RBSE Solutions for Class 6 Maths Chapter 14 Perimeter and Area Additional Questions

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

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 6
Subject	Maths
Chapter	Chapter 14
Chapter Name	Perimeter and Area
Exercise	Additional Questions
Number of Questions	18
Category	RBSE Solutions

Rajasthan Board RBSE Class 6 Maths Chapter 14 Perimeter and Area Additional Questions

Multiple Choice Questions

Question 1.
Length of a rectangular park is L and (RBSESolutions.com) Breadth is B, then its area will be
(i) 2 × (L + B)
(ii) 2 × (L – B)
(iii) 2 × (L ÷ B)
(iv) L × B

Question 2.
Distance covered in a complete round of a squared field will be
(i) 4 × side
(ii) (side)²
(iii) side ÷ 4
(iv) 4 side² l

Question 3.
Length of a rectangular lawn is 25 m and (RBSESolutions.com) breadth is 15 m, then its area will be
(i) 80 sq. cm
(ii) 375 sq. cm
(iii) 40 sq. cm
(iv) 100 sq. cm

Question 4.
Side of a square figure is 15 cm, then its area will be
(i) 225 sq. cm
(ii) 60 sq. cm
(iii) 75 sq. cm
(iv) 100 sq. cm

Question 5.
Length of a rectangular field is 10 m and breadth is 8 m. We (RBSESolutions.com) have to fence this field with wire, then total length of required wire will be
(i) 80 m
(ii) 160 m
(iii) 36 m
(iv) 18 m

Question 6.
Sides of a triangle are 10 cm, 8 cm and 5 cm, then perimeter of triangle will be
(i) 40.5 cm
(ii) 23 cm
(iii) 22 cm
(iv) 20 cm

Question 7.
Perimeter of (RBSESolutions.com) square
(i) 4 × (length of side)
(ii) 2 + (l + b)
(iii) 2 × (l + b)
(iv) Product of four sides

Answers:
1. (iv)
2. (i)
3. (ii)
4. (i)
5. (iii)
6. (ii)
7. (i)

Question 1.
Fill in the blanks :
(i) Area of square is …………….. .
(ii) Perimeter of (RBSESolutions.com) equilateral triangle is 3x …………….. length of
(iii) A figure which has all sides and angles equal, is called …………….. .
(iv) Area of rectangle is …………….. .
Answer:
(i) (Side)²
(ii) Sides
(iii) Closed regular figure
(iv) l × b

Very Short Answer Type Questions

Question 1.
What is Area?
Solution:
The amount of surface (RBSESolutions.com) enclosed by a closed figure is called its area.

Question 2.
What are regular figures?
Solution:
Such figures, whose all sides and angles are equal are called regular figures.

Question 3.
What is Perimeter?
Solution:
Perimeter is the distance covered along the boundary forming a closed figure when we go round the figure once.

Question 4.
Find the perimeter of a regular pentagon, (RBSESolutions.com) whose each side is of length 3 cm.
Solution:
Regular pentagon has 5 sides
each side of this is of length 3 cm.
Perimeter of regular pentagon = 5 × 3 cm = 15 cm.

Short/Long Answer Type Questions

Question 1.
Find the area of triangle given in figure.
Solution:
We can find area of triangle in figure, by (RBSESolutions.com) counting the boxes under it.

Number of complete boxes = 27
Number of half or more than half boxes = 9
∴ Area of triangle = 27 + 9 = 36 boxes or 36 sq. unit

Question 2.
Length and breadth of a farmer, field is 240 m and 180 m respectively. He want of fence his (RBSESolutions.com) field thrice by a rope. What is the total length of rope he must use?
Solution:
The farmer has to cover 3 times of perimeter of his field with rope. Therefore, required length of rope will b thrice of perimeter of field.
Perimeter of field = 2 × (l + b)
= 2 × (240 m + 180 m)
= 2 × 420 m = 840 m
∴ Total length of rope required = 3 × 840 m = 2520 m

Question 3.
Find the area of an iron rod in sq. m, whose (RBSESolutions.com) length is 2 m 30 cm and breadth is 1 m 20 cm.
Solution:
Length of iron rod = 2 m 30 cm = 2.30 m
Breadth of iron rod = 1 m 20 cm = 1.20 m
Area of iron rod = l × b = 2.30 × 1.20 = 2.76 sq.m

Question 4.
Length and breadth of a rectangular field is 0.7 km and 0.5 km respectively. We have to fence this field with a wire in 4 rows. What is the total length of wire we must use?
Solution:
Perimeter (RBSESolutions.com) of field = 2 × (l + b)
= 2 × (0.7 km + 0.5 km)
= 2 × 1.2 km = 2.4 km
Required length of wire will be 4 times of the Perimeter of field.
Thus, length of wire = 4 × Perimeter
= 4 × (2.4 km) = 9.6 km

Question 5.
Length of a piece of string is 30 cm. What will be the length of each side, if we made by string
(a) a square
(b) an equilateral triangle
Solution:
(a) Length (RBSESolutions.com) of string = Perimeter = 30 cm
∵ A square has 4 equal sides.

(b) Length of string = Perimeter = 30 cm
∵ An equilateral triangle has 3 equal sides

Question 6.
Broke the rectangles from following (RBSESolutions.com) figure, and 3 find its area. (Measurement of sides are given in cm.)

Solution:
We find the area of given figure, by (RBSESolutions.com) dividing it in A, B and C rectangles :

Area of rectangle (A) = (2 × 1) = 2 sq. cm
Area of rectangle (B) = (5 × 1) = 5 sq. cm
Area of rectangle (C) = (2 × 1) = 2 sq. cm
Total area of figure = Sum of area (RBSESolutions.com) of three rectangles = (2 + 5 + 2) = 9 sq. cm.

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