ML Aggarwal CBSE Solutions Class 6 Math Eleventh Chapter Mensuration Exercise 11.1

ML Aggarwal CBSE Solutions Class 6 Math 11th Chapter Mensuration Exercise 11.1

Exercise 11.1

(1) Find the perimeter of each of the following figures:

Solution 1: Perimeter = 5 cm + 3 cm + 2 cm + 7 cm = 17 cm

Solution 2: 31 cm + 38 cm + 48 cm + 38 cm = 155 cm

Solution 3: 4 x 19 cm = 76 cm.

Solution 4: 5 x 7cm = 35 cm.

(2) Find the perimeter of each of the following shapes:

(i) A triangle of sides 3 cm, 4 cm, and 6 cm

Solution: Perimeter of triangle: 3 cm + 4 cm + 6 cm = 13 cm.

(ii) A triangle of sides 7 cm, 5.4 cm and 10.2 cm.

Solution: Perimeter of triangle: 7 cm + 5.4 cm + 10.2 cm = 22.6 cm

(iii) An equilateral triangle of side 11 cm.

Solution: Perimeter of equilateral triangle = 3 x 11 cm = 33 cm.

(iv) An isosceles triangle with equal sides 10 cm each and third side 7 cm.

Solution: Perimeter of isosceles triangle = 10 cm + 10 cm + 7 cm  = 27 cm.

(3) 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 tape required  = 40 cm x 10 cm = 400 cm.

(4) A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table top?

Solution: The perimeter of the table top = 2 m 25 cm x 1 m 50 cm

= 2 x 100 cm + 25 cm + 1 x 100 cm + 50 cm

= 200 cm + 25 cm + 100 cm + 50 cm

= 225 cm + 150 cm

= 375 cm

= 3 m 75 cm

(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: Perimeter of a rectangular piece of land measures  = 2 x (0.7 + 0.5)

= 2 x 1.2 km

= 2.4 km.

Each side is to be fenced with 4 rows of wires.

∴ The length of the wire needed = 4 x 2.4 km = 9.6 km.

(6) Find the perimeter of a regular hexagon with each side measuring 7.5 m.

Solution: In regular polygon all sides are equal.

Therefore, The perimeter of a regular hexagon = 6 x 7.5 m = 45.0 m.

(7) The lengths of two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangles is 36 cm. What is the length of its third side?

Solution: Let, the length of the third side be x cm

Then, 12 cm  + 14 cm + x cm = 36 cm

=> x cm = 36 – 12 – 14

=> x cm = 10

Ans. The length of the third side is 10 cm.

(8) The perimeter of a regular pentagon is 100 cm. How long is its side?

Solution: A regular pentagon all sides are equal.

A pentagon has 5 sides.

Its sides = 100 cm / 5 = 20 cm.

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

(i) a square? (ii) an equilateral triangle? (iii) a regular hexagon.

Solution: (i) A square has 4 sides.

Therefore, the length of each side of a square = 30 / 4 = 7.5 cm

(ii) An equilateral triangle has 3 sides and all sides are equal.

Therefore the length of each side of an equilateral triangle = 30 / 3 = 10 cm

(iii) A regular hexagon has 6 sides and all the sides are equal.

Therefore the length of each side of an regular hexagon = 30 / 6 = 5 cm.

(10) Find the cost of fencing a rectangular park of length 225 m and breadth 115 m at the rate of Rs. 13 per metre.

Solution: Here the perimeter of a rectangular park = 2 x (225 + 115)

= 2 x 340 m

= 680 m

The total cost required for fencing = 680 m x 13 = Rs. 8840

Ans. Total cost required = Rs. 8840/-

(11) Meera want to a rectangular park 140 m long and 90 m wide. She took 5 complete rounds on its boundary. What is the distance covered by her.

Solution: Here the perimeter of a rectangular park = 2 x (140 + 90)

= 2 x 230

= 460 m

She took 5 complete rounds on its boundary.

Total Distance covered by her = 5 x 460 = 2300 m.

(12) Pinky runs 8 times around a rectangular park with length 80 m and breadth 55 m while Pankaj runs 7 times around a square park of side 75 m. `Who covers more distance and by how much?

Solution: Pinky’s Case,

Perimeter of rectangular park 2 x (80 m + 55 m)

= 2 x 135 m = 270 m

She runs 8 times around the field.

Total distance he covered = 270 m x 8 = 2160 m

Pankaj case,

Perimeter of the square park = 75 m x 4 = 300 m.

He runs 7 times around a square park.

Total distance he covered = 7 x 300 m = 2100 m

Therefore, Pinky runs more distance by (2160 – 2100 = 60 m).


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