Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1

Question 1.
Fill in the blanks:
(i) The ratio between the circumference and diameter of any circle is _________.
(ii) A line segment which joins any two points on a circle is a ______.
(iii) The longest chord of a circle is _______.
(iv) The radius of a circle of diameter 24 cm is ______.
(v) A part of circumference of a circle is called as _____.
Solution:
(i) π
(ii) chord
(iii) diameter
(iv) 12 cm
(v) an arc

Question 2.
Match the following

Solution:
(i) 3
(ii) 4
(iii) 5
(iv) 2
(v) 1

Question 3.
Find the central angle of the shaded sectors (each circle is divided into equal sectors)

Solution:

Question 4.
For the sectors with given measures, find the length of the arc, area and perimeter, (π = 3.14)
(i) central angle 45°, r = 16 cm
(ii) central angle 120°, d = 12.6 cm
(iii) central angle 60°, r = 36 cm
(iv) central angle 72°, d = 10 cm
Solution:
(i) Central angle 45°, r = 16 cm
Length of the arc l = \(\frac{\theta^{\circ}}{360^{\circ}}\) × 2πr units
l = \(\frac{45^{\circ}}{360^{\circ}}\) × 2 × 3.14 × 16 cm
l = \(\frac{1}{8}\) × 2 × 3.14 × 16 cm
l = 12.56 cm
Area of the sector = \(\frac{\theta^{\circ}}{360^{\circ}}\) × πr²
A = \(\frac{45^{\circ}}{360^{\circ}}\) × 3.14 × 16 × 16
A = 100.48 cm²
Perimeter of the sector
P = l + 2r units
P = 12.56 + 2(16) cm
P = 44.56 cm

(ii) Central angle 120°, d = 12.6 cm
∴ r = \(\frac{12.6}{2}\) cm
r = 6.3 cm
Length of the arc


Area of the sector missing
Perimeter of the sector
P = l + 2r units
P = 6.28 + 2(5) cm
P = 6.28 + 10 cm
P = 16.28 cm

Question 5.
From the measures given below, find the area of the sectors.

Solution:
(i) Area of the sector
A = \(\frac{l r}{2}\) sq. units
l = 48 m
r = 10 m
= \(\frac{48 \times 10}{2}\) m²
= 24 × 10 m²
= 240 m²
Area of the sector = 240 m²

(ii) length of the arc l = 12.5 cm
Radius r = 6 cm
Area of the sector
A = \(\frac{l r}{2}\) sq. units
= \(\frac{12.5 \times 6}{2}\)
= 12.5 × 3 cm² cm²
= 37.5 cm²
Area of the sector = 37.5 cm²

(iii) length of the arc l = 50 cm
Radius r = 13.5 cm
Area of the sector
A = \(\frac{l r}{2}\) sq. units
= \(\frac{50 \times 13.5}{2}\)
= 25 × 13.5 cm² cm²
= 337.5 cm²
Area of the sector = 337.5 cm²

Question 6.
Find the central angle of each of the sectors whose measures are given below (π = \(\frac{22}{7}\))

Solution:
(i) Radius of the sector = 21 cm
Area of the sector = 462 cm²
\(\frac{l r}{2}\) = 462
\(\frac{l \times 21}{2}\) = 462

∴ Central angle of the sector = 120°

(ii) Radius of the sector = 8.4 cm
Area of the sector = 18.48 cm²

(iii) Radius of the sector = 35 m
Length of the arc l = 44 m

Question 7.
Answer the following questions:
(i) A circle of radius 120 m is divided into 8 equal sectors. Find the length of the arc of each of the sectors.
(ii) A circle of radius 70 cm is divided into 5 equal sectors. Find the area of each of the sectors.
Solution:
(i) Radius of the circle r = 120 m
Number of equal sectors = 8

(ii) Radius of the sector r = 70 cm
Number of equal sectors = 5

Note: We can solve this problem using A = \(\frac{1}{n}\) πr² sq. units also.

Question 8.
Find the area of a sector whose length of the arc is 50 mm and radius is 14 mm.
Solution:
Length of the arc of the sector l = 50 mm
Radius r = 14 mm
Area of the sector = \(\frac{l r}{2}\) sq. units = \(\frac{50 \times 14}{2}\) mm² = 50 × 7 mm² = 350 mm²
Area of the sector = 350 mm²

Question 9.
Find the area of a sector whose perimeter is 64 cm and length of the arc is 44 cm.
Solution:
Length of the arc of the sector l = 44 cm
Perimeter of the sector P = 64 cm
l + 2r = 64 cm
44 + 2 r = 64 .
2 r = 64 – 44
2 r = 20
r = \(\frac{20}{2}\) = 10 cm²
Area of the sector = \(\frac{l r}{2}\) sq. units = \(\frac{44 \times 10}{2}\) cm² = 22 × 10 cm² = 220 cm²
Area of the sector = 220 cm²

Question 10.
A sector of radius 4.2 cm has an area 9.24 cm². Find its perimeter
Solution:
Radius of the sector r = 4.2 cm
‘ Area of the sector = 9.24 cm²
\(\frac{l r}{2}\) = 9.24
\(\frac{l \times 4.2}{2}\) = 9.24
l × 2.1 = 9.24
l = \(\frac{9.24}{2.1}\)
l = 4.4 cm
Perimeter of the sector = 1 + 2r units = 4.4 + 2(4.2) cm
= 4.4 + 8.4 cm = 12. 8 cm
Perimeter of the sector = 12.8 cm

Question 11.
Infront of a house, flower plants are grown in a circular quadrant shaped pot whose radius is 2 feet. Find the area of the pot in which the plants grow. (π = 3.14)

Solution:
Central angle of the quadrant = 90°
Radius of the circle = 2 feet

Area of the quadrant = 3.14 sq. feet (approximately)

Question 12.
Dhamu fixes a square tile of 30 cm on the floor. The tile has a sector design on it as shown in the figure. Find the area of the sector, (π = 3.14).

Solution:
Side of the square = 30 cm
∴ Radius of the sector design = 30 cm
Given design in the design of a circular quadrant.
Area of the quadrant = \(\frac{1}{4}\) πr² sq. units
= \(\frac{1}{4}\) × 3.14 × 30 × 30 cm²
= 3.14 × 15 × 15 cm²
∴ Area of the sector design = 706.5 cm² (approximately)

Question 13.
A circle is formed with 8 equal granite stones as shown in the figure each of radius 56 cm and whose central angle is 45°. Find the area of each of the granite. (π = \(\frac{22}{7}\))

Solution:
Number of equal sectors ‘n’ = 8
Radius of the sector ‘r’ = 56 cm
Area of the each sector = \(\frac{1}{n}\) πr² sq. units
= \(\frac{1}{8} \times \frac{22}{7}\) × 56 × 56 cm² = 1232 cm²
Area of each sector = 1232 cm² (approximately)