You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.2

Miscellaneous Practice Problems

Question 1.

A piece of wire is 36 cm long. What will be the length of each side if we form

i) a square

ii) an equilateral triangle.

Solution:

Given the length of the wire = 36 cm

i) When a square is formed out of it

The perimeter of the square = 36 cm

4 × side = 36

side = \(\frac{36}{4}\) = 9 cm

Side of the square

ii) When an equilateral triangle is formed out of it, its perimeter = 36 cm

i.e., side + side + side = 36 cm .

3 × side = 36 cm

side = \(\frac{36}{3}\) = 12 cm

One side of an equilateral triangle = 12 cm

Question 2.

From one vertex of an equilateral triangle with side 40 cm, an equilateral triangle with 6 cm side is removed. What is the perimeter of the remaining portion?

Solution:

If an equilateral triangle of side 6 cm is removed the perimeter = (40 + 34 + 6 +34) cm = 114 cm

Perimeter of the remaining portion = 114 cm

Question 3.

Rahim and peter go for a morning walk, Rahim walks around a square path of side 50 m and Peter walks around a rectangular path with length 40 m and breadth 30 m. If both of them walk 2 rounds each, who covers more distance and by how much?

Solution:

Perimeter of the square = (4 × sides) units

side of the square = 50 m

∴ Perimeter = 4 × 50 m = 200 m

Rahim walked twice around the park.

∴ Total distance walked by Rahim = 2 × 200m

Total distance covered by Rahim = 400m

Perimeter of a rectangle = 2 (length + breadth) unit

Length = 40 m

Breadth = 30 m

∴ perimeter = 2 × (40 + 30) m = (2 × 70) m = 140 m

Peter walked around twice

∴ Distance covered by Peter = 2 × 140 m = 280 m

∴ Distance covered by Peter = 280 m

400 m > 280 m

Difference = 480 – 280 = 120 m

∴ Rahim covers 120m more than Peter

Question 4.

The length of a rectangular park is 14 m more than its breadth. If the perimeter of the park is 200 m, what is its length? Find the area of the park?

Solution:

Given length of rectangular park is 14m more than its breadth.

Let the breadth be b m .

∴ Length of the park will be l = b + 14 m

Given perimeter = 200 m

2 × (l + b) = 200 m

2 × (b + 14 + b) = 200 m [∵ l = b + 14]

2 × (2b + 14) = 200 m

2b + 14 = \(\frac{200}{2}\) m

2b + 14 = 100 m

2b = 100 – 14 m

2b = 86 m

b = \(\frac{86}{2}\) m

b 43 m

Length Length of the park = 57 m

Area of a rectangle = (length × breadth) unit^{2}

= (57 × 43) m^{2} = 2,451 m^{2}

Area of the park = 2,451 m^{2}

Question 5.

Your garden is in the shape of a square of side 5 m. Each side is to be fenced with 2 rows of wire. Find how much amount is needed to fence the garden at Rs 10 per meter.

Solution:Perimeter of a square = (4 × side) units

side = 5 m

∴ Perimeter = (4 × 5) m = 20 m

Each side is fenced with 2 rows of wire

∴ Distance to be fenced = 2 × 20 m = 40 m

Cost of fencing per metre = ₹ 10

∴ Cost of fencing 40 m = 40 × 10 = ₹ 400

Cost of fencing the garden = ₹ 400

Challenge Problems

Question 6.

A closed shape has 20 equal sides and one of its sides is 3 cm. Find its perimeter.

Solution:

Number of equal sides in the shape = 20

One of its side = 3 cm

Perimeter = length of one side × Number of equal sides

∴ Perimeter = (3 × 20) cm = 60 cm

∴ Perimeter = 60 cm

Question 7.

A rectangle has length 40 cm and breadth 20 cm. How many squares with side 10 cm can be formed from it.

Solution:

l = 40 cm, b = 20 cm

Area of the rectangle = l × b sq units

= 40 × 20 cm²

= 800 cm²

a = 10 cm

Area of the square = a × a sq. units

= 10 × 10 cm²

= 100 m²

No of squares formed = \(\frac{800}{100}\) cm²

= 8

Question 8.

The length of a rectangle is three times its breadth. If its perimeter is 64 cm, find the sides of the rectangle.

Solution:

Length l = 3b

breadth = b

perimeter = 64 cm

2 (l + b) = 64 cm

2(3b + b) = 64

2 (4b) = 64

8b = 64

b = \(\frac{64}{8}\)

breadth b = 8 cm

length l = 3b

l = 3 × 8

= 24 cm

Question 9.

How many different rectangles can be made with a 48 cm long string? Find the possible pairs of length and breadth of the rectangles.

Solution:

12 rectangles

(1, 23), (2, 22), (3, 21), (4, 20), (5, 19), (6, 18), (7, 17), (8, 16), (9, 15), (10, 14), (11, 13), (12, 12)

Question 10.

Draw a square B whose side is twice of the square A. Calculate the perimeters of the squares A and B.

Solution:

The perimeter of square B = 4 × 4 m = 16 m

The perimeter of square A = 4 × 2 m = 8 m

The perimeter of square B is twice that of square A.

Question 11.

What will be the area of a new square formed if the side of a square is made one – fourth?

Solution:

Let the side of square is s units then area = (s × s) units^{2}

If the side of the new square is made one fourth then side = \(\left(\frac{1 \times s}{4}\right)\) units

Then area = \(\left(\frac{1 \times s}{4} \times \frac{1 \times s}{4}\right)\) units^{2} = \(\frac{s \times s}{16}=\frac{1}{16}\) (s × s) units^{2}

Area of the new square is reduced to \(\frac{1}{16}\) times to that of original area.

Question 12.

Two plots have the same perimeter. One is a square of side 10 m and another is a rectangle of breadth 8 m. Which plot has the greater area and by how much?

Solution:

Given perimeter of square = perimeter of rectangle

4 × side = 2 (length + breadth)

(4 × 10) m = 2(l + 8)m

\(\frac{4 \times 10}{2}\) = l + 8

20 = l + 8

l = 20 – 8

l = 12 m

∴ length of the rectangle = 12 m

Area of the square plot – side × side = 10 × 10 m^{2} = 100 m^{2}

Area of the rectangular plot = length × breadth = (12 × 8) m^{2} = 96 m^{2}

100 m^{2} > 96 m^{2}

∴ Square plot has greater area by 4m^{2}

Question 13.

Look at the picture of the house given and find the total area of the shaded portion.

Solution:

Area of a square = a × a sq.units

= 6 x 6 cm²

= 36 cm²

Area of a rectangle = l × b sq units

= 9 x 6 cm²

= 54 cm²

Area of a triangle = \(\frac{1}{2}\) × b × h sq.units

= \(\frac{1}{2}\) × 4² × 6 cm²

= 12 cm²

Total area of the shaded portion

= (36 + 54 + 12) cm²

= 102 cm²

Question 14.

Find the approximate area of the flower in the given square grid.

Solution:

Approximate area = Number of full squares + Number of more than half squares + \(\frac{1}{2}\) × Number of half squares.

= 10 + 5 + (\(\frac{1}{2}\) × 1) Sq units. = 10 + 5 + \(\frac{1}{2}\) sq. units

= 15 \(\frac{1}{2}\) sq. units = 15.5 sq. units.