Chapter 9 : Areas of Parallelograms and Triangles - Ncert Solutions for Class 9 Maths CBSE
If you love numbers & have a deep interest in statistics, then Mathematics is probably one of the scoring subject ever. A Class 9th CBSE Mathematics is one of the challenging subject one can undergo as, it involves a lot of crunching of complex numbers, geometrical formulae, diagrams & much more. Hence, to simplify the mathematical complexity, TopperLearning has framed a customise solution that involves Test preparation notes, Textbook solutions, Videos, & several other study material that help the student to memorise the concepts quickly. We have bifurcated our CBSE Class 9 Study Material (Solution) into 3 Different parts namely:
- CBSE Class 9 Practice Test includes Revision notes, Question bank & Sample Paper
- TopperLearning Classroom that includes Videos, Ask the Experts & Preparation Tips
- Text Book Solutions, etc
TopperLearning packages involves all the ingredients of CBSE Class 9 Study Material that includes Solved Question Papers, High Animated Videos, Solutions by SME (Subject Matter Expert), CBSE class 9 Preparation Tips, Update to CBSE 9th Mathematics syllabus, Practice books, Reference Materials, etc that help you to score great marks with glorious marks.
Getting a good score is now relatively easy if you prefer TopperLearning solutions for CBSE Class 9th Mathematics subject. By purchasing our package, you will be accessed to guaranteed success in your examination!
Chapter 9 - Areas of Parallelograms and Triangles Excercise Ex. 9.1
Chapter 9 - Areas of Parallelograms and Triangles Excercise Ex. 9.2
We know that,
Area of parallelogram = Base corresponding attitude
Area of parallelogram ABCD = CD AE = AD CF
16 cm 8 cm = AD 10 cm
In parallelogram ABCD
AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)
AB = CD [Opposite sides of a parallelogram are equal]
Since HEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF.
area (BQC) = area (APB)
In parallelogram ABCD we find that
AB || EF (By construction) ...(1)
ABCD is a parallelogram
AB || EF and AE || BF
So, quadrilateral ABFE is a parallelogram
Now, we may observe that APB and parallelogram AB || FE are lying on the same base AB and between the same
area (PCD) = area (EFCD) ... (4)
Adding equations (3) and (4), we have
In parallelogram ABCD we may observe that
MN || AD (By construction) ...(6)
ABCD is a parallelogram
MN || AD and AM || DN
So, quadrilateral AMND is a parallelogram
area (PCB) = area (MNCB) ... (9)
Adding equations (8) and (9), we have
On comparing equations (5) and (10), we have
Area (APD) + area (PBC) = area (APB) + area (PCD)
and also these are in between same parallel lines SR and PB.
As these lie on the same base and are between same parallel lines AS and BR
Area of PSA + Area of PAQ + Area of QRA = area of PQRS ... (1)
We know that if a parallelogram and triangle are on the same base and between the same parallels, the area of triangle is half the area of the parallelogram.
Area (PSA) + area (QRA) = area (PQRS) ... (3)
Chapter 9 - Areas of Parallelograms and Triangles Excercise Ex. 9.3
Now the line segments BF and DE are joining two parallel lines EF and BD of same length.
So, line segments BF and DE will also be parallel to each other and also these will be equal in length.
Now as each pair of opposite sides are equal in length and parallel to each other.
Therefore BDEF is a parallelogram
We know that diagonal of a parallelogram divides it into two triangles of equal area.
Area (AFE) + area (BDF) + area (CDE) + area (DEF) = area (ABC)
DNO = BMO (By construction)
DON = BOM (Vertically opposite angles)
OD = OB (Given)
By AAS congruence rule
Area (DON) + area (DNC) = area (BOM) + area (BMA)
So, area (DOC) = area (AOB)
Area (DOC) = area (AOB)
Area (DCB) = area (ACB)
Now if two triangles are having same base and equal areas, these will be between same parallels
(DA || CB).
Since, BCE and BCD are lying on a common base BC and also having equal areas, so, BCE and BCD will lie between the same parallel lines.
XY || BC EY || BC
BE || AC BE || CY
So, EBCY is a parallelogram.
It is given that
XY || BC XF || BC
FC || AB FC || XB
Now parallelogram EBCY and parallelogram BCFX are on same base BC and between same parallels BC and EF
These are on same base BE and are between same parallels BE and AC
Area (ABE) = area (ACF)
the same parallels AC and BF
Let it meet the extended side CD of ABCD at point E. Join BE and AD. Let them intersect each other at O. Now portion AOB can be cut from the original field so that new shape of field will be BCE.
Now we have to prove that the area of AOB (portion that was cut so as to construct Health Centre) is equal to the area of the DEO (portion added to the field so as to make the area of new field so formed equal to the area of original field)
CBSE Class 9 Maths Homework Help
Clear all your doubts instantly at our “Ask Doubt” section. Get expert help and guidance at your comfort. To know the syllabus in detail, click here.
NCERT Textbooks are the rich stimulant for those students who want to score better in the CBSE examinations. By solving our papers the students have achieved a better and higher result. One of the primary objectives of creating ncert solution for class 9 is that the student gets access to an easy solution; which acts as a strong catalyst in improving the marks. We usually focus that the students don’t find any difficulty in understanding the concepts and can memorize them easily.
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number