RBSE Solutions for Class 5 Maths Chapter 14 Perimeter and Area

Rajasthan Board RBSE Solutions for Class 5 Maths Chapter 14 Perimeter and Area Textbook Exercise Questions and Answers.

RBSE Class 5 Maths Solutions Chapter 14 Perimeter and Area

Intext Questions:

Question 1. 
Nirmals math book contains the question of finding the perimeter of some figures.

 
Can you find the perimeters of these three figures?
Solution:
We know that the perimeter is only of a closed figure, so here only perimeter of the figure (a) can be found. Figure (b) and (c) are not a closed figure so we can not find the perimeter of figure (b) and (c). 

Question 2.
Some shapes given below :

Find the perimeter of these shapes.
Solution:
(a) We know the perimeter of regular shapes = No. of sides x length of one side
perimeter = 3 × 10 = 30 cm
(b) perimeter = 4 × 12 = 48 cm
(c) perimeter = 5 × 7 = 35 cm

 

Question 3. 
Think, is rectangle a regular shape? If you want to find a formula for the perimeter of a rectangle, what will you do?
Solution:
Given figure is a rectangle whose length (l) and breadth (b) (15 and 5 m respectively) are equal. So, it is not a regular figure. Rule for perimeter of rectangle
= l + b + l + b 
= 2 × l + 2 × l
= 2(1 + b)
Thus, rule is :
Perimeter of rectangle = 2 (l + b)
= 2 (15 + 5)
= 2 (20) = 40 m

Question 4. 
Find the area of figures (ii) and (iii) given at the following grid paper.

Solution:
Ist Method - Area of two figures is such that in figures, by counting number of squares we get area

(ii)

(iii)
Thus, area of figure (ii) is 16 sq. cm and area of of figure (iii) is 21 sq. cm.

Second Method - Figure (ii) is a square whose each side = 4 cm
Thus, area of square = Side × Side 
= 4 × 4 = 16 sq cm.

Figure (iii) is a rectangle whose length is 7 cm and breadth is 3 cm.
Thus, area of rectangle = Length × Breadth
= 7 × 3 = 21 sq. cm

 

Exercise:

Question 1. 
Find the perimeter of the following figures.
 

Solution:
(a) Perimeter = Sum of sides
= 7 + 10 + 7+ 10 + 8 + 7 + 9 + 7 = 65 cm

(b)    Perimeter = Sum of sides 
= 10 + 5 + 4 + 5 + 5 + 4 + 5 = 38 cm 
(c) Perimeter = Sum of sides = 7 + 7 + 7 + 7 = 28cm

Question 2. 
Find out the perimeter of rectangular shapes with given measurements.
(a) Length = 30 cm Breadth = 48 cm
(b) Length = 20 cm Breadth = 34 cm
(c) Length = 60 cm Breadth = 20 cm
(d) Length = 30 cm Breadth = 12 cm 
Solution:
(a) Perimeter = 2(l + b)
= 2(30 + 48) 
= 2(78) = 156 cm

(b)    Perimeter = 2(l + b)
= 2(20 + 34) 
= 2(54) = 108 cm

(c) Perimeter = 2(1 + b)
= 2(60 + 20) 
= 2(80) = 160 cm

(d)    Perimeter = 2(1 + b)
= 2(30 + 12) 
= 2(42) = 84 cm

 

Question 3. 
Find out the perimeter of following regular shapes with the help of formula.

Solution:
(a) Perimeter = No. of sides × Measure of side 
= 3 × 14 cm = 42 cm 

(b) Perimeter = No. of sides × Measure of side 
= 4 × 6 cm = 24 cm
 
(c) Perimeter = No. of sides × Measure of side 
= 6 × 6 cm = 36 cm

(d) Perimeter = No. of sides × Measure of side 
= 5 × 9 cm = 45 cm 

 

Question 4. 
Vijay has made a rectangle. Find its perimeter and area. Can you extend its length and width in such a way, so that it has equal perimeter and area ?
Solution:
Perimeter of rectangle = Sum of sides
= 5 + 3 + 5 + 3
= 8 + 8 = 16 cm 

Area of Rectangle = Length × Breadth 
= 5 × 3 = 15 sq. cm
 
If length of rectangle became 4 cm and breadth 4 cm then
Perimeter = 4 × side 
= 4 × 4 = 16 cm 

Area = (Side) × (Side)
= 4 × 4 = 16 sq. cm 
Thus, perimeter and area both will be same.

Question 5. 
Length and width of a rectanglular field are 25 meter and 30 meter respectively. Find its area. 
Solution:
Length of field = 25 m And breadth = 30 m 
∴ Area of field = Length × Breath 
= 25 × 30 = 750 sq. cm 

 

Question 6. 
Length and width of a rectangular towel are 125 cm and 60 cm respectively. What will be the perimeter of the towel?
Solution:
Length of towel = 125 cm 
And breadth = 60 cm 
∴ Perimeter of towel = 2(l + b)
= 2(125 + 60) 
= 2(185) = 370 cm

Question 7. 
A square field needs 260 meter long barbed wire for throughout fencing. Find its one side.
Solution:
Length of wire = Perimeter of field = 260 m
∴ Perimeter of field = 260 m
⇒ 4 × Side = 260 m
⇒ Side = \(\frac{260}{4}\) = 65 m
Thus, side of field in 65 m.

 

Question 8. 
The length and width of a room floor is 8 meter and 7 meter respectively. In this room one mat covers the entire floor. Find out the area of this mat.
Solution:
Length of floor = 8 m
Breadth of floor = 7 m 
Required area of carpet = Area of floor
= Length x Breadth 
= 8 × 7 = 56 sq. m 
Thus, area of carpet is 56 sq. m.

Question 9. 
Find out the perimeter of a square stool, whose side is 60 centimeter.
Solution:
Side of small table = 60 cm 
Perimeter of small table = 4 × Side 
= 4 × 60 = 240 cm
Thus, perimeter of small tables is 240 cm.

 

Question 10. 
To take a roud of a square field. Dev had to walk 40 meter. Find the side of the square field.
Solution:
Distance covered in two round = 40 m
Distance covered in 1 round = \(\frac{40}{2}\) = 20 m
All four sides of square field are same. 
∴ 4 × Side = Perimeter of field
⇒ 4 × Side = 20 m
⇒ Side = \(\frac{20}{4}\) m = 5 m
Thus, each side of field is 5 m.

Important Questions:

Multiple Choice Questions:

Question 1.    
A rectangular park has length L and breadth B, then its area will be :
(a) 2 × (L + B) 
(b) 2 × (L - B)
(c) 2 × (L ÷ B) 
(d) L × B
Solution: 
(d) L × B

Question 2.    
Distance travelled in 1 round of a square field will be :
(a) 4 × Side 
(b) (Side)² 
(c) Side ÷ 4    
(d) 4 Side² 
Solution: 
(a) 4 × Side 

 

Question 3.    
A rectangular wooden plank has length 25 m and breadth 15 m. Then, its area will be :
(a) 80 sq. m. 
(b) 375 sq. cm.
(c) 40 sq. m. 
(d) 100 sq. cm.
Solution:
(b) 375 sq. cm.

Question 4.    
The side of a square figure is 15 cm, then its area will be :
(a) 225 sq. cm. 
(b) 60 sq. cm.
(c) 75 sq. cm. 
(d) 100 sq. cm.
Solution:
(a) 225 sq. cm. 

 

Question 5.    
A rectangular field has length 10 m and breadth 8m. Perimeter wiring is to be fitted around the field, then total length of wire will be :
(a) 80 m
(b) 160 m
(c) 36 m    
(d) 18 m
Solution:
(c) 36 m    

Question 6.    
A triangle whose sides are 10 cm, 8 cm and 5 cm, then perimeter of triangle will be :
(a) 40.5 cm 
(b) 23 cm 
(c) 22 cm    
(d) 20 cm
Solution:
(b) 23 cm 

Question 7.    
Perimeter of square is :
(a) 4 × (Length of ane side)
(b) 2 + (l × b)
(c) 2 × (l + b)
(d) Product of four sides
Solution:
(a) 4 × (Length of ane side)

 

Very Short Answer Type Questions:

Question 1. 
What is area?
Solution:
Any geometrical figure occupy how much space on places that is called area.

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

Question 3. 
What is perimeter?
Solution:
Distance travelled around the closed figure one time is called perimeter of that figure.

Question 4. 
Find perimeter of a regular pentagon whose each side is of length 3 cm.
Solution:
In regular pentagon, number of sides are 5, in which each side of length is 3 cm.
Perimeter of regular pentagon = 5 × 3 cm = 15 cm

 

Short Answer Type Questions:

Question 1. 
In given figure find area of triangle.

Solution:
In figure, by counting number of squares in triangle we find its area 
No. of complete square = 27 
No. of incomplete square = 9 
Thus, area of triangle = 27 + 9
= 36 square or 36 sq. units
 
Question 2. 
Find area of iron rod of length 2m 30 cm and breadth 1 m. 20 cm in sQuestion m.
Solution:
Length of rod = 2 m 30 cm = 2.30 m 
Breath of rod = 1 m 20 cm = 1.20 m 
Area of iron = l × b
= 2.30 × 1.20 = 2.76 sq. m

 

Question 3. 
Draw any circle on graph paper. By counting number of squares in circle, find estimated area of circular area.
Solution:
Draw a circle on graph paper whose each square is of measure 1 cm x 1 cm.

No. of complete square in circle = 1
No. of incomplete square in circle = 4
No. of incomplete square (less than half) in circle = 4
Area enclosed by complete squares = (1 × 1) sq. cm = 1 sq. cm
Area enclosed by squares (more than half) = (4 × 1) sq.cm. = 4 sq.cm. 
Area enclosed by square (less than half) = 4 × 0 sq. cm. = 0 sq. cm.
∴ Total area = 1 + 4 + 0 sq. cm.
= 5 sq. m.

Question 4. 
A photo of length 32 cm and breadth 21 cm has to be wooden frame. Find the length of wooden sheet required.
Solution:
Photo frame is rectangular.
∴ Perimeter of wooden sheet required for frame = 2 × (l + b)
= 2 × (32 cm + 21 cm)
= 2 × (53 cm) = 106 cm
Thus, length of required wooden sheet = 106 cm

 

Question 5. 
Find perimeter of the following figures :

Solution:
(a) Perimeter = Sum of length of sides
= 5 cm + 1 cm + 2 cm + 4 cm = 12 cm 

(b) Perimeter = Sum of length of sides 
= 40 cm + 35 cm + 23 cm + 35 cm = 133 cm . 

(c) Perimeter = 4 × Length of one side 
= 4 × 15 cm = 60 cm

(d) Perimeter = 5 × Length of one side 
= 5 × 3 cm = 15 cm

(e) Perimeter = Sum of length of sides 
= 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm + 1 cm = 15 cm
 
(f) Perimeter = Sum of length of sides 
= 4 cm + 3 cm + 2 cm + 3 cm + 1 cm + 4 cm + 3 cm + 2 cm + 3 cm + 1 cm + 4 cm + 3 cm + 2 cm + 3 cm + 1 cm + 4 cm + 3 cm + 2 cm + 3 cm + 1 cm = 52 cm

 

Question 6. 
Find perimeter of a triangle whose sides are 10 cm, 14 cm and 15 cm.
Solution:
Perimeter of triangle = Sum of length of sides 
= 10 cm + 14 cm + 15 cm = 39 cm

Question 7. 
Find perimeter of a regular hexagon, whose each side is of measure 8 m.
Solution:
In a regular hexagon, there are 6 sides.
∴ Perimeter = 6 × Length of side
= 6 × (8m) = 48 m

Question 8. 
Find side of a square, whose perimeter is 20 m.
Solution:
Perimeter of square = 20 m  
∴ Square has 4 equal sides.
∴ Side of square = Perimeter/4
= \(\frac{20}{4}\) = 5 m.

 

Question 9. 
Perimeter of a regular pentagon is 100 cm. Find length of each side.
Solution:
Perimeter of regular pentagon = 100 cm
∴ Regular pentagon has 5 equal sides.
Length of one side = \(\frac{\text { perimeter }}{5}=\frac{100}{5}\)
= 20 cm