NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 9 Areas of Parallelograms and Triangles Ex 9.2.



Ex 9.2 Class 9 Maths Question 1.
In the given figure, ABCD is a parallelogram, AE DC and CF AD. If AB = 16 cm,
AE = 8 cm and CF = 10 cm, find AD.

Solution.

Given: AB = 16 cm, AE = 8 cm and CF = 10 cm

To find: AD
We know that
Area of a parallelogram = Base × Height = DC × AE
                                           = 16 × 8 = 128 cm2           …………….(i)

[  opposite sides of a parallelogram are equal, i.e., AB = DC = 16 cm]
Area of a parallelogram = AD × CF
                                           = AD × 10                        ……………(ii)
From Eqs. (i) and (ii), we have
AD × 10 = 128
AD = 128/10 = 12.8 cm
Hence, the length of AD is 12.8 cm.

 

Ex 9.2 Class 9 Maths Question 2.
If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = ½ ar (||gm ABCD).

Solution.

Given: E, F, G and H are respectively mid-points of the sides AB, BC, CD and AD.
To prove: ar (||gm EFGH) = ½ ar (||gm ABCD)
Construction: Join HF.
Proof: Since H and F are mid-points of AD and BC, respectively.
DH = ½ AD and CF = ½ BC              ……..(i)
Now, as ABCD is a parallelogram.
  AD = BC and AD || BC
½ AD = ½ BC and DH || CF
DH = CF and DH || CF
So, HDCF is a parallelogram. [
a pair of opposite sides are equal and parallel]
Now, as the parallelogram HDCF and ΔHGF lie on the same base HF and between the same parallels DC and HF.
Therefore, ar (ΔHGF) = ½ ar ( ||gm  HDCF)        ……….(ii)
Similarly, ar (ΔHEF) = ½ ar ( ||gm ABFH)             ……….(iii)

On adding eqs. (ii) and (iii), we get
ar ( ΔHGF) + ar (ΔHEF) = ½ [ar (||gm HDCF) + ar (||gm ABFH)]
ar (EFGH) = ½ ar( ||gm ABCD)
Hence Proved.

 

Ex 9.2 Class 9 Maths Question 3.
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (ΔAPB) = ar (ΔBQC).

Solution.

Given: In parallelogram ABCD, P and Q are any two points lying on the sides DC and AD, respectively.
To prove: ar (ΔAPB) = ar (ΔBQC)
Proof: Here, parallelogram ABCD and ΔBQC lie on the same base BC and between the same parallels BC and AD.
ar (
ΔBQC) = ½ ar (||gm ABCD)        …..… (i)
Similarly, 
ΔAPB and parallelogram ABCD lie on the same base AB and between the same parallels AB and DC.
ar (ΔAPB) = ½ ar (||gm ABCD)     …..… (ii)
From eqs. (i) and (ii), we get
ar (ΔAPB) = ar (ΔBQC)

Hence Proved.

 

Ex 9.2 Class 9 Maths Question 4.
In the figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar (
ΔAPB) + ar (ΔPCD) = ½ ar (||gm ABCD)
(ii) ar (
ΔAPD) + ar (ΔPBC) = ar (ΔAPB) + ar (ΔPCD)
[Hint: Through P, draw a line parallel to AB]


Solution.

Given: ABCD is a parallelogram, i.e., AB || DC and AD || BC
To prove:
(i) ar (
ΔAPB) + ar (ΔPCD) = ½ ar (||gm ABCD)
(ii) ar (
ΔAPD) + ar (ΔPBC) = ar (ΔAPB) + ar (ΔPCD)

Proof:

(i) Through the point P, draw MR parallel to AB.
MR || AB and AM || BR           [ AD || BC]
So, ABRM is a parallelogram.
Now, as 
ΔAPB and parallelogram ABRM lie on the same base AB and between the same parallels AB and MR.
ar (ΔAPB) = ½ ar (||gm ABRM)                        …………….(i)
and ar (
ΔPCD) = ½ ar (||gm MRCD)                   ……………..(ii)
Adding eqs. (i) and (ii), we get

ar (ΔAPB) + ar (ΔPCD) = ½ ar (||gm ABRM) + ½ ar (||gm MRCD)
ar (
ΔAPB) + ar (ΔPCD) = ½ ar (||gm ABCD)        ……………(iii)

(ii) Clearly, ar (||gm ABCD) = ar (ΔAPD) + ar (ΔPBC) + ar (ΔAPB) + ar (ΔPCD)
                                              = ar (
ΔAPD) + ar (ΔPBC) + ½ ar (||gm ABCD)     [From eq. (iii)]
ar (ΔAPD) + ar (ΔPBC) = ar (||gm ABCD) – ½ ar (||gm ABCD)
ar (ΔAPD) + ar (ΔPBC) = ar (||gm ABCD)       ………..… (iv)
From eqs. (iii) and (iv), we get
ar (
ΔAPD) + ar (ΔPBC) = ar (ΔAPB) + ar (ΔPCD)

Hence proved.

 

Ex 9.2 Class 9 Maths Question 5.
In the figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (||gm PQRS) = ar (||gm ABRS)
(ii) ar (
ΔAXS) = ½ ar (||gm PQRS)


Solution.
Given: PQRS and ABRS both are parallelograms and X is any point on BR.
To prove:

(i) ar (||gm PQRS) = ar (||gm ABRS)
(ii) ar (
ΔAXS) = ½ ar (||gm PQRS)

 

Proof:
(i)
 Here, parallelograms PQRS and ABRS lie on the same base SR and between the same parallels SR and PB.
ar (||gm PQRS) = ar (||gm ABRS)         ……..… (i)

(ii) Again, ΔAXS and parallelogram ABRS lie on the same base AS and between the same parallels AS and BR.
 ar (ΔAXS) = ½ ar (||gm ABRS)              …….… (ii)
From eqs. (i) and (ii), we get
ar (
ΔAXS) = ½ ar (||gm PQRS)
Hence proved.

 

Ex 9.2 Class 9 Maths Question 6.
A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

Solution.

Given, PQRS is a parallelogram and A is any point on RS. Now, join PA and PQ. Thus, the field will be divided into three parts and each part is in the shape of triangle.
Since, the ΔAPQ and parallelogram PQRS lie on the same base PQ and between the same parallels PQ and SR.

ar (ΔAPQ) = ½ ar (||gm PQRS)           ……..…(i)
Then, remaining
ar (ΔASP) + ar (ΔARQ) = ½ ar (||gm PQRS)
Now, from eqs. (i) and (ii), we get
ar (ΔAPQ) = ar (ΔASP) + ar (ΔARQ)
So, farmer has two options.
Thus, the farmer can sow wheat in (∆PAQ) and pulses in [(∆APS) + (∆AQR)] or wheat in [(∆APS) + (∆AQR)] and pulses in (∆PAQ).


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