Chapter 10. Mensuration | CBSE Class 6th (VI) Mathematics | Exercise 10.1 - Solutions
Things to remember...
Solution.
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.
Question 3. A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
Solution.
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.
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.
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.
(b) An equilateral triangle of side 9 cm.
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
Question 7. Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Solution.
Question 8. Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution. A regular hexagon has 6 equal sides.
Question 9.Find the side of the square whose perimeter is 20 m.
Solution. Given, Perimeter of square = 20 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.
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?
(b) If the string is used to form an equilateral triangle?
(b) If the string is used to form a regular hexagon?
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.
Question 13. Find the cost of fencing a square park of side 250 m at the rate of Rs 20 per metre.
Solution.
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.
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.
Question 16. What is the perimeter of each of the following figures? What do you infer from the answers?
Solution.
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.
(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.
(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
- 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
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
