NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles in Hindi and English Medium prepared for the academic session 2022-2023.

**Class 9 Maths Chapter 9 Solutions in English Medium**

Class 9 Maths Exercise 9.1 in English

Class 9 Maths Exercise 9.2 in English

Class 9 Maths Exercise 9.3 in English

Class 9 Maths Exercise 9.4 in English

**Class 9 Maths Chapter 9 Solutions in Hindi Medium**

Class 9 Maths Exercise 9.1 in Hindi

Class 9 Maths Exercise 9.2 in Hindi

Class 9 Maths Exercise 9.3 in Hindi

Class 9 Maths Exercise 9.4 in Hindi

## NCERT Solutions for Class 9 Maths Chapter 9

NCERT Textbook Solutions of Prashnavali 9.1, 9.2, 9.3 and Prashnavali 9.4 of Samantar Chaturbhujon aur Tribhujon ka kshetrfal in Hindi Medium is also available to study online or download in PDF free based on new CBSE Curriculum 2022-2023. All the textbook solutions for 2022-23 and Offline Apps are applicable for CBSE Board, Gujrat Board, Uttarakhand Board, UP Board, MP Board and all other Boards using NCERT (https://ncert.nic.in/) Books 2022-2023 for their course book of the academic year 2022-2023.

### Study Material for Class 9 Maths Chapter 9

Class 9: | Mathematics |

Chapter 9: | Areas of Parallelograms and Triangles |

Content: | NCERT Exercises Solution |

Mode of Content: | Text and Vidoes Format |

Medium: | Hindi and English Medium |

#### 9th Maths Chapter 9 Solutions in English and Hindi Medium

CBSE NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles all exercises are given below for new session 2022-2023. For other questions visit to Class 9 Maths main page. Join the Discussion Forum to share your knowledge and to solve your doubts.

### Important Questions on 9th Maths Chapter 9

### E is any point on median AD of a Î”ABC. Show that ar (ABE) = ar (ACE).

In Î”ABC, AD is median. [âˆµ Given] Hence, ar(ABD) = ar(ACD) â€¦ (1) [âˆµ A median of a triangle divides it into two triangles of equal areas.] Similarly, in Î”EBC, ED is median. [âˆµ Given] Hence, ar(EBD) = ar(ECD) â€¦ (2) Subtracting equation (2) from (1), we get ar(ABD) â€“ ar(EBD) = ar(ACD) â€“ ar(ECD) â‡’ ar(ABE) = ar(ACE)

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

Diagonals of parallelogram bisect each other.

Therefore, PO = OR and SO = OQ

In Î”PQS, PO is median. [âˆµ SO = OQ]

Hence, ar(PSO) = ar(PQO) â€¦ (1) [âˆµ A median of a triangle divides it into two triangles of equal areas.]

Similarly, in Î”PQR, QO is median. [âˆµ PO = OR] Hence, ar(PQO) = ar(QRO) â€¦ (2)

And in Î”QRS, RO is median. [âˆµ SO = OQ]

Hence, ar(QRO) = ar(RSO) â€¦ (3)

From the equations (1), (2) and (3), we get

ar(PSO) = ar(PQO) = ar(QRO) = ar(RSO)

Hence, in parallelogram PQRS, diagonals PR and QS divide it into four triangles in equal area.

### D and E are points on sides AB and AC respectively of Î”ABC such that ar (DBC) = ar (EBC). Prove that DE||BC.

Î”DBC and Î”EBC are on the same base BC and ar(DBC)=ar(EBC).

Therefore, DE || BC [âˆµTriangles on the same base (or equal bases) and having equal areas lie between the same parallels.]

### Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

Triangles ABD and ABC are on the same base AB and between same parallels,

AB || CD.

Hence, ar(ABD) = ar(ABC) [âˆµ Triangles on the same base (or equal bases) and between the same parallels are equal in area.]

Subtracting ar(ABO) form both the sides

ar(ABD) â€“ ar(ABO) = ar(ABC) â€“ ar(ABO)

â‡’ ar(AOD) = ar(BOC)

#### Important Notes on 9th Maths Chapter 9

1. Area of a figure is a number associated with the part of the plane enclosed by that figure.

2. Two congruent figures have equal areas but the converse need not be true.

3. A diagonal of a parallelogram divides it into two triangles of equal areas.

4. Two figures are said to be on the same base and between same parallels, if they have a common base and vertices opposite to the common base of each figure lie on a line parallel to the base.

5. Parallelograms on the same base and between same parallels are equal in area.

6. A parallelogram and a rectangle on the same base and between same parallels are equal in area.

##### Do you know?

1. Parallelograms on the same base and having equal areas lie between the same parallels.

2. Triangles on the same base and between the same parallels are equal in area.

3. If a parallelogram and a triangle are on the same bases and between the same parallels, them area of the triangle is half the area of parallelogram.

4. Area of a triangle is half the product of its base and the corresponding altitude.

5. Triangles with equal bases and equal areas have equal corresponding altitudes.

6. Triangles on the same base and having equal areas lie between the same parallels.

7. A median of a triangle divides it into two triangles of equal areas.

### What are the main points of chapter 9 of class 9th Maths that students should remember at the final exam?

There are 12 main points of chapter 9 of class 9th Maths that students should remember at the time of the final exam:

- Area of a figure is a number (in some unit) associated with the part of the plane enclosed by that figure.
- Two congruent figures have equal areas but the converse need not be true.
- If a planar region formed by a figure T is made up of two non-overlapping planar regions formed by figures P and Q, then ar (T) = ar (P) + ar (Q), where ar (X) denotes the area of figure X.
- Two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices, (or the vertex) opposite to the common base of each figure lie on a line parallel to the base.
- Parallelograms on the same base (or equal bases) and between the same parallels are equal in area.
- Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels.
- Triangles on the same base (or equal bases) and between the same parallels are equal in area.
- Triangles on the same base (or equal bases) and having equal areas lie between the same parallels.
- A median of a triangle divides it into two triangles of equal areas.

### Which exercise of chapter 9 of class 9th Maths is the easiest and has the least number of sums?

In chapter 9 of class 9th Maths, there are four exercises:

- The first exercise (Ex 9.1) has one question with 6 parts.
- The second exercise (Ex 9.2) contains 6 questions and two examples (examples 1, 2).
- The third exercise (Ex 9.3) contains 16 questions and two examples (examples 3, 4).
- The fourth (optional) exercise (Ex 9.4) contains eight questions.
- The first exercise (Ex 9.1) is the easiest exercise of chapter 9 and has the least number of sums.

### Does chapter 9 of grade 9th Maths contain any optional exercise? If yes, which exercise of chapter 9 is optional?

Yes, chapter 9 of grade 9th Maths contains an optional exercise. There are four exercises in chapter 9 of class 9th Maths, and the last exercise (Exercise 9.4) is the optional exercise of chapter 9 of grade 9th Maths.

### Is chapter 9 of class 9th Maths easy or complicated to understand and solve?

Chapter 9 of class 9th Maths is not easy and not complicated. It lies in the middle of easy and difficult because some examples and questions of this chapter are easy, and some are complicated. However, the difficulty level of any topic varies from child to child. Some children find it difficult, some find it easy, and some find it in the middle of easy and difficult.

### How many theorems are there in chapter 9 of class 9th Maths?

There are only three theorems in chapter 9 of class 9th Maths. All the theorems are compulsory to do.