Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Additional Questions

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Additional Questions

Additional Questions and Answers

Exercise 2.1

Question 1.
A circular disc of radius 28 cm is divided into two equal parts. What is the perimeter of each semicircular shape disc? Also find the perimeter of the circular disc.
Answer:
Radius of the circular disc = 28 cm
∴ circumference of the circle = 2 π r units

= 2 × \(\frac { 22 }{ 7 } \) × 28 cm
= 2 × 22 × 4 = 44 × 4 = 176 cm
Perimeter of the circular disc = 176 cm
Now circumference of the semicircular disc = \(\frac { 1 }{ 2 } \) × (2 π r) = π r units
Perimeter of the semicircular disk = π r + r + r = \(\frac { 22 }{ 7 } \) × 28 + 28 + 28
= 22 × 4 + 28 + 28 = 88 + 28 + 28 = 144
Perimeter of each semicircular disc = 144 cm

Question 2.
A gardener wants to fence a circular garden of diameter 14m. Find the length of the rope he needs to purchase if he makes 3 rounds of fence. Also find the cost of the rope, if it costs ₹ 4 per metre.
Solution:
Diameter of the circular garden (d) = 14 m
Circumference = π d = \(\frac { 22 }{ 7 } \) × 14 = 44 m
Since the rope makes 3 rounds of fence,
length of the rope needed = 3 × circumference of the garden
= 3 × 44 m = 132 m
Cost of rope per meter = ₹ 4 × 132 = ₹ 528

Question 3.
Latha wants to put a lace on the edge of a circular table cover of diameter 3 m. Find the length of the lace required and also find the cost if one metre of the lace costs ₹ 30 (Take π = 3.15 )
Solution:
Diameter of the circular table cover = 3m
Circumference C = π d units = 3.15 × 3 m = 9.45 m
Length of the lace required = 9.45 m
Cost of lace per meter = ₹ 30
∴ Cost of 9.45m lace = ₹ 30 × 9.45 = ₹ 283.50

Exercise 2.2

Question 1.
The circumference of two circles are on the ratio 5 : 6. Find the ratio of their areas.
Solution:
Let the radii of the given circles be r₁ and r₂
Let their circumference be C₁ and C₂ respectively
C₁ = 2πr₁ and C₂ = 2πr₂

Now let the area of the given circles the A₁ and A₂

∴ The areas of two circles are in the ratio 25 : 36.

Question 2.
Find the area of the circle whose circumference is 88 cm.
Solution:
The circumference of the circle = 88 cm
2πr = 88 cm
2 × \(\frac { 22 }{ 7 } \) × r = 88 cm
r = \(\frac{88 \times 7}{2 \times 22}\) = 14 cm
r = 14 cm
Area of the circle A = πr²
= \(\frac { 22 }{ 7 } \) × 14 × 14 cm²
= 22 × 2 × 14 cm²
= 616 cm²

Question 3.
Find the cost of polishing a circular table top of diameter 16 dm if the rate of polishing is ₹ 15 per dm².
Solution:
Diameter of the circular table top = 16 dm
Radius (r) = \(\frac { 16 }{ 2 } \) = 8cm
Area of the circular table top
= πr² sq.units
= \(\frac { 22 }{ 7 } \) × 8 × 8 dm² = \(\frac { 1408 }{ 7 } \) dm²
= 201.14 dm²
Rate of the polishing per dm² = ₹ 15
∴ Rate of the polishing 201.14 dm²
= ₹ 15 × 201.14
= ₹ 3,017

Exercise 2.3

Question 1.
From a circular sheet of radius 5 cm a circle of radius 3 cm is removed. Find the area of the remaining sheet.
Solution:
Radius of the outer circle R = 5 cm
Radius of the inner circle r = 3 cm
Area of the remaining sheet = Area of the outer circle
– Area of the inner circle
= πR² – πr² sq. units = π(R² – r²) sq. units
= \(\frac { 22 }{ 7 } \) (5² – 3²) cm² = \(\frac { 22 }{ 7 } \) (5² – 3²) cm²
= \(\frac { 22 }{ 7 } \) × (5 + 3) (5 – 3) = \(\frac { 22 }{ 7 } \) × (8) (2)
= \(\frac { 352 }{ 7 } \) = 50.28 cm²

Question 2.
A picture is painted on a cardborad 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
Solution:
Length of the outer rectangle L = 8 cm
Breadth of the outer rectangle B = 5 cm
Area of the outer rectangle = L × B sq. units = 8 × 5 cm² = 40 cm²
Length of the inner rectangle l = L – 2W = 8 – 2(1.5)cm = 8 – 3cm
l = 5cm
Breadth of thqnner rectangle b = B – 2W = 5 – 2(1.5) cm
= 5 – 3cm = 2cm
∴ Area of the margin = Area of the outer rectangle
– Area of the inner rectangle
= (40 – 10) cm² = 30cm²

Question 3.
A circular piece of radius 2 cm is cut from a rectangle sheet of length 5 cm and breadth 3 cm. Find the area left in the sheet.
Solution:
Radius of the portion removed r = 2 cm
Area of the circular sheet = πr² sq.units
= \(\frac { 22 }{ 7 } \) × 2 × 2 cm² = \(\frac { 88 }{ 7 } \) cm² = 12.57 cm²
Length of the rectangular sheet L = 5 cm
Breadth of the rectangular sheet B = 3 cm
Area of the rectangle = L × B sq. units = 5 × 3 cm² = 15 cm²
Area of the sheet left over = Area of the rectangle – Area of the circle
= 15 cm² – 12.57 cm² = 2.43 cm²