RBSE Solutions for Class 6 Maths Chapter 14 Perimeter and Area Ex 14.2 is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 14 Perimeter and Area Exercise 14.2.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 6 |

Subject |
Maths |

Chapter |
Chapter 14 |

Chapter Name |
Perimeter and Area |

Exercise |
Ex 14.2 |

Number of Questions |
10 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 6 Maths Chapter 14 Perimeter and Area Ex 14.2

Question 1.

By counting squares, estimate the (RBSESolutions.com) areas of the figures

Solution:

To find the areas of figures given in question we (RBSESolutions.com) count the number of squares filled by figures. Which is the required area we ignore ‘less than half – filled squares’ and count ‘more than half – filled squares’, just as one square. Area of given figures are given in following table

Question 2.

Find the areas of the following figures. What (RBSESolutions.com) do you infer from this?

Solution:

Area of first figure (This is a square) = (side)^{2}

= (4)^{2} = 16 sq. cm

Area of second figure (This is a rectangle) = l × b

= 8 × 2 = 16 sq. cm

Area of third figure (This a rectangle) = l × b

= 16 × 1 sq. cm = 16 sq. cm

**Conclusion :** Different figures can also have same area.

Question 3.

By splitting the following figures into rectangles, find (RBSESolutions.com) their areas. (The measures are given in centimeters)

Solution:

We can split all figures (RBSESolutions.com) into rectangles as below

Question 4.

A room is 10 m long and 8 m wide. How many (RBSESolutions.com) square meters of carpet is required to cover the floor of the room?

Solution:

Required carpet to cover the floor of room = Area of the floor of the room

= l × b = 10 × 8 = 80 sq. m

Thus, 80 sq. m carpet will be required.

Question 5.

Find the area of a square (RBSESolutions.com) frame of side 9 cm.

Solution:

Area of squared frame,

= (side) × (side) = 9 × 9 = 81 sq.m

Thus, Area of frame is 81 sq. cm.

Question 6.

Find the areas of the following rectangles. Which rectangle has (RBSESolutions.com) least area and which one has greatest area ?

(i) l = 2 m, b = 80 cm

(ii) l = 180 cm, b = 70 cm

(iii) l = 200 cm, b = 1 m

(iv) l = 190 cm, b = 1 m

Solution:

(i) l = 2 m = 200 cm

b = 80 cm

Area = l × b = 200 × 80

= 16000 sq. cm

(ii) l = 180 cm

b = 70 cm

Area = l × b = 180 × 70

= 12600 sq. cm

(iii) l = 200 cm

b = 1 m = 100 cm

Area = 1 × b = 200 × 100

= 20000 sq. cm

(iv) l = 190 cm

b = 1 m = 100 cm

Area = I × b = 190 × 100

= 19000 sq.cm

∵ 20000 > 19000 > 16000 > 12600

Thus, (iii) rectangle has greatest and (ii) rectangle has least area.

Question 7.

The area of a rectangular garden 50 m long is 300 sq. m. Find (RBSESolutions.com) the width of the garden.

Solution:

Area of Garden = 300 sq.m

l = 50 m, b = ?

∴ l × b = area

⇒ 50 × b = 300

⇒ b = [latex s=1]\frac { 300 }{ 50 } [/latex] = 6 m

Thus, breadth of garden is 6 m.

Question 8.

Six square flower beds each of side 1 m are dug on a piece of (RBSESolutions.com) land 8 m long 6 m wide. What is the area of the remaining part of the land?

Solution:

Length of land = 8 m and b = 6 m

Area of land = 8 × 6 = 48 sq. m

Area of six squared flower beds each of sides 1 m = (side x side) × Number of flower beds

= (1 × 1) × 6 = 6 sq. m

remaining area = 48 – 6 = 42 sq. m

Thus, Area of remaining land will be 42 sq. m.

Question 9.

What will be the change in the (RBSESolutions.com) area of a rectangle if its

(i) Length and breadth are both doubled?

(ii) Length is tripled and breadth is doubled twice?

Solution:

(i) Let,

Length of rectangle = a and

breadth = b

Then, area = a × b = ab

New length of rectangle = 2a and new breadth = 2b

Then

New area = 2a × 2b = 4ab

= 4 × (ab) = 4 (initial area)

Thus, area of rectangle became 4 times of initial area.

(ii) Let, length of (RBSESolutions.com) rectangle = a,

and breadth = b

Then, area = a × b = ab

New length of rectangle = 3a and

new breadth = 45

Then New area = 3a × 4b = 12 ab

= 12 (ab)

= 12 (initial area)

Thus, area of rectangle became 12 times of initial area.

Question 10.

What will be the change (RBSESolutions.com) in the area of a square if its side is

(i) Halved?

(ii) Doubled?

Solution:

(i) (let) Side of square = a

Then, Area = a × a = a^{2}

If length of side = [latex s=1]\frac { 1 }{ 2 } [/latex] × a = [latex s=1]\frac { a }{ 2 } [/latex]

Then Area = [latex s=1]\frac { a }{ 2 } [/latex] [latex s=1]\frac { a }{ 2 } [/latex] = [latex s=1]\frac { a }{ 2 } [/latex]

= [latex s=1]\frac { 1 }{ 4 } [/latex] a^{2 }= [latex s=1]\frac { 1 }{ 4 } [/latex] (initial area)

Thus, New area, remains one fourth of initial area.

(ii) Side of (RBSESolutions.com) square (let)= a

then. Area = a × a = a^{2}

if length doubled, then side = 2a

and Area = 2a × 2a = 4a^{2}

= 4(a^{2}) = 4 (initial area)

Thus, New area, became four times of initial area.

We hope the RBSE Solutions for Class 6 Maths Chapter 14 Perimeter and Area Ex 14.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 6 Maths Chapter 14 Perimeter and Area Exercise 14.2, drop a comment below and we will get back to you at the earliest.

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