NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2.

- Areas of Parallelograms and Triangles Class 9 Ex 9.1
- Areas of Parallelograms and Triangles Class 9 Ex 9.2
- Areas of Parallelograms and Triangles Class 9 Ex 9.3
- Areas of Parallelograms and Triangles Class 9 Ex 9.4

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 9 |

Chapter Name |
Areas of Parallelograms and Triangles |

Exercise |
Ex 9.2 |

Number of Questions Solved |
6 |

Category |
NCERT Solutions |

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

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, D E

We know that

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

[ ∵ opposite sides of a parallelogram are equal i.e., AB = DC = 16 cm]

= 16 x 8 = 128 cm^{2}…………….(i)

Area of a parallelogram = AD x CF

= AD x 10 …(ii)

From Eqs. (i) and (ii), we have

AD x 10 = 128

⇒ AD == 12.8 cm.

Hence, the value of AD is 12.8 cm.

Question 2.

If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (||^{gm} EFGH) = 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) = (||^{gm }ABCD)

Construction : Join HF.

Proof : Since, H and F are mid-points of AD and BC, respectively.

∴ DH = AD and CF = BC ……..(i)

Now, as ABCD is a parallelogram.

∴ AD = DC and AD || BC

AD = 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 Δ HGF stand on the same base HF and lie between the same parallel lines DC and HF.

Similarly, ar (Δ HGF) = ar ( ||^{gm} HBFH)……….(ii)

similarly, ar (Δ HGF) = ar ( ||^{gm} ABFH)……….(iii)

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

ar ( Δ HGF) + ar (Δ HEF) = i [ar (||^{gm} HDCF) + ar (||^{gm} ABFH)]

⇒ ar (EFGH) = ar( ||^{gm} ABCD)

Hence Proved

Question 3.

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (Δ APB) ar (Δ BQC).

Solution.

Given: In parallelogram ABCD, P and Q are any two points lying on the sides DC, and AD, respectively.

To prove : ar (Δ APB) = ar (Δ BQC)

Proof : Here, parallelogram ABCD and ABQC stand on the same base BC and lie between the same parallel BC and AD.

ar(ΔBQC) = ar (||^{gm} ABCD) … (i)

Similarly, Δ APB and parallelogram ABCD stand on the same base AB and lie between the same parallel AB and CD.

∴ ar(Δ APB) = ar (||^{gm} ABCD) … (ii)

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

ar(Δ APB) = ar(Δ BQC)

Hence Proved.

Question 4.

In the figure, P is a point in the interior of a parallelogram ABCD. Show that

(i) ar (Δ APB) + ar(Δ PCD) = ar(||^{gm} ABCD)

(ii) ar (Δ APD) + ar (Δ APBC) = ar (Δ APB) + ar(Δ PCD).

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

Solution.

Given : ABCD is a parallelogram.

i.e., AB || CD and AD || BC

To prove :

(i) ar (Δ APB) + ar (Δ PCD) = ar(||^{gm} ABCD)

(ii) ar(Δ APD) + ar (Δ PBC) = ar (Δ APB) + ar (Δ 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 Δ APB and parallelogram ABRM are on the same base AB and between the same parallels AB and MR.

∴ ar (Δ APB) = ar (||^{gm} ABRM)

ar(Δ PCD) = ar (||^{gm} MRCD)

Now, ar (ΔAPB) + ar (ΔPCD) = ar (IP ABRM) + ar (||^{gm} MRCD)

= ar (||^{gm} ABCD) …(i)

**(ii)** Clearly, ar (||^{gm} ABCD) = ar (ΔAPD) +ar (ΔPBC) + ar (Δ APB) + ar(Δ PCD)

= ar (ΔAPD) + ar (ΔPBC) + ar (||^{gm} ABCD) [from eq. (i)]

∴ ar(Δ APD) + ar (Δ PBC) = ar (||^{gm} ABCD) – ar (||^{gm} ABCD)

ar (||^{gm} ABCD)()

⇒ ar(Δ APD) + ar(Δ PBC) = – ar (||^{gm} ABCD) … (ii)

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

ar (Δ APD) + ar (Δ PBC) = ar (Δ APB) + ar (Δ PCD) Hence proved.

Question 5.

In the figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that

(i) ar (||^{9m} PQRS) = ar (||^{gm} ABRS)

(ii) ar (Δ AXS) = 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 (Δ AXS) = ar (||^{gm} PQRS)

Proof:

**(i)** Here, parallelograms PQRS and ABRS lie on the same base SR and between the same parallel lines SR and

∴ ar (||^{gm} PQRS) = ar (||^{gm} ABRS) … (i)

**(ii)** Again, Δ AXS and parallelogram ABRS lie on the same base AS and between the same parallel lines AS and

∴ ar (Δ AXS) = ar (||^{gm} ABRS) … (ii)

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

ar (Δ AXS) = ar (||^{gm} PQRS)

Hence proved.

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 fields 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 fie divided into three parts and each part is in the shape of triangle.

Since, the Δ APQ and parallelogram PQRS lie on the same base PQ and between same parallel lines PQ and SR.

ar(Δ APQ) = ar( ||^{gm} PQRS) …(i)

Then, remaining

ar(Δ ASP) + ar(Δ ARQ) = ar(||^{gm} PQRS)

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

ar(Δ APQ) = ar(Δ ASP) + ar(Δ ARQ)

So, farmer has two options.

Either the farmer should sow wheat and pulses in Δ APS and Δ AQR or in ar [Δ AQP and (Δ APS and Δ AQR)] separately.

We hope the NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths 9 Areas of Parallelograms and Triangles Ex 9.2, drop a comment below and we will get back to you at the earliest.

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