Monday, 10 November 2014

Chapter 10. Mensuration | CBSE Class 6th (VI) Mathematics | Exercise 10.1 - Solutions

Things to remember...
  • Perimeter is the distance covered once along the boundary of a closed figure .
  • Perimeter of a rectangle = 2 × (length + breadth)
  • Perimeter of a square = 4 × length of a side
  • Perimeter of an equilateral triangle = 3 × length of a side
Question 1. Find the perimeter of each of the following figures :



Solution.


(a). Perimeter of figure (a)= 5 cm + 1 cm + 2cm + 4cm = 12 cm

(b). Perimeter of figure (b)= 40 cm + 35 cm + 23cm + 35 cm = 133 cm

(c). Perimeter of figure (c)= 15 cm + 15 cm + 15 cm + 15 cm = 60 cm

(d). Perimeter of figure (d)= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm = 20 cm

(e). Perimeter of figure (e)= 4 cm + 0.5 cm + 2.5 cm + 2.5 cm+ + 0.5 cm + 4 cm + 1 cm = 15 cm

(f). Perimeter of figure (f)= 4 cm + 3 cm + 2cm + 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 2. The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?

Solution.
The length of the rectangular box= 40 cm
The breadth of the rectangular box= 10 cm
Length of the tape required= Perimeter of rectangular lid of the box
= 2 × (length + breadth)
=2 × (40 + 10) cm
=2 × 50 cm
=100 cm


Question 3. A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?

Solution.
The length of the table-top= 2 m 25 cm
= 2.25 m
The breadth of the table-top= 1 m 50 cm
= 1.5 m
Perimeter of the table-top= 2 × (length + breadth)
=2 × (2.25 + 1.5) m
=2 × 3.75 m
= 7.5 m


Question 4. What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?

Solution.
The length of the photograph= 32 cm
The breadth of the photograph= 21 cm
Length of the wooden strip required to frame the photograph= Perimeter of the photograph
= 2 × (length + breadth)
=2 × (32 + 21) cm
=2 × 55 cm
=110 cm


Question 5. A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?

Solution.
The length of the rectangular piece of land= 0.7 km
The breadth of the rectangular piece of land= 0.5 km
Length of the one row of wire= Perimeter of land
= 2 × (length + breadth)
=2 × (0.7 + 0.5) km
=2 × 1.2 km
=2.4 cm
Total length of wire=4 × 2.4 km
=9.6 km


Question 6. Find the perimeter of each of the following shapes :

(a) A triangle of sides 3 cm, 4 cm and 5 cm.
(b) An equilateral triangle of side 9 cm.
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.

Solution.
(a) A triangle of sides 3 cm, 4 cm and 5 cm.

Perimeter of triangle with unequal sides= side1 + side2 + side3
= 3 cm + 4 cm + 5 cm
= 12 cm


(b) An equilateral triangle of side 9 cm.

Perimeter of equilateral triangle= 3 × any side of triangle
= 3 × 9 cm
= 27 cm


(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.

Perimeter of isosceles triangle= 2 × any one of two equal sides + Third side
= 2 × 8 cm + 6 cm
= 16 cm + 6 cm
= 22 cm


Question 7. Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.

Solution.
Perimeter of triangle with unequal sides= side1 + side2 + side3
= 10 cm + 14 cm + 15 cm
= 39 cm


Question 8. Find the perimeter of a regular hexagon with each side measuring 8 m.

Solution. A regular hexagon has 6 equal sides.
∴ Perimeter of regular hexagon= 6 × Side of hexagon
= 6 × 8 m
= 48 m


Question 9.Find the side of the square whose perimeter is 20 m.

Solution. Given, Perimeter of square = 20 m
As we know, the perimeter of the square= 4 × Side of the square
∴ Side of the square= Perimeter of the square/4
= 20m/4
= 5 m

Chapter 10. Mensuration | CBSE Class 6th (VI) Mathematics |  EXERCISE 10.1 - Solutions

Question 10. The perimeter of a regular pentagon is 100 cm. How long is its each side?

Solution. Given, Perimeter of a regular pentagon = 100 cm
As we know, a regular pentagon has 5 equal sides.
Perimeter of a regular pentagon= 5 × Side of pentagon
∴ Side of a regular pentagon= Perimeter of regular pentagon/5 cm
= 100 cm/5
= 20 cm


Question 11. A piece of string is 30 cm long. What will be the length of each side if the string is used to form :

(a) a square?

(b) an equilateral triangle?

(c) a regular hexagon?

Solution.

(a) If the string is used to form a square?
Perimeter of a square= Length of the string
4 × Side of the Square= 30 cm
∴ Side of the Square= 30 cm/4
= 7.5 cm


(b) If the string is used to form an equilateral triangle?
Perimeter of an equilateral triangle= Length of the string
3 × Side of equilateral triangle= 30 cm
∴ Side of equilateral triangle= 30 cm/10
= 10 cm


(b) If the string is used to form a regular hexagon?
Perimeter of a regular hexagon= Length of the string
6 × Side of regular hexagon= 30 cm
∴ Side of regular hexagon= 30 cm/6
= 5 cm


Question 12. Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?

Solution.
Perimeter of the triangle= 36 cm
Side1 + Side2 +Side3= 36 cm
12 cm + 14 cm +Side3= 36 cm
∴ Side3= 36 cm - 12 cm -14 cm
= 10 cm




Question 13. Find the cost of fencing a square park of side 250 m at the rate of Rs 20 per metre.

Solution.
Side of the square park= 250 m
Length of the Park fencing= Perimeter of the square park
= 4 × side of the square park
= 4 × 250 m
= 1000 m
The cost of 1 metre of fencing= Rs 20
∴ Total cost of 1000 metre of fencing= Rs 20 × 1000
= Rs 20,000


Question 14. Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs 12 per metre.

Solution.
Length of the rectangular park= 175 m
Breadth of the rectangular park= 125 m
Length of the Park fencing= Perimeter of the rectangular park
= 2 × (Length + Breadth)
= 2 × (175 + 125 ) m
= 2 × 300 m
= 600 m
The cost of 1 metre of fencing= Rs 12
∴ Total cost of 600 metre long fencing= Rs 12 × 600
= Rs 7200


Question 15. Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance? 
 

Solution.
Distance covered by Sweety= Perimeter of the square park
= 4 × Side of the square park
= 4 × 75 m
= 300 m
Distance covered by Bulbul= Perimeter of the rectangular park
= 2 × (Length + Breadth)
= 2 × (60 + 45) m
= 2 × 105 m
= 210 m
Hence, Bulbul covers less distance


Question 16. What is the perimeter of each of the following figures? What do you infer from the answers?


Solution.
Figure (a) is a square of side 25 cm
∴ Perimeter of the Figure (a)= 4 × Side of the square
= 4 × 25 m
= 100 m
Figure (b) is a rectangle with length 40 cm and breadth 10 cm
∴ Perimeter of the Figure (b)= 2 × (Length + Breadth)
= 2 × (40 + 10) cm
= 2 × 50 cm
= 100 cm
Figure (c) is again a a rectangle with length 30 cm and breadth 20 cm
∴ Perimeter of the Figure (c)= 2 × (Length + Breadth)
= 2 × (30 + 20) cm
= 2 × 50 cm
= 100 cm
Figure (d) is isosceles triangle with equal sides 30 cm each and third side 40 cm
∴ Perimeter of the Figure (c)= 2 × (any one of two equal sides) + Third side
∴ Perimeter of the Figure (c)= 2 × 30 cm + 40 cm
= 60 cm + 40 cm
= 100 cm
Hence, from the answers above, We can infer that all the figures have same perimeter


Question 17. Avneet buys 9 square paving slabs, each with a side of 1/2 m. He lays them in the form of a square.

(a) What is the perimeter of his arrangement [Fig (i)]?

(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [(Fig (ii)]?

(c) Which has greater perimeter?

(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e. they cannot be broken.)

(a) What is the perimeter of his arrangement [Fig (i)]?

Solution.
The arrangement so formed in Fig (i) is a square of side equal to 3 × side of the square slab
∴ Perimeter of his arrangement in Fig (i)= 4 × ( 3 × side of the square slab)
= 4 × ( 3 × 0.5) m
= 4 × 1.5 m
= 6 m


(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [(Fig (ii)]?

Solution.
The cross arrangement so formed in Fig (ii) by 9 square slabs, has perimeter, which consists of 20 length segments each equal to side of a square slab i.e. 0.5 m
∴ Perimeter of her arrangement in Fig (ii)= 20 × ( side of the square slab)
= 20 × 0.5 m
= 10 m


(c) Which has greater perimeter?

Solution. The Cross arrangement has greater perimeter

(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e. they cannot be broken.)

Solution. No, there is no way of getting perimeter greater than 10 m from any other possible arrangement formed by 9 square slabs when placed edge by edge completely.

CBSE Class 6th (VI) Mathematics | Chapter 10. Mensuration : EXERCISE 10.1 - Solutions


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