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 State Syllabus 10th Class Maths 9th Lesson Important Questions and Answers Tangents and Secants to a Circle

Question 1.

What do we call the part a and b in the below circle ?

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°

OA^{2} = OB^{2} + AB^{2}

⇒ AB^{2} = OA^{2} – OB^{2} = 25^{2} – 7^{2}

⇒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 cm^{2}

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:

ΔOAB is a right triangle.

OA^{2} = 0B^{2} + AB^{2}

5^{2} = OB^{2} + 4^{2}

OB^{2} = 25 -16 = 9

OB = \(\sqrt{9}\) = 3 cm.

Question 5.

Find the area of the shaded part in the given figure.

Solution:

Area of shaded part

= Area of semi-circle – Area of triangle

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

Distance between centre to point be = 13 cm

13^{2} = 5^{2} + x^{2}

x^{2} = 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:

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:

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.

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 cm^{2}

Area of the Minor Segment = Area of sector OACB – Area of Δ OAB

= 78.5 – 50.0 = 28.5 cm^{2}

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.

Area of sector OACB = \(\frac{\mathrm{x}}{360}\) × πr^{2}

= \(\frac{90^{\circ}}{360^{\circ}}\) × 3.14 × 10 × 10

= 78.5 Sq. cm

Area of ∆ AOB = \(\frac { 1 }{ 2 }\) × 10 × 10

= 50 cm^{2}

Area of the Minor Segment = Area of sector OACB – Area of ∆ OAB

= 78.5 – 50.0 = 28.5 cm^{2}

Area of Major Segment = Area of the circle – Area of the Minor Segment

= (3.14 × 10 × 10) – 28.5

= 314-28.5 = 285.5 cm^{2}

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:

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.

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.

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:

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

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

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 = \(\frac{22}{7}\))

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 cm^{2} ……………(1)

Now the area of shaded part = area of semicircle – area of ΔPQR

= \(\frac { 1 }{ 2 }\)πrcm^{2} – 30 cm^{2} …………….(2)

In ΔPQR, QR^{2} = PQ^{2} + PR^{2}

(from Pythagoras theorem)

QR^{2} = 12^{2} + 5^{2}

= 144 + 25 = 169 = 13^{2}

∴ QR =13 then

Radius of the circle (r) = QO = \(\frac{\mathrm{Q} R}{2}\)

= \(\frac{13}{2}\) = 6.5 cm

then area of semicircle

= \(\frac { 1 }{ 2 }\)πrcm^{2}

= \(\frac{1}{2} \times \frac{22}{7} \times \frac{13}{2} \times \frac{13}{2}\) = 66.39 cm^{2} …………….(3)

Now putting the values of (1) and (3) in (2) we get

Area of shaded part = (66.39 – 30)

= 36.39 cm^{2}.

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.

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.

Solution:

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 πrcm^{2}

= \(\frac{120}{360} \times \frac{22}{7}\) × 21 × 21

= 22 × 21 = 462 cm^{2} ……………………… (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° = \(\frac{\mathrm{AD}}{\mathrm{OA}}\)

∴ area of ΔOAB = \(\frac { 1 }{ 2 }\) bh

= \(\frac { 1 }{ 2 }\) × AB × OD = AD × OD

= \(\) = 190.95 cm^{2} ………….. (2)

∴ area of segment \(\widehat{\mathrm{AYB}}\)

= 462 – 190.95

= 271.05 cm^{2}

= 271.05 cm^{2}

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.

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 = \(\frac{77}{3}\) cm^{2} = 25.66 cm^{2}

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

∠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. \(\frac{\sqrt{3}}{4}\) a^{2}

= 6. \(\frac{\sqrt{3}}{4}\) × 14 × 14

Area of hexagon = 294\(\sqrt{3}\) cm^{2}

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)

Now side of square ABCD = 3 + 3 = 6

Then area of square ABCD

= 6 × 6

= 36 cm^{2}

and now area of sector = \(\frac{x}{360}\) × πr^{2}

Area of 4 sectors

= 4 × \(\frac{90}{360} \times \frac{22}{7}\) × 3 × 3

= \(\frac{198}{7}\) = 28.3 cm^{2}

∴ Area in between 4 pans i.e., shaded

= 36 – 28.3 = 7.7 cm^{2}