## 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 × 53 cm =106 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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