RBSE Solutions for Class 6 Maths Chapter 14 Perimeter and Area Ex 14.3 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 14.3.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 6 |

Subject |
Maths |

Chapter |
Chapter 14 |

Chapter Name |
Perimeter and Area |

Exercise |
Ex 14.3 |

Number of Questions |
6 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 6 Maths Chapter 14 Perimeter and Area Ex 14.3

Question 1.

Find the areas and perimeters of the following (RBSESolutions.com) rectangular figures. Which of them have same perimeter but different area ?

Solution:

(i) l = 40 cm, b = 15 cm

Perimeter = 2 (l + b)

= 2 (40 + 15) = 2 (55) = 110 cm

Area = l × b = 40 × 15 = 600 sq. cm

(i) l= 45 cm, b = 20 cm

Perimeter = 2 (l + b)

= 2 (45 + 20) = 2 × 65 = 130 cm

Area = l × b = 45 × 20 = 900 sq. cm

(iii) l = 8 cm, b = 7 cm

Perimeter = 2 (l + b)

= 2 (8 + 7) = 2 (15) = 30 cm

Area = l × b

= 8 × 7 = 56 sq. cm

(iv) l = 10 cm, b = 5 cm

Perimeter = 2 (l + b)

= 2 (10 + 5) = 2 × 15 = 30 cm

Area = l × b

= 10 × 5 = 50 sq. cm

Rectangles with (RBSESolutions.com) same perimeter = (iii) and (iv)

No. rectangle is same in area Thus, rectangle (iii) and (iv) have same perimeter and different area.

Question 2.

Gopi has a square field of side 75 m. Narayan has a rectangular field of length 85 m. If the perimeters of both the fields are same, whose field has greater area and by how much?

Solution:

Side of squared field of Gopi = 75 m

and area = 75 × 75 = 5625 sq. m

and perimeter = 4 × side = 4 × 75 = 300 m

Length of (RBSESolutions.com) rectangular field of Narayan 85 m

and b = a m (let)

then Area = 85 × a = 85a sq. m

and Perimeter= 2 (l + b) = 2 (85 + a) m

According to question, both perimeter are equal

∴ 2 (85 + a) = 300

⇒ (85 + a) = [latex s=1]\frac { 300 }{ 2 } [/latex] = 150

⇒ a = 150 – 85 = 65

Thus, breadth of field = 65 m

then, area = 85 × 65 = 5525 m

∵ 5625 > 5525 and 5625 – 5525 = 100

∴ Area of Gopi’s field is more by 100 sq. m

Question 3.

Area of a square is 65 sq. cm. Perimeter of a (RBSESolutions.com) rectangle is equal to the perimeter of this square. Find the length of the rectangle if its breadth is 6.5 cm. Which figure has greater area?

Solution:

Area of square = 64 sq.m

∴ Area = side × side = (side)^{2}

⇒ Side = √area = √64 = √8^{2} = 8

Thus, side is 8 cm.

then, its perimeter= 4 × side = 4 × 8 = 32 cm

breadth of rectangle = 6.5 cm and l = a cm

then perimeter = 2 (6.5 + a) cm

According to (RBSESolutions.com) question, both perimeter are equal

∴ 2 (6.5 + a) = 32

⇒ 6.5 + a = 16

⇒ a = 16 – 6.5 = 9.5 cm

Area = l × b = 9.5 × 6.5 = 61.75 sq. cm

∵ 64 > 61.75

Thus, length of rectangle = 9.5 cm

Area of square will be more by 2.25 sq. cm.

Question 4.

A rectangular piece of 20 cm × 15 cm is cut off from a bigger (RBSESolutions.com) rectangle as shown in the figures below. In each case, find the difference is perimeter before and after the piece is cut off.

Solution:

(i) Perimeter of (RBSESolutions.com) square (when rectangular piece cut off) = 80 + 80 + 30 + 15 + 20 + 15 + 30 + 80 = 350 cm

Perimeter of complete square = 4 (80) = 320 cm

∵ 350 > 320

∴ Perimeter will be increased.

Thus, increment in perimeter = 350 – 320 = 30 cm

(ii) Initial perimeter of square = 4 (60) = 240 cm

Perimeter, after piece cut off = 60 + 60 + 45 + 20 + 15 + 40

= 60 + 60 + (45 + 15) + (20 + 40) = 60 + 60 + 60 + 60 = 240

∵ 240 = 240

Thus, no change in perimeter of square.

Question 5.

On a centimeter squared paper, make (RBSESolutions.com) as many rectangle as you can, such that the area of the rectangle is 64 sq.cm (consider only natural number lengths).

(i) Which rectangle has the greatest perimeter?

(ii) Which rectangle has the least perimeter?

(iii) Find the change in the width of rectangle with decreasing perimeter of the rectangle.

Solution:

Following be the length and breadth of rectangle to find 64 sq. cm.

**Area:**

(a) Perimeter = 2 (64 + 1 ) = 2 (65) = 130 cm

(b) Perimeter = 2 (32 + 2) = 2 (34) = 68 cm

(c) Perimeter = 2 (16 + 4) = 2 (20) = 40 cm

(d) Perimeter = 2 (8 + 8) = 2 (16) = 32 cm

- Perimeter of rectangle (a) is greatest (130 cm) whose length is 64 cm and breadth is 1
- Perimeter of rectangle (d) is least (32 cm). Whose length is 8 and breadth is 8 cm. Its also called square.
- Perimeter decreased as breadth of rectangle increased.

Question 6.

On a centimeter squared paper, make as (RBSESolutions.com) many rectangles as you can, such that the perimeter of the rectangle is 16 cm (consider only natural number lengths).

(i) Which rectangle has the greatest area?

(ii) Which rectangle has the least area?

(iii) Find the change in length of rectangle with increasing area of the rectangle.

Solution:

Following be the length and breadth of (RBSESolutions.com) rectangle to find 16 cm perimeter

(a) Area = l × b = 7 × 1 = 7 sq. cm

(b) Area = l × b = 6 × 2 = 12 sq. cm

(c) Area = l × b = 5 × 3 = 15 sq. cm

(d) Area = l × b = 4 × 4 = 16 sq. cm

- Area of rectangle (d) is greatest (16 sq. cm) whose (RBSESolutions.com) length is 4 cm and breadth is 4 cm. Its also called square.
- Area of rectangle (a) is least (7 sq. cm), whose length is 7 cm and breadth is 1 cm.
- Area increased as length of rectangle decreased.

We hope the RBSE Solutions for Class 6 Maths Chapter 14 Perimeter and Area Ex 14.3 will help you. If you have any query regarding Rajasthan Board RBSE Class 6 Maths Chapter 14 Perimeter and Area Exercise 14.3, drop a comment below and we will get back to you at the earliest.

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