AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle

These AP 10th Class Maths Chapter Wise Important Questions Chapter 9 Tangents and Secants to a Circle will help students prepare well for the exams.

AP SSC 10th Class Maths Important Questions 9th Lesson Tangents and Secants to a Circle

Question 1.
What do we call the part a and b in the below circle ?
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 1
Solution:
‘a’ is minor segment and ‘b’ is major segment.

Question 2.
Find the length of the tangent to a circle of radius 7 cm at a point from a distance 25 cm from the centre.
Solution:
Given OA = 25 cm, OB = r = 7 cm
In ΔAOB, ∠B = 90°
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 2
OA2 = OB2 + AB2
⇒ AB2 = OA2 – OB2 = 252 – 72
⇒AB = \(\sqrt{25^{2}-7^{2}}=\sqrt{625-49}\)
= \(\sqrt{576}\)
= 24 cm

Question 3.
Find the area of a sector of a circle whose radius is 7 cm and angle at the centre is 60°.
Solution:
Radius = 7 cm, Angle at centre = 60°
Area of the sector = \(\frac{\mathrm{x}}{360}\) x πr²
= \(\frac{60 \times \frac{22}{7} \times 7 \times 7}{360}=\frac{154}{6}\) = 25.66 cm2

AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle

Question 4.
A tangent is drawn to a circle of radius 4 cm. from a point that lies at a distance of 5 cm. from the centre. Find the measure of length of the tangent.
Solution:
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 3
ΔOAB is a right triangle.
OA2 = 0B2 + AB2
52 = OB2 + 42
OB2 = 25 -16 = 9
OB = \(\sqrt{9}\) = 3 cm.

Question 5.
Find the area of the shaded part in the given figure.
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 4
Solution:
Area of shaded part
= Area of semi-circle – Area of triangle
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 5

Question 1.
Find the length of the tangent from a point 13 cm away from the centre of the circle of radius 5 cm.
Solution:
Radius of the circle = 5 cm
Length of the tangent = x cm
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 6
Distance between centre to point be = 13 cm
132 = 52 + x2
x2 = 169 – 25 = 144 ⇒ x = 12 cm

Question 2.
If tangents PA and PB from a point P to a circle with centre ‘O’ are inclined to each other at angle of 80°, then find the measured of ∠POA.
Solution:
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 7
Between ΔA0P and ΔB0P
∠OAP = ∠OBP = 90° (∵ tangent and line from origin meet at 90° to each other)
0A=0B = radius of the circle = r (say)
and OP is the common side
Hence, we can say ΔAOP = ΔBOP
Therefore, we can say
∠OPA = ∠OPB = \(\frac { 1 }{ 2 }\)(∠APB)
= \(\frac { 1 }{ 2 }\)(80°) = 40°
(∵ given that tangents PA and PB are inclined to each other by 80°)
Now, in ΔAOP,
∠POA + ∠OPA + ∠A = 180°
⇒ ∠POA + 40° + 90° = 180°
⇒ ∠POA = 50°

Question 3.
Draw a circle of radius 3 cm, mark a point ‘P’ on the circle and draw a tan gent at ‘P’.
Solution:
Steps of construction:
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 8

1) Draw a circle of radius ‘3’ cm from the centre “O” and pick a point ‘P’ on the circle. Join \(\overline{\mathrm{OP}}\)
2) Now draw a perpendicular at the point ‘P’ to the line segment \(\overline{\mathrm{OP}}\)
such that XY ⊥ \(\overline{\mathrm{OP}}\).
3) Then \(\overline{\mathrm{XY}}\) is the desired tangent at ‘P’ to the given circle.

Question 4.
A chord of a circle of radius 10 cm. subtends a right angle at the centre. Find the area of the corresponding minor segment (use π = 3.14),
Solution:
Radius of circle (r) = 10 cm
Sector angle (x) = 90°
Radius of sector (r) = 10 cm.
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 9
Area of sector OACB = \(\frac{\mathrm{x}}{360}\) × πr²
= \(\frac{90^{\circ}}{360^{\circ}}\) × 3.14 × 10 × 10
= 78.5 Sq. cm
Area of Δ AOB = \(\frac { 1 }{ 2 }\) × 10 × 10
= 50 cm2
Area of the Minor Segment = Area of sector OACB – Area of Δ OAB
= 78.5 – 50.0 = 28.5 cm2

AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle

Question 1.
A chord of circle of radius 10 cm sub-tends a right angle at the centre. Find the area of the corresponding :
i) Minor segment ii) Major segment (use π = 3. 14)
Solution:
Radius of circle (r) = 10 cm
Sector angle (x) = 90°
Radius of sector (r) = 10 cm.
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 10
Area of sector OACB = \(\frac{\mathrm{x}}{360}\) × πr2
= \(\frac{90^{\circ}}{360^{\circ}}\) × 3.14 × 10 × 10
= 78.5 Sq. cm
Area of ∆ AOB = \(\frac { 1 }{ 2 }\) × 10 × 10
= 50 cm2
Area of the Minor Segment = Area of sector OACB – Area of ∆ OAB
= 78.5 – 50.0 = 28.5 cm2
Area of Major Segment = Area of the circle – Area of the Minor Segment
= (3.14 × 10 × 10) – 28.5
= 314-28.5 = 285.5 cm2

Question 2.
Draw a circle of radius 3 cm. Take a point ’P’ at a distance of 5 cm from the centre of the circle. From P, draw 2 tangents to the circle.
Solution:
Award marks for construction as follows:
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 11
i) To draw a circle with radius 3 cm
ii) To plot a point P such that OP = 5 cm.
iii) To Bisect QP at M and draw circle with radius OM br MP
iv) To draw tangents from intersecting points of two circles.

Question 3.
Draw a Circle of radius 4 cm. From a point 7.5 cm away from its centre, construct the pair of tangents to the circle.
Solution:
1) Draw a circle with:radius 4 cm with centre O.
2) Locate a point P such that OP 7.5 cm
3) Bisect OP and draw circle with radius MO or MP with centre M.
4) Draw tangents PA and PB from external point P to the given circle.
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 12

Question 4.
Draw a circle of radius 5 cm. From a point 8 cm away from its centre, con-struct a pair of tangents to the circle. Find the lengths of tangents.
Solution:
Steps of construction :
1) Construct a circle with a radius of 5 cm.
2) Trace the point ‘p’ in the exterior of the circle which is at a distance of ‘8’ cm from its centre.
3) Construct a perpendicular bisector to OP which meets at M.
4) The draw a circle with a radius of MP or MO from the point M. This circle cuts the previous circle drawn from the centre ‘O’ at the points A and B.
5) Now join the points PA and then PB.
6) PA, PB are the required tangents which are measured 6.2 cm long.
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 13
OA = 5 cm ; OP = 8 cm
AP = PB = 6.2 cm.

Question 5.
Draw a circle of radius 4 cm and draw a pair of tangents to the circle, which are intersecting each other 6 cm away from the centre.
Solution:
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 14
Steps of Construction:
1) Draw a circle with centre ‘O’ and radius 4 cm.
Take a point ‘P’ outside the circle such that OP = 6 cm. Join OP.
Draw the perpendicular bisector to OP which bisects it at M.
4) Taking M as centre and PM or MO as radius draw a circle.
Let the circle intersects the given circle at ’A’ and ‘B’.
Join P to A and B.
PA and PB are the required tangents of lengths.

Question 6.
Two tangents TP and TQ are drawn to a circle with centre ‘O’ from an external point T, then prove that ∠PTQ = 2. ∠OPQ.
Solution:
Given a circle with centre 0.
Two tangents TP, TQ are drawn to the circle from an external point T.
We need to prove ∠PTQ = 2∠OPQ
Let ∠PTQ = θ
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 15
TP = TQ (The lengths of tangents drawn from an external point to a circle are equal)
So ΔTPQ is an isosceles triangle
∴ ∠TPQ + ∠TQP + ∠PTQ = 180°
(Sum of three angles in a triangle)
∠TPQ = ∠TQP = \(\frac { 1 }{ 2 }\)(180° – θ)
= 90°- \(\frac{\theta}{2}\)
∠OPQ = ∠OPT – ∠TPQ
= 90° – 0(90° – \(\frac{\theta}{2}\)) = \(\frac{\theta}{2}\)
= \(\frac { 1 }{ 2 }\) ∠PTQ
∴∠PTQ = 2∠OPQ

AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle

Question 7.
Find the area of the segment shaded in the figure in which PQ = 12 cm, PR = 5 cm and QR is the diameter of the circle with centre ‘O’. (Take π = \(\frac{22}{7}\))
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 16
Solution:
To find the area of the segment shaded in the given figure.
Here ‘PQ’ = 12 cm; ‘PR’ = 5 cm; ‘QR’ is diameter
Now PQOR is a semicircle then angle in a semicircle is 90°.
then ∠QPR = 90°
∴ ΔPQR is a right angled triangle
∴ Area of ΔPQR = \(\frac { 1 }{ 2 }\) bh
= \(\frac { 1 }{ 2 }\) × PQ × PR
= \(\frac { 1 }{ 2 }\) × 12 × 5 =30 cm2 ……………(1)
Now the area of shaded part = area of semicircle – area of ΔPQR
= \(\frac { 1 }{ 2 }\)πrcm2 – 30 cm2 …………….(2)
In ΔPQR, QR2 = PQ2 + PR2
(from Pythagoras theorem)
QR2 = 122 + 52
= 144 + 25 = 169 = 132
∴ QR =13 then
Radius of the circle (r) = QO = \(\frac{\mathrm{Q} R}{2}\)
= \(\frac{13}{2}\) = 6.5 cm
then area of semicircle
= \(\frac { 1 }{ 2 }\)πrcm2
= \(\frac{1}{2} \times \frac{22}{7} \times \frac{13}{2} \times \frac{13}{2}\) = 66.39 cm2 …………….(3)
Now putting the values of (1) and (3) in (2) we get
Area of shaded part = (66.39 – 30)
= 36.39 cm2.

Question 8.
Draw two tangents to a circle of radius 2.5 cm, from a point ‘P’ at a distance of 7 cm from its centre.
Solution:
1. Draw a line segment PO = 7 cm.
2. From the point O, draw a circle of radius = 2.5 cm.
3. Draw a perpendicular bisector of PO. Let M be the mid-point of PO.
4. Taking M as centre and OM as radius, draw a circle.
5. Let this circle intersects the given circle at the point Q and R.
6. Join PQ and PR.
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 17

Question 1.
As shown in the figure, radius of the given circle is 21 cm and ∠AOB = 120°. Then find the area of segment AYB.
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 18
Solution:
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 19
Given radius of area = OA = OB = 21 cm
Now3 the angle at centre for the sector \(\widehat{\mathrm{OAB}}\) = 120°.
Formula for area of sector = \(\frac{x}{360}\) x πrcm2
= \(\frac{120}{360} \times \frac{22}{7}\) × 21 × 21
= 22 × 21 = 462 cm2 ……………………… (1)

Now area of segment \(\widehat{\mathrm{AYB}}\)
= Area of sector – area of Δ OAB.
Let \(\overline{\mathrm{OD}}\) is perpendicular to AB, then
∠AOB = \(\frac{120}{2}\) = 60°
∴ sin 60° = \(\frac{\mathrm{AD}}{\mathrm{OA}}\)
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 20
∴ area of ΔOAB = \(\frac { 1 }{ 2 }\) bh
= \(\frac { 1 }{ 2 }\) × AB × OD = AD × OD
= \(\) = 190.95 cm2 ………….. (2)
∴ area of segment \(\widehat{\mathrm{AYB}}\)
= 462 – 190.95
= 271.05 cm2
= 271.05 cm2

AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle

Question 2.
In a wall clock, length of minutes needle is 7 cm. Then find the area covered by it in 10 minutes of time.
Solution:
Length of minutes needle =(r) = 7 cm
We need to calculate the area covered by in 10 minutes of time = Area of sector.
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 21
Now the angle covered by it in 60 minutes = 360°
∴ In 10 minutes = \(\frac{360}{60} \) × 10 = 60°
Area of sector = \(\) × 7 × 7
= \(\frac{154}{6}=\frac{77}{3}\)
∴ Area covered by it in 10 minutes of time = \(\frac{77}{3}\) cm2 = 25.66 cm2

Question 3.
Find the area of a right hexagon inscribed in a circle having 14 cm of radius.
Solution:
Radius of circle = OA = OB = OC = OD = OE = OF = 14 cm
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 22
∠AOB = \(\frac{360}{6} \) = 60°
∴ In ΔAOB AO = BO = 14 cm
∠AOB = 60°
then ∠OAB = ∠OBA
(∵ Opposite to equal sides)
And ∠OAB + ∠OBA + 60 = 180
=> ∠OAB = ∠OBA = ∠AOB = 60°
Hence it is an equilateral triangle.
∴ OA = OB = AB = 14 cm
∴ Area of hexagon = 6 (Area of ΔAOB)
= 6. \(\frac{\sqrt{3}}{4}\) a2
= 6. \(\frac{\sqrt{3}}{4}\) × 14 × 14
Area of hexagon = 294\(\sqrt{3}\) cm2

Question 4.
Four carrorp board pans are arranged as shown in figure. Radius of the pan is 3 cm each. Then find the area in between of them.
Solution:
Area in between 4 pans = Area of square ABCD formed by joining their centres – 4 (area of sector)
AP 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle 23
Now side of square ABCD = 3 + 3 = 6
Then area of square ABCD
= 6 × 6
= 36 cm2
and now area of sector = \(\frac{x}{360}\) × πr2
Area of 4 sectors
= 4 × \(\frac{90}{360} \times \frac{22}{7}\) × 3 × 3
= \(\frac{198}{7}\) = 28.3 cm2
∴ Area in between 4 pans i.e., shaded
= 36 – 28.3 = 7.7 cm2