Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 2 Measurements Ex 2.1 Text Book Back Questions and Answers, Notes.

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

Question 1.

Fill in the blanks:

(i) The ratio between the circumference and diameter of any circle is _______ .

Answer:

π

(ii) A line segment which joins any two points on a circle is a _______ .

Answer:

Chord

(iii) The longest chord of a circle is _______ .

Answer:

Diameter

(iv) The radius of a circle of diameter 24 cm is _______ .

Answer:

12 cm

(v) A part of circumference of a circle is called as _______ .

Answer:

an arc

Question 2.

Match the following

Answer:

(i) – c

(ii) – d

(iii) – e

(iv) – b

(v) – a

Question 3.

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

Answer:

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

Answer:

(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} sq. units

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

Answer:

∴ r = \(\frac{12.6}{2}\) cm

r = 6.3cm

Length of the arc l = \(\frac{\theta^{\circ}}{360^{\circ}}\) × 2πr units

l = \(\frac{120^{\circ}}{360^{\circ}}\) × 2 × 3.14 × 63 cm

l = 13.188cm

I = 13.19cm

Area of the sector A = \(\frac{\theta^{\circ}}{360^{\circ}}\) × πr^{2} sq. units

A = \(\frac{120^{\circ}}{360^{\circ}}\) × 3 14 × 6.3 × 6.3 cm^{2}

A = 3.14 × 6.3 × 2.1 cm^{2}

A = 41.54 cm^{2}

Perimeter of the sector P = l + 2r cm

P = 13.19 + 2(6.3) cm

= 13.19 + 1.2.6 cm

P = 25.79 cm

Question 5.

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

(i) Length of the arc = 48 m, r = 10 m

Answer:

Area of the sector A = \(\frac{l r}{2}\) sq. units

l = 48m

r = 10m

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

= 24 × 10m^{2}

= 240 m^{2}

Area of the sector = 240 m^{2}

(ii) length of the arc = 50 cm, r = 13.5 cm

Answer:

Length of the arc l = 12.5 cm

Radius r = 6 cm

Area of the sector A = \(\frac{l r}{2}\) sq. units

A = \(\frac{12.5 \times 6}{2}\)

A = 12.5 × 3cm^{2}

A = 37.5 cm^{2}

Area of the sector A = 37.5 cm^{2}

Question 6.

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

(i) area = 462 cm^{2}, r = 21 cm

Answer:

area = 462 cm^{2}, r = 21 cm

Radius of the Sector = 21 cm

Area of the sector = 462 cm^{2}

\(\frac{l r}{2}\) = 462

\(\frac{l \times 21}{2}\) = 462

l = \(\frac{462 \times 2}{21}\)

l = 22 × 2

Length of the arc l = 44 cm

θ° = 120°

∴ Central angle of the sector = 120°

(ii) length of the arc = 44 m, r = 35 m

Answer:

Radius of the sector = 8.4cm

Area of the sector = 18.48 cm^{2}

\(\frac{l r}{2}\) = 18.48

\(\frac{1 \times 8.4}{2}\) = 18.48

l = \(\frac{18.48 \times 2}{8.4}\)

Hint:

Length of the arc l = 4.4 cm

Hint:

θ° = 30°

Central angle = 30°

Question 7.

A circle of radius 120 m is divided into 8 equal sectors. Find the length of the arc of each of the sectors.

Answer:

Radius of the circle r = 120 m

Number of equal sectors = 8

∴ Central angle of each sector = \(\frac{360^{\circ}}{n}\)

θ° = \(\frac{360^{\circ}}{8}\)

θ° = 45°

Length of the arc l = \(\frac{\theta^{\circ}}{360^{\circ}}\) × 2πr units

= \(\frac{45^{\circ}}{360^{\circ}}\) × 2π × 120 m

Length of the arc = 30 × πm

Another method:

l = \(\frac{1}{n}\) × 2πr = \(\frac{1}{8}\) × 2 × π × 120 = 30 π m

Length of the arc = 30 π m

Question 8.

A circle of radius 70 cm is divided into 5 equal sectors. Find the area of each of the sectors.

Answer:

Radius of the sector r = 70 cm

Number of equal sectors = 5

∴ Central angle of each sector = \(\frac{360^{\circ}}{n}\)

θ° = 360°

θ° = 72°

Area of the sector = \(\frac{\theta^{\circ}}{360^{\circ}}\) × πr^{2} sq.units

= \(\frac{72^{\circ}}{360^{\circ}}\) × π × 70 × 70cm^{2}

Hint:

= 14 × 70 × πcm^{2}

= 980 πcm^{2}

Note: We can solve this problem using A = \(\frac{1}{n}\) πr^{2} sq. units also.

Question 9.

Dhamu fixes a square tile of 30cm 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).

Answer:

Side of the square = 30 cm

∴ Radius of the sector design = 30 cm

Given the design of a circular quadrant.

Area of the quadrant = \(\frac{1}{4}\) πr^{2} sq.units

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

= 3.14 × 15 × 15cm^{2}

∴ Area of the sector design = 706.5 cm^{2} (approximately)

Question 10.

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 stones. (π = \(\frac { 22 }{ 7 }\))

Answer:

Number of equal sectors ‘n’ = 8

Radius of the sector ‘r’ = 56 cm

Area of each sector = \(\frac{1}{n}\) πr^{2} sq. units

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

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