NCERT Solutions for Class 9 Maths Chapter 9
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.1, 9.2, 9.3 and Exercise 9.4 in English Medium prepared and updated for the academic session 2021-2022.
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 2021-2022. All the textbook solutions for 2021-22 and Offline Apps are applicable for CBSE Board, Gujrat Board, Uttarakhand Board, UP Board, MP Board and all other Boards using NCERT Books 2021-2022 for their course book of the academic year 2021-2022.NCERT Solutions for Class 9 Maths Chapter 9
Class 9th Maths Exercise 9.1 Solutions
Class 9th Maths Exercise 9.2 Solutions
Class 9th Maths Exercise 9.3 Solutions
Class 9th Maths Exercise 9.4 Solutions
Study Material for for 9th Maths 2021-22
| Class 9: | Mathematics |
| Chapter 9: | Areas of Parallelograms and Triangles |
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 2021-2022. For other questions visit to Class 9 Maths main page. Join the Discussion Forum to share your knowledge and to solve your doubts.
Class 9 Maths Exercise 9.1 and 9.2 Explanation in Videos
Class 9 Maths Exercise 9.3 and 9.4 Explanation in Videos
Class 9 Maths Exercise 9.4 Question 8 Solution
Class 9 Maths Exercise 9.1 and 9.2 Solutions in Video
Class 9 Maths Exercise 9.3 and 9.4 Solutions in Video
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.
Important Questions on 9th Maths Chapter 9
E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE).
In ΔABC, AD is median. [∵ Given] Hence, ar(ABD) = ar(ACD) … (1) [∵ A median of a triangle divides it into two triangles of equal areas.] Similarly, in ΔEBC, ED is median. [∵ Given] Hence, ar(EBD) = ar(ECD) … (2) Subtracting equation (2) from (1), we get ar(ABD) – ar(EBD) = ar(ACD) – ar(ECD) ⇒ 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 ΔPQS, PO is median. [∵ SO = OQ]
Hence, ar(PSO) = ar(PQO) … (1) [∵ A median of a triangle divides it into two triangles of equal areas.]
Similarly, in ΔPQR, QO is median. [∵ PO = OR] Hence, ar(PQO) = ar(QRO) … (2)
And in ΔQRS, RO is median. [∵ SO = OQ]
Hence, ar(QRO) = ar(RSO) … (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 ΔABC such that ar (DBC) = ar (EBC). Prove that DE||BC.
ΔDBC and ΔEBC are on the same base BC and ar(DBC)=ar(EBC).
Therefore, DE || BC [∵Triangles 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) [∵ 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) – ar(ABO) = ar(ABC) – ar(ABO)
⇒ ar(AOD) = ar(BOC)
