Class 9 NCERT Solutions Maths Chapter 9 - Areas of Parallelograms and Triangles
Explore the NCERT Solutions for CBSE Class 9 Mathematics Chapter 9 Areas of Parallelograms and Triangles at TopperLearning for Maths practice. These solutions are carefully written by some of the respected Maths experts in the country to assist you with revision of concepts. This chapter gives you complete solutions for questions based on calculating the area of a parallelogram or a triangle according to the given data.
Learn to use constructions such as line segments for finding correct answers in your Maths exam. Our NCERT textbook solutions for CBSE Class 9 Maths will give you clarity on how to go about the constructions according to the given Maths question. In addition, attempt chapter-related questions in our online practice tests and sample question papers for more practice.
Areas of Parallelograms and Triangles Exercise Ex. 9.1
Areas of Parallelograms and Triangles Exercise 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)
Areas of Parallelograms and Triangles Exercise 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)