**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.**

To
find: AD

We know that

Area of a parallelogram = Base × Height = DC ×
AE

= 16
× 8 = 128 cm^{2 }…………….(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.**

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 Î”HGF lie on the same base HF and between the same parallels DC and HF.

Therefore, ar (Î”HGF) = ½ ar ( ||

^{gm}HDCF) ……….(ii)

Similarly, ar (Î”HEF) = ½ ar ( ||

^{gm}ABFH) ……….(iii)

On
adding eqs. (ii) and (iii), we get

ar ( Î”HGF) + ar (Î”HEF) = 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 (Î”APB) = ar (Î”BQC).

**Solution.**

To prove: ar (Î”APB) = ar (Î”BQC)

Proof: Here, parallelogram ABCD and Î”BQC lie on the same base BC and between the same parallels BC and AD.

ar (Î”BQC) = ½ ar (||

^{gm}ABCD) …..… (i)

Similarly, Î”APB and parallelogram ABCD lie on the same base AB and between the same parallels AB and DC.

∴ ar (Î”APB) = ½ ar (||

^{gm}ABCD) …..… (ii)

From eqs. (i) and (ii), we get

ar (Î”APB) = ar (Î”BQC)

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 (Î”APB) + ar (Î”PCD) = ½ ar (||

^{gm}ABCD)

**(ii)**ar (Î”APD) + ar (Î”PBC) = ar (Î”APB) + ar (Î”PCD)

[

**Hint:**Through P, draw a line parallel to AB]

Given: ABCD is a parallelogram, i.e., AB || DC and AD || BC

To prove:

**(i)**ar (Î”APB) + ar (Î”PCD) = ½ ar (||

^{gm}ABCD)

**(ii)**ar (Î”APD) + ar (Î”PBC) = ar (Î”APB) + ar (Î”PCD)

**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 Î”APB and parallelogram ABRM lie on the same base
AB and between the same parallels AB and MR.

∴ ar (Î”APB) = ½ ar (||^{gm} ABRM) …………….(i)

and ar (Î”PCD) = ½ ar (||^{gm} MRCD) ……………..(ii)

Adding eqs. (i) and (ii), we get

ar
(Î”APB) + ar (Î”PCD) = ½ ar (||^{gm}
ABRM) + ½ ar (||^{gm} MRCD)

ar (Î”APB) + ar (Î”PCD) = ½ ar (||^{gm} ABCD)
……………(iii)

**(ii)** Clearly, ar (||^{gm} ABCD) = ar (Î”APD) + ar (Î”PBC) + ar (Î”APB) + ar (Î”PCD)

= ar (Î”APD) + ar (Î”PBC) + ½ ar (||^{gm} ABCD)
[From eq. (iii)]

∴ ar (Î”APD) + ar (Î”PBC) = ar (||^{gm} ABCD) – ½ ar (||^{gm} ABCD)

⇒ ar (Î”APD) + ar (Î”PBC) = ar (||^{gm} ABCD) ………..… (iv)

From eqs. (iii) and (iv), we get

ar (Î”APD) + ar (Î”PBC) = ar (Î”APB) + ar (Î”PCD)

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 (Î”AXS) = 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 (Î”AXS) = 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, Î”AXS and parallelogram ABRS lie on the same base AS and between
the same parallels AS and BR.

∴ ar (Î”AXS) = ½ ar (||^{gm} ABRS)
…….… (ii)

From eqs. (i) and (ii), we get

ar (Î”AXS) = 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.**

Since, the Î”APQ and parallelogram PQRS lie on the same base PQ and between the same parallels PQ and SR.

∴ ar (Î”APQ) = ½ ar (||^{gm} PQRS) ……..…(i)

Then, remaining

ar (Î”ASP) + ar (Î”ARQ) = ½ ar (||^{gm} PQRS)

Now, from eqs. (i) and (ii), we get

ar (Î”APQ) = ar (Î”ASP) + ar (Î”ARQ)

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).

**NCERT Solutions for Maths Class 10**