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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 = \(\frac{\theta^{\circ}}{360^{\circ}}\) × πr^{2}

A = \(\frac{45^{\circ}}{360^{\circ}}\) × 3.14 × 16 × 16

A = 100.48 cm^{2}

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

= \(\frac{48 \times 10}{2}\) m^{2}

= 24 × 10 m^{2}

= 240 m^{2}

Area of the sector = 240 m^{2}

(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 cm^{2} cm^{2}

= 37.5 cm^{2}

Area of the sector = 37.5 cm^{2}

(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 cm^{2} cm^{2}

= 337.5 cm^{2}

Area of the sector = 337.5 cm^{2}

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 cm^{2}

\(\frac{l r}{2}\) = 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 cm^{2}

(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 = \(\frac{1}{n}\) πr^{2} 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 = \(\frac{50 \times 14}{2}\) mm^{2} = 50 × 7 mm^{2} = 350 mm^{2}

Area of the sector = 350 mm^{2}

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 cm^{2}

Area of the sector = \(\frac{l r}{2}\) sq. units = \(\frac{44 \times 10}{2}\) cm^{2} = 22 × 10 cm^{2} = 220 cm^{2}

Area of the sector = 220 cm^{2}

Question 10.

A sector of radius 4.2 cm has an area 9.24 cm^{2}. Find its perimeter

Solution:

Radius of the sector r = 4.2 cm

‘ Area of the sector = 9.24 cm^{2}

\(\frac{l r}{2}\) = 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 = \(\frac{1}{4}\) πr^{2} sq. units

= \(\frac{1}{4}\) × 3.14 × 30 × 30 cm^{2}

= 3.14 × 15 × 15 cm^{2}

∴ Area of the sector design = 706.5 cm^{2} (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 = \(\frac{1}{n}\) πr^{2} sq. units

= \(\frac{1}{8} \times \frac{22}{7}\) × 56 × 56 cm^{2} = 1232 cm^{2}

Area of each sector = 1232 cm^{2} (approximately)