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

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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) unit²
= (57 × 43) m² = 2,451 m²
Area of the park = 2,451 m²

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) units²
If the side of the new square is made one fourth then side = \(\left(\frac{1 \times s}{4}\right)\) units
Then area = \(\left(\frac{1 \times s}{4} \times \frac{1 \times s}{4}\right)\) units² = \(\frac{s \times s}{16}=\frac{1}{16}\) (s × s) units²
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 m² = 100 m²
Area of the rectangular plot = length × breadth = (12 × 8) m² = 96 m²
100 m² > 96 m²
∴ Square plot has greater area by 4m²

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.