TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3

Students can practice 10th Class Maths Study Material Telangana Chapter 9 Tangents and Secants to a Circle Ex 9.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Exercise 9.3

Question 1.
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (use π = 3.14)
(i) Minor segment
(ii) Major segment. (A.P. Mar. ’16, June ’15)
Solution:

Given :
Angle subtended by the chord = 90°
Radius of the circle = 10 cm
Area of the minor segment = Area of the sector POQ – Area of ∆POQ
Area of the sector = \(\frac{\mathrm{x}^{\circ}}{360^{\circ}}\) × πr²
= \(\frac{90}{360}\) × 3.14 × 10 × 10 = 78.5
Area of the triangle POQ
= \(\frac{1}{2}\) × Base × Height
= \(\frac{1}{2}\) × 10 × 10 = 50
∴ Area of the minor segement
⇒ 78.5 – 50 = 28.5 cm²
Area of the major segment = Area of the circle – Area of the minor segment
⇒ 3.14 × 10 × 10 – 28.5
⇒ 314 – 28.5 cm²
⇒ 285.5 cm²

Question 2.
A chord of a circle of radius 12 cm. subtends an angle of 120° at the centre. Find the area of the corresponding minor segment of the circle, (use π = 3.14 and \(\sqrt{3}\) = 1.732).
Solution:
Radius of the circle r = 12 cm
Area of the sector = \(\frac{\mathrm{x}^{\circ}}{360^{\circ}}\) πr²
Here x = 120°

= \(\frac{120^{\circ}}{360^{\circ}}\) × 3.14 × 12 × 12
= 150.72
Drop a perpendicular from ‘O’ to the chord ‘PQ’
∆OPM = ∆OQM [∵ OP = OQ, ∠P = ∠Q; angles opposite to equal sides OP & OQ, ∠OMP = ∠OMQ by A.A.S]
∴ ∆OPQ = ∆OPM + ∆OQM
= 2(∆OPM)
Area of ∆OPM = \(\frac{1}{2}\) × PM × OM
But cos 30° = \(\frac{\mathrm{PM}}{\mathrm{OP}}\)
[∴ In ∆OPQ ∠POQ = 120° ∠OPQ = ∠OQP = \(\frac{180-120^{\circ}}{2}\) = 120°]
∴ PM = \(\frac{12 \times \sqrt{3}}{2}\) = 6\(\sqrt{3}\)
Also sin 30° = \(\frac{\mathrm{OM}}{\mathrm{OP}}\)
⇒ \(\frac{1}{2}\) = \(\frac{\mathrm{OM}}{12}\) ⇒ OM = \(\frac{12}{2}\) = 6
∴ ∆OPM = \(\frac{1}{2}\) × 6\(\sqrt{3}\) × 6
= 18 × 1.732 = 31.176 cm²
∴ ∆OPQ = 2 × 31.176 = 62.352 cm²
Area of the minor segment PQ = (Area of the sector) – (Area of the ∆OPQ)
= 150.72 – 62.352 = 88.368 cm²

Question 3.
A car has two wipers which do not overlap. Each wiper has a blade of length 25cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades (use π = \(\frac{22}{7}\))
Solution:

Angle made by the each blade = 115°
It is evident that each wiper sweeps a sector of a circle of radius 25cm and sector angle 115°.
Area of a sector = \(\frac{\theta^{\circ}}{360^{\circ}}\) × πr²
Total area cleaned at each sweep of the blades
= 2 × \(\frac{\theta^{\circ}}{360^{\circ}}\) × πr²
= 2 × \(\frac{115^{\circ}}{360^{\circ}}\) × \(\frac{22}{7}\) × 25 × 25
= 2 × \(\frac{23}{72}\) × \(\frac{22}{7}\) × 25 × 25
= \(\frac{23}{36}\) × \(\frac{22}{7}\) × 25 × 25
= 1254.96 cm²

Question 4.
Find the area of the shaded region in the figure, where ABCD is a square of side 10cm and semicircles are drawn with each side of the square as diameter. (use π = 3.14).

Solution:
Let us mark the four unshaded regions as I, II, III and IV.

Area of I + Area of III = Area of ABCD – Areas of 2 semicircles with radius 5 cm.
= 10 × 10 – 2 × \(\frac{1}{2}\) × 7π × 5²
= 100 – 3.14 × 25
⇒ 100 – 78.5 ⇒ 21.5 cm²
Similarly, Area of II + Area of IV = 21.5 cm²
So, area of the shaded region
= are a of ABCD – Area of unshaded region
⇒ 100 – 2 × 21.5
= 100 – 43 = 57 cm²

Question 5.
Find the area of the shaded region in the figure, if ABCD is a square of side 7cm and APD and BPC are semi-circles. (use π = \(\frac{22}{7}\))

Solution:
Given,
ABCD is a square of side 7 cm
Area of the shaded region
= Area of ABCD – Area of 2 semi-circles with 7 radius (\(\frac{7}{2}\) = 3.5 cm)
APD and BPC are semi-circles 1 22
= 7 × 7 – 2 × \(\frac{1}{2}\) × \(\frac{22}{7}\) × 3.5 × 3.5
= 49 – 38.5 = 10.5 cm²
∴ Area of shaded region = 10.5 cm²

Question 6.
In figure, OACB is a quadrant of a circle with centre ‘O’ and radius 3.5 cm. If OD = 2 cm, find the area of the shaded region, (use π = \(\frac{22}{7}\)).

Solution:
Given OACB is a quadrant of a circle radius 3.5 cm OD = 2 cm
Area of the shaded region = Area of the sector – Area of ABOD
= \(\frac{\mathrm{x}^{\circ}}{360^{\circ}}\) × πr² – \(\frac{1}{2}\) OB . OD
= \(\frac{99^{\circ}}{360^{\circ}}\) × \(\frac{22}{7}\) × 3.5 × 3.5 – \(\frac{1}{2}\) × 3.5 × 2
= \(\frac{1}{4}\) × \(\frac{22}{7}\) × 3.5 × 3.5 – 3.5
= \(\frac{1}{4}\) × \(\frac{22}{7}\) × 12.25 – 3.5
= 9.625 – 3.5 = 6.125 cm²
Area of shaded region = 6.125 cm²

Question 7.
AB and CD are respectively arcs of two concentric circles of radii 21cm and 7cm with centre ‘O’ (see figure) If ∠AOB = 30°, find the area of the shaded region. (Use π = \(\frac{22}{7}\))

Solution:
Area of the shaded region
= Area of sector OAB – Area of the sector COD
= \(\frac{30^{\circ}}{360^{\circ}} \) × \(\frac{22}{7}\) × 21 × 21 – \(\frac{30^{\circ}}{360^{\circ}} \) × \(\frac{22}{7}\) × 7 × 7
= \(\frac{30^{\circ}}{360^{\circ}} \) × \(\frac{22}{7}\) (21 × 21 – 7 × 7)
= \(\frac{1}{12}\) × \(\frac{22}{7}\) (441 – 49)
= \(\frac{1}{6}\) × \(\frac{11}{7}\) × 392
= \(\frac{1}{6}\) × 11 × 56
= \(\frac{11 \times 28}{3}\)
= 102.67 cm²

Question 8.
Calculate the area of the designed region in figure, common between the two quadrants of the circles of radius 10 cm. each, (use π = 3.14)

Solution:
Radius of the circle (r) = 10 cm
Area of the designed region
= 2 [Area of quadrant ABYD – Area of ∆ABD]
= 2 [\(\frac{1}{4}\) × πr² – \(\frac{1}{2}\) × Base × Height]
= 2 [(\(\frac{1}{4}\) × 3.14 × 10 × 10) – (\(\frac{1}{2}\) × 10 × 10)]
= 2 [78.5 – 50]
⇒ 2 × 28.5
= 57 cm²