# Class9 Maths Ncert Exemplar with solution.ch13(Mensuration)

Hello students this is the full NCERT exemplar questions and answer with full details.

So you can download the pdf any time .

Class 9 Maths Ncert solution Exemplar will help you to give a overview of the chapter.

It will help a lot in the exams.

#### Perimeter of a closed figure is the distance covered in one round

along the boundary of the figure.

called a

• A closed figure in which all sides and angles are equal

regular polygon.

#### • Perimeter of a rectangle = 2 x (length + breadth)

#### • Perimeter of a square = 4x length of its side • Perimeter of an equilateral triangle = 3 x length of a side

#### The amount of region enclosed by a plane closed figure is called its area.

#### • Area of a rectangle = length x breadth

#### • Area of a square= side x side

#### (B) Solved Examples

Example 1:

#### Choose the correct answer from the given four options:

In Fig. 6.1, a square of side

1 em is joined to a square of

side 3 cm. The perimeter of the

new figure is

(A) 13cm

(B) 14cm

(C) 15cm

(D) 16cm

Solution:

Correct answer is (B)

Fig. 6.1

11.4.2018

UNIT-6

Example 2:

#### Which of the following statements are true or false? (a) Geeta wants to raise a boundary wall around her

house. For this, she must find the area of the land of

her house.

(b) A person preparing a track to conduct sports must

find the perimeter of the sports ground. (a) False Boundry wall is around her house, so she must know the perimeter of the land and not the

Solution:

area. (b) True - Track is prepared along the boundary of the sports ground. àª›ે.

#### Example 3:

#### Fill in the blanks to make the statements true:

(a) Perimeter of a triangle with sides 4.5cm, 6.02cm and 5.38cm is. OUીàª®ાં

(b)

Area of a square of side 5cm

Solution:

(a) 15.9cm.

(b) 25sqcm.

#### Example 4:

#### Bhavna runs 10 times around a square field of side

80m. Her sister Sushmita runs 8 times around a rectangular field with length 150m and breadth 60m.

Who covers more distance? By how much? of the square field

Solution:

Distance covered by Bhavana in one round Perimeter

4 x side of square field = 4 x 80m = 320m

Distance covered in 10 rounds (320 x 10)m = 3200m not Distance covered by Sushmita in one round

= Perimeter of the rectangular field

= 2 x (length + breadth). = 2 x (150+60)m= 2 x 210 = 420m

Distance covered in 8 rounds 420 x 8 = 3360m

Distance covered in 8 rounds = 420 × 8 = 3360m

Hence, Sushmita has covered 160m more than the

distance coverd by Bhavna.

#### Example 5: The length of a rectangular field is thrice its breadth. If the eperimeter of this field is 800m, what is the length of the field?

#### Solution: Perimeter of a rectangle = 2 (length + breadth)

Length of the rectangular field = 3 × breadth

Therefore perimeter of field = 2 (3 × breadth +

breadth)

= 2 ( 4 × breadth)

= 8 × breadth

The given perimeter = 800m

Therefore 8 × breadth = 800

Or, breadth = 800 ÷ 8 = 100m

So, length = 3 × 100m = 300m

#### Example 6: Cost of fencing around a square field is Rs. 12000. If the cost of fencing per metre is Rs. 30, find the area of the square field.

Solution: Cost of fencing per metre = Rs 30

Total cost of fencing = Rs 12000

So, the length of fencing (perimeter) =

Total cost

Cost per metre

= = 400m

Now, length of fencing = Perimeter of the square field

= 4 × side of the field

Therefore, 4 × side of the field = 400m

or, side of the field = m = 100m

So, area of the field = 100m × 100m

= 10000sq m.

#### Example 7: Sabina wants to cover the floor of her room whose length is 4 m and breadth is 3m by square tiles. If each square tile is of side 20cm, then find the number of tiles required to cover the floor of her room.

Solution: Length of the room = 4m = 400cm

Breadth of the room = 3m = 300cm

Area of the floor of the room = Length × Breadth

400 × 300sqcm = 120000sqcm

Side of the square tile = 20cm

Area of the square tile = Side × Side = 20 × 20sqcm

= 400sqcm

So, number of tiles required = =

#### Example 8: By splitting the figure into rectangles, find its area.

Solution: By splitting the figure into four rectangles, we get

Fig. 6.3

Area of the figure = Area AJIY + Area YWCB

+ Area DWUE + Area FUHG.

Area A J I Y = A J × J I = 3 × 3 = 9

Now, B Y = A B – Y A = 4 – 3 = 1

So, Area Y W C B = B Y × B C = 1 × 2 = 2

Next, D W = D C + C W = 2 + 1 = 3

Therefore, area D W U E = D W × D E = 3 × 3 = 9

Similarly, U H = I H – I U = 4 – 2 = 2

G H = F U and F U = E U + F E

= D W + F E = 3 + 1= 4

Area F U H G = U H × G H = 2 × 4 = 8

Therefore, the area of the figure = 9 + 2 + 9 + 8

= 28sq unit

EXERCISE

#### In questions 1 to 6, out of the four options only one is correct. Write the correct answer.

1. Following figures are formed by joining six unit squares. Which figure

has the smallest perimeter in Fig. 6.4?

(i) (ii) (iii) (iv)

2)A square shaped park ABCD of side 100m has

two equal rectangular flower beds each of size

10m × 5m (Fig. 6.5). Length of the boundary of

the remaining park is

(A) 360m (B) 400m

(C) 340m (D) 460m

3. The side of a square is 10cm. How many times will the new perimeter

become if the side of the square is doubled?

(A) 2 times (B) 4 times (C) 6 times (D) 8 times

4. Length and breadth of a rectangular sheet of paper are 20cm and

10cm, respectively. A rectangular piece is cut from the sheet as shown

in Fig. 6.6. Which of the following statements is correct for the

remaining sheet?

10cm

20cm

5cm

2cm

Fig. 6.6

(A) Perimeter remains same but area changes.

Thanks for your valuable time.

You can check our MCQ sets.

Thanks.

By Sahil Patel

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