Exercise 10.1

1. Find the perimeter of each of the following figures.

(a) Perimeter of the given figure = sum of all the sides = 4 cm + 5 cm + cm + cm = cm

(b) Perimeter of the given figure = sum of all the sides = 23 cm + 35 cm + cm + cm = cm

(c) Perimeter of the given figure = sum of all the sides = 15 cm + 15 cm + cm + cm = cm

(d) Perimeter of the given figure = sum of all the sides = 4 cm + 4 cm + cm + cm + cm = cm

(e) Perimeter of the given figure = sum of all the sides = 4 cm + 3 cm + 2 cm + 1 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 = cm

(f) Perimeter of the given figure = sum of all the sides = 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm = 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:

Length = cm, Breadth = cm.

Thus, by substituting in the formula,

Perimeter = 2 (l + b)

= 2( + )

= 2 × = cm

Thus, the tape length required is 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:

Length = 2 m 25 cm, Breadth = 1 m 50 cm

Let's convert the length and breadth to have the same units.

Length = 2 m + 25 cm = 2 × cm + 25 cm = cm + 25 cm = cm

Breadth = 1 m + 50 cm = cm + 50 cm = cm

Thus, Length = 225 cm while Breadth = 150 cm

Substitute these values in the formula of the perimeter of a rectangle.

Perimeter of table - top = 2 (l + b)

= 2 ( + )

= 2 ×

= cm

Therefore, the perimeter of the table-top is 750 cm or 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:

The length of the wooden strip = of the photograph

Length = cm, Breadth = cm

Now, substituting the values of length and breadth in the perimeter formula we get = 2(l+b)

= 2 ( + )

= 2 ×

= cm

Thus, the length of the wooden strip required will be 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:

Length = km, Breadth = km

Substituting the values we get, Length of the wire required = = × 2(l+b)

= 4 × 2 ( + )

= 4 × 2 × = km

Therefore, the length of the wire required to fence the park is 9.6 km.

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

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

Solution:

Instructions

(b) An equilateral triangle of side 9 cm.

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

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

Solution:

We will be using the formula of the perimeter of a triangle which is equal to the of all its sides.

Perimeter = + + = cm

Therefore, the required perimeter of the triangle is 39 cm.

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

Solution:

We will use the formula for the perimeter of a regular hexagon which is × side of a regular hexagon

Here, AB = BC = CD = DE = EF = FA = m

Hence, the perimeter of a regular hexagon = 6 × 8 = m

Thus, 48 m is the perimeter of the regular hexagon.

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

Solution:

The perimeter of square = m

The perimeter of a square = × Side of the square

20 = 4 × Side of the square

Side of the square = 204420

Thus, side of the square = m

Therefore, 5 m is the side length of the square.

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

Solution:

Perimeter of regular pentagon = cm

Perimeter of regular pentagon = × side

Perimeter = 5 × side = 100 cm

Side = 10055100 = cm

Therefore, each side of the regular pentagon is 20 cm.

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?

(a) a square?

Solution:

Instructions

(b) an equilateral triangle?

(c) a regular hexagon?

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:

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

Length of the square park = m

Cost of fencing = Rs. per meter

Therefore, the Perimeter of the square park = × 250 = m = km

So, Total cost = × 10001 = Rs.

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

Instructions

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?

Instructions

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

(a) Perimeter of the square = × cm = cm

(b) Perimeter of the rectangle = ( cm + cm) = 2 × cm = cm

(c) Perimeter of the rectangle = ( cm + cm) = 2 × cm = cm

(d) Perimeter of the triangle = 30 cm + cm + cm = cm

Avneet buys 9 square paving slabs, each with a side of 12 m. He lays them in the form of a square.

(a) What is the perimeter of his arrangement?

From the figure, we know that the sqaure arrangement has a side length = (12 m + 12 m + 12 m) =

Thus, Perimeter of the square arrangement = × side = 4 × 32 m = 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 ?

Perimeter of cross-arrangement = m + 1 m + 1 m + m + 1 m + 1 m + m + m + m + m + m + m = m.

(c) Which has greater perimeter?

Since 10 m >< 6 m, crosssquare 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)

Total number of tiles =

We have the following arrangement having the perimeter equal to m.

The above arrangement doesn'tdoes have a greater perimeter than the previously considered arrangements.