{ - ar}\left( \text{DRC} \right)\text{ = ar}\left( \text{BDP} \right)\text{ - ar}\left( \text{DPC} \right)$

$\Rightarrow \text{ar}\left( \text{ADC} \right)\text{ = ar}\left( \text{BDC} \right)$

Here, we can see that $\text{ }\!\!\Delta\!\!\text{ ADC}$ and $\text{ }\!\!\Delta\!\!\text{ BDC}$ lie on the same base $\text{DC}$. So, they must lie between two parallel lines.

$\therefore \text{AB}\parallel \text{DC}$

Hence, we can say that $\text{ABCD}$ is a trapezium.

Therefore, we have shown that $\text{DCPR}$ and $\text{ABCD}$ are trapeziums.

### Weightage of NCERT Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

Chapter 9 of Class 9 Maths NCERT is a Geometry chapter that introduces students to concepts like how to find the areas of various figures under different conditions. The Areas of Parallelograms and Triangles chapter carries a total of 14 marks in the exam. While from the total unit of Menstruation, one can expect 2 multiple choice questions of 1 mark each, 2 short questions of 6 marks each and 1 long question of 6 marks.

## NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3

Opting for the NCERT solutions for Ex 9.3 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 9.3 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 9 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 9 Maths Chapter 9 Exercise 9.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 9 Maths Chapter 9 Exercise 9.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 9 Maths Chapter 9 Exercise 9.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

### Access Other Exercise Solutions of Class 9 Maths Chapter 9 - Area of Parallelograms and Triangles

 Chapter 8 Area of Parallelograms and Triangles All Exercises in PDF Format Exercise 9.1 1 Question and Solutions Exercise 9.2 6 Questions and Solutions Exercise 9.4 8 Questions and Solutions

## FAQs on NCERT Solutions for Class 9 Maths Chapter 9: Areas of Parallelograms and Triangles - Exercise 9.3

1. What is Class 9 Maths Chapter 9 consist of?

Class 9 Maths Chapter 9 deals with the various concepts and theorems regarding the parallelograms and triangles. The in-house subject matter experts at Vedantu have created the NCERT solutions for Class 9 Maths Chapter 9 – Area of Parallelogram and Triangles. This chapter also deals with Triangles on the Same Base and Between Same Parallels.

2. Why should I choose Vedantu’s NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.3?

All the solutions provided by Vedantu are designed as per the latest NCERT syllabus and guidelines. The NCERT Solutions for Class 9 Maths Chapter 9 – Area of Parallelogram and Triangles consists of detailed step by step explanation to the questions that are given in the exercises at the end of chapter 9. It helps students to be thorough with the concept and also gain practice solving questions.

3. Is it possible to access the best quality NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.3 in PDF format?

Yes, it is very much possible to access the top quality NCERT Solutions for Class 9 Maths Chapter 9 Exercises 9.3 in PDF format. To do that, you can visit our website any time as per your convenience or download Vedantu app from google play store to access the solutions in PDF format. All these solutions are undoubtedly top quality and can be downloaded at absolutely no cost.

4. Give me a brief idea about the Class 9 Maths Chapter 9 Exercise 9.3?

All the questions in Exercise 9.3 are based on two main theorems that are discussed in Chapter 9 of the Class 9 textbooks. NCERT Solutions for class 9 maths explain these theorems very well. And, to explain these concepts, there are easy to understand NCERT Solutions for Class 9 Maths designed by our experts to various questions of exercise 9.3 that follows the latest CBSE guidelines This exercise consists of 16 questions. Among which, there are 12 short answer type questions along with 4 long answer type questions.

However, the theorems are as follows:

Theorem 9.2: Two triangles on the same base( or equal bases) and between the same parallels are equal in area.

Theorem 9.3: Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.

5. What are the topics that are learnt in  Chapter 9 Exercise 9.3 of Class 9 Maths?

The topics that the students learn from this Exercise 9.3 are the diagonals of parallelograms, triangles and also different geometrical figures. The proofs and the theorems based on these figures are explained in detail. Students will learn about the area and its meaning in detail and the method of calculation.  The exercise also covers the figures having the same base and between the same parallels and their areas will be the same if the required conditions are fulfilled.

6. When can it be said that the triangles are on the common base and are between the same parallels according to Chapter 9 of Class 9 Mths?

If two triangles have a common side and the vertice opposite of the common side lies on a straight line then they are on the common base and are between the same parallels. Also to be known that the triangles which have the common base and are between the same parallels have the areas also equal. These are the basic points to be kept in mind so that while proving the theorems and answering the questions of Exercise 9.3 of Chapter 9 of Class 9 Maths, it will be easy and will be more relevant.

7. When can you say that a Parallelogram and a Triangle are on the same base and are between the same parallels?

A Triangle and a Parallelogram can be on the same base and between the same parallels only if the common side and the vertices of the common side lie on a straight line that is parallel to the common side. If the Triangle and the Parallelogram are on the common base then we can say that the area of a triangle is equal to half the area of the parallelogram. Students should practice more of the theorems so the concept of each geometrical figure is cleared.

8. What can students learn from Chapter 9 Exercise 9.3 of Class 9 Maths?

Class 9 Exercise 9.3 is very important and it is essential to learn each topic accurately and with full focus. Practise has to be done every day in solving the theorems and the numericals.  Practising the derivation and the proofs in Exercise 9.3 of different geometrical figures gives you conceptual clarity which is an important factor to score high in the subject.

Class 9 Maths has a total of 15 chapters.

The chapters are Number Systems, Polynomials, Coordinate Geometry, Linear Equations in two Variables, Introduction of Euclid’s Geometry, Lines and Angles, Triangle, Quadrilaterals, Areas of Parallelograms and Triangles, Circles, Constructions, Heron’s formula, Surface Areas and Volumes. Statistics, Probabality.

9. How can Vedantu’s NCERT solutions help in the preparation of Exercise 9.3 of Chapter 9 of Class 9 Maths?

Exercise 9.3 of Chapter 9 of Class 9 Maths has several theorems and proofs. These questions are of good weightage and have to be learnt very appropriately. Students should not have any misconceptions about this topic. Vedantu will be the best guide and also can give you the thorough practice of the variety of numericals from the chapter. The solutions from Exercise 9.3 of Chapter 9 of Class 9 Maths which contains a maximum number of theorems are given in easy steps which will help the students to have a quick and thorough understanding. Solutions by Vedantu which are available on Vedantu website(vedantu.com) or Vedantu Mobile app, are free of cost.