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

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:
Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.2 1
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:
Distance covered by Rahim
= 50 × 4 m
= 200 m
If he walks 2 rounds, distance covered = 2 × 200 m
= 400 m
Distance covered by peter
= 2 (40 + 30) m
= 2(70)m
= 140 m
If he walks 2 rounds, distance covered = 2 × 140 m
= 280 m
∴ Rahim covers more distance by (400 – 280) = 120 m

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

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:
a = 5 m
Perimeter of the garden
= 4 a units
= 4 × 5 m
= 20 m
For 1 row
Amount needed to fence l m= Rs 10
Amount needed to fence 20 m
= Rs 10 × 20
= Rs 200
For 2 rows
Total amount needed = 2 × Rs 200
= Rs 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:
Given perimeter of a rectangle = 64 cm
Also given length is three times its breadth.
Let the breadth of the rectangle = b cm
∴ Length = 3 × b cm
Perimeter = 64 m
i.e., 2 × (l + b) = 64 m
2 × (3b + b) = 64 m
2 × 4b = 64m
4b = \(\frac{64}{2}\) = 32 m
b = \(\frac{32}{4}\) = 8 m
l = 3 × b = 3 × 8 = 24 m
∴ Breadth of the rectangle = 8 m
Length of the rectangle = 24 m

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)

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

Question 10.
Draw a square B whose side is twice of the square A. Calculate the perimeters of the squares A and B.
Solution:
Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.2 2
Perimeter of A = s + s + s + s units = 4 s units
Perimeter of B = (2s + 2s + 2s + 2s) units
= 8s units = 2 (4s) units.
∴ Perimeter of B is twice perimeter of A

Question 11.
What will be the area of a new square formed if the side of a square is made one – fourth?
Solution:
Area of the new square is reduced to \(\frac{1}{16}\) th times to that of the 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:
Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.2 3
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.
Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.2 4
Solution:
Total area of the shaded region = Area of a right triangle + Area of a rectangle
= (\(\frac{1}{2}\) × b × h) + (l × b) cm2
= [(\(\frac{1}{2}\) × 3 × 4) + (9 × 6)] cm2
= (6 + 54) cm2 = 60 cm2

Question 14.
Find the approximate area of the flower in the given square grid.
Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.2 5
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.

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

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