## NCERT Class 6 Mathematics Tenth Chapter Mensuration Exercise 10.1 Solution

## EXERCISE 10.1

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

Solution: (a) (4 + 2 + 1 + 5) cm = 12 cm

(b) (35 + 23 + 35 + 40) = 133 cm

(c) (15 + 15 + 15 + 15) = 60 cm

(d) (4 + 4 + 4 + 4 + 4) = 20 cm

(e) (1 + 4 + 4 + 0.5 + 0.5 + 2.5 + 2.5)

= 9 + 1 + 5 = 15 cm

(f) (1 + 1 + 1 + 1 + 4 + 4 + 4 + 4 + 2 + 2 + 2 + 2 + 3 + 3 + 3+ 3 + 3+ 3+ 3 + 3)

= (4 + 16 + 8 + 24) = 52 cm

**(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: Perimeter of rectangular box = 2 × (40 + 10) = 2 × 50 = 100 cm

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

Solution: Here, 2 m 25 cm = 225 cm and 1 m 50 cm = 150 cm

Perimeter of Table-top = 2 × (225 + 150) = 2 × 375 = 750 cm or 7.5 m

**(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: Perimeter of wooden strip = 2 × (32 + 21) = 2 × 53 = 106 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 rectangular piece = 2 × (0.7 + 0.5) = 2 × 1.2 = 2.4 cm

Therefore the length of the wire = 4 × 2.4 = 9.6 cm

**(6) Find the perimeter of each of the following shapes: **

(a) A triangle of sides 3 cm, 4 cm and 5 cm. = (3 + 4 + 5) = 12 cm

(b) An equilateral triangle of side 9 cm. = (3 × 9) = 27 cm

(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm. = (8 + 8 + 6) = 22 cm

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

Solution: The perimeter of triangle = (10 + 14 + 15) = 39 cm

**(8) Find the perimeter of a regular hexagon with each side measuring 8** m.

Solution: Perimeter of hexagon = (6 × 8) = 48 cm

**(9) Find the side of the square whose perimeter is 20 m.**

Solution: Side of the square = (20 ÷ 4) = 5 m

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

Solution: Side of the regular pentagon = (100 ÷ 5) = 20 cm

**In case you are missed :- NCERT Solution for Playing with Numbers Exercise 3.7**

**(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? = (30 ÷ 4) = 7.5 cm

(b) An equilateral triangle? = (30 ÷ 3) = 10 cm

(c) A regular hexagon? = (30 ÷ 6) = 5 cm

**(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: The sum of two side of the triangle = (12 + 14) = 26 cm

Therefore the other side of the triangle = (36 – 26) = 10 cm

**(13) Find the cost of fencing a square park of side 250 m at the rate of Rs 20 per** metre.`

Solution: Perimeter of square park = (4 × 250) = 1000 m

The cost of fencing the square park = (1000 × 20) rupees = 20000 rupees.

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

Solution: Perimeter of rectangular park = 2 × (175 + 125) = 2 × 300 = 600 m

The cost of fencing = Rs (600 × 12) = Rs 7200

**(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: Perimeter of square park which Sweety covers = (4 × 75) = 300 m

Perimeter of rectangular park which Bulbul covers = 2 × (60 + 45) = 2 × 105 = 210 m

**(16) What is the perimeter of each of the following figures? What do you infer from the answers?**

Solution: (a) (4 × 25) = 100 cm

(b) 2 × (20 + 30) = 2 × 50 = 100 cm

(c) 2 × (10 + 40) = 2 × 50 = 100 cm

(d) (2 × 30) + 40 = 60 + 40 = 100 cm

All the figure have same perimeter.

**In case you are missed :- Next Chapter Exercise Solution**