Tangents And Secants To A Circle Class 10th Mathematics AP Board Solution

Class 10th Mathematics AP Board Solution
Exercise 9.1
  1. Fill in the blanks i. A tangent to a circle intersects it in …………….. point (s).…
  2. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the…
  3. Draw a circle and two lines parallel to a given line such that one is a tangent…
  4. Calculate the length of tangent from a point 15 cm. away from the circle of a…
  5. Prove that the tangents to a circle at the end points of a diameter are…
Exercise 9.2
  1. The angle between a tangent to a circle and the radius drawn at the point of…
  2. From a point Q, the length of the tangent to a circle is 24 cm. and the…
  3. If AP and AQ are the two tangents a circle with centre O so that ∠POQ = 110°,…
  4. If tangents PA and PB from a point P to a circle with centre O are inclined to…
  5. In the figure XY and X^1 Y^1 are two parallel tangents to a circle with centre…
  6. Two concentric circles of radii 5 cm and 3 cm are drawn. Find the length of the…
  7. Prove that the parallelogram circumscribing a circle is a rhombus.…
  8. A triangle ABC is drawn to circumscribe a circle of radius 3 cm. such that the…
  9. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct…
  10. Construct a tangent to a circle of radius 4 cm from a point on the concentric…
  11. Draw a circle with the help of a bangle, Take a point outside the circle.…
  12. In a right triangle ABC, a circle with a side AB as diameter is drawn to…
  13. Draw a tangent to a given circle with center O from a point ‘R’ outside the…
Exercise 9.3
  1. A chord of a circle of radius 10 cm. subtends a right angle at the centre. Find…
  2. A chord of a circle of radius 12 cm. subtends an angle of 120o at the centre.…
  3. A car has two wipers which do not overlap. Each wiper has a blade of length 25…
  4. Find the area of the shaded region in figure, where ABCD is a square of side 10…
  5. Find the area of the shaded region in figure, if ABCD is a square of side 7 cm.…
  6. In figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD…
  7. AB and CD are respectively arcs of two concentric circles of radii 21 cm. and 7…
  8. Calculate the area of the designed region in figure, common between the two…

Exercise 9.1
Question 1.

Fill in the blanks

i. A tangent to a circle intersects it in …………….. point (s).

ii. A line intersecting a circle in two points is called a ……………..

iii. A circle can have ……………… parallel tangents at the most.

iv. The common point of a tangent to a circle and the circle is called ……………

v. We can draw ……………….. tangents to a given circle.


Answer:

i. One


Property- a tangent to a circle touches it at only one point called the common point.


ii. Secant


Secant- a line that touches the circle at two different points


ii. Two


A circle can have two tangents that are parallel, these two tangents touch the circle on the opposite sides, and distance between the point of contacts is the diameter.


iv. Point of contact


The point where tangent touches the circle is called point of contact.


v. Infinite


We can draw infinite tangents to a circle, as a circle can be assumed to be curve having infinite points and one tangent can be drawn from each point on the circle.



Question 2.

A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that P OQ = 12 cm. Find length of PQ.


Answer:


We know that tangent to a circle makes a right angle with radius.


∠OPQ = 90°


Applying Pythagoras


PQ2 = OP2 + OQ2


PQ2 = 52 + 122


PQ2 = 25 + 144


PQ2 = 169


PQ = 13cm



Question 3.

Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.


Answer:


Step1: Draw circle of random radius with BC as diameter.


Step2: Draw line AD perpendicular to BC to touch the circle at D


Such that ∠D = 90°


Step3: Draw line through D parallel to BC that will form a tangent


Step4: Take a random point on line AD as F and draw a line through F parallel to BC to intersect circle in two points that forms a secant.



Question 4.

Calculate the length of tangent from a point 15 cm. away from the circle of a circle of radius 9 cm.


Answer:


We know that tangent to a circle makes a right angle with radius.


∠OPQ = 90°


Applying Pythagoras


PQ2 = OP2 + OQ2


PQ2 = 92 + 152


PQ2 = 81 + 225


PQ2 = 306


PQ =  cm


Length of tangent =  cm



Question 5.

Prove that the tangents to a circle at the end points of a diameter are parallel.


Answer:


To prove: DE ∥ FG


Proof:


We know that tangent to a circle makes a right angle with the radius.


Let DE and FG be tangent at B and C respectively.


BC forms the diameter.


∴ ∠OBE = ∠OBD = ∠OCG = ∠OCF = 90°


Also, ∠OBD = ∠OCG and ∠OBE = ∠OCF as alternate angles


∴ DE and FG make 90° to same line BC which is the diameter.


Thus DE ∥ FG




Exercise 9.2
Question 1.

Choose the correct answer and give justification for each.

The angle between a tangent to a circle and the radius drawn at the point of contact is

A. 60o

B. 30o

C. 45o

D. 90o


Answer:

Property of tangent- tangent to a circle makes right angle with radius at point of contact


Question 2.

Choose the correct answer and give justification for each.

From a point Q, the length of the tangent to a circle is 24 cm. and the distance of Q from the centre is 25 cm. The radius of the circle is

A. 7 cm

B. 12 cm

C. 15 cm

D. 24.5 cm


Answer:

Let O be center, OP be radius.


Applying Pythagoras


PQ = 24, OQ = 25


OQ2 = OP2 + PQ2


252 = OP2 + 242


625 = OP2 + 576


OP2 = 49


OP = 7 cm


Radius of Circle = 7 cm


Question 3.

Choose the correct answer and give justification for each.

If AP and AQ are the two tangents a circle with centre O so that ∠POQ = 110°, then ∠PAQ is equal to



A. 60o

B. 70o

C. 80o

D. 90o


Answer:

.


Question 4.

Choose the correct answer and give justification for each.

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80o, then ∠POA is equal to

A. 50o

B. 60o

C. 70o

D. 80o


Answer:


We know that tangent to a circle makes right angle with radius.


∴ ∠A = ∠B = 90° and ∠P = 80°


Sum of all angles of quadrilateral = 180°


∴ ∠A + ∠B + ∠P + ∠O = 360°


∴ 90° + 90° + 80° + ∠O = 360°


∠O = 100°


∠ AOP = ∠ BOP


∴ ∠POA = 50°


Question 5.

Choose the correct answer and give justification for each.

In the figure XY and X1Y1 are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X1Y1 at B then ∠AOB =



A. 80o

B. 100o

C. 90o

D. 60o


Answer:


Construct Line OC = radius


OP = OQ = OC = radius


OC∥AP and AC∥OP and


Thus AP = OP = radius


As AP = OP and ∠P = 90°


ΔOAP is isosceles triangle


∴ ∠ PAO = ∠POA = 45°


Also ∠ PAC = 90°


∠OAC = 45° ---1


Also AB is perpendicular to OC


∠OCA = 90°


In ΔAOC


∠OCA + ∠OAC + ∠COA = 180°


45 + 90 + ∠COA = 180°


∠COA = 45°


Similarly


∠BOC = 45°


∴ ∠AOB = 90°


Question 6.

Two concentric circles of radii 5 cm and 3 cm are drawn. Find the length of the chord of the larger circle which touches the smaller circle.


Answer:


AB = AF = 5cm radius of larger circle


AD = AC = 3cm radius of smaller circle


Pythagoras theorem


AF2 = AD2 + DF2


52 = 32 + DF2


252 = 92 + DF2


DF = 4 units


Length of chord = 2 × DF = 2 × 4 = 8cm



Question 7.

Prove that the parallelogram circumscribing a circle is a rhombus.


Answer:


FGHI is a parallelogram


∴ HI = FG and FI = GH----1


Also


IB = IE tangents to circle from I


And similarly


HE = HD, CG = GD and CF = BF


Adding all the equations


IE + HE + GC + CF = BF + BI + GD + DH


IH + GF = IF + GH


∴ 2HI = 2HG


HI = HG---2


∴ GF = GH = HI = IF


From 1 and 2


Thus FGHI is a rhombus



Question 8.

A triangle ABC is drawn to circumscribe a circle of radius 3 cm. such that the segments BD and DC into which BC is divided by the point of contact D are of length 9 cm. and 3 cm. respectively (See adjacent figure). Find the sides AB and AC.



Answer:


Construction : Draw radius OB = 3cm


Proof: AC, BC and AB are tangents


BF = BD = 9cm—tangents from B


AF = AB—tangent from A


∴ CD = CB tangents from C


OD = OB = 3cm radius


OBCD forms a square of side 3cm


OD = OB = BC = CD = 3cm


∴ ∠BCD = 90°


BD = 9cm and DC = 3cm


OB = 3cm radius of circle


Let AF = AB = x


Applying Pythagoras


AB = BC + AC


(9 + x)2 = (12)2 + (3 + x)2


81 + 18x + x2 = 144 + 9 + 6x + x2


12x = 72


X = 6cm


∴ AB = 9 + 6 = 15cm


AC = 6 + 3 = 9cm



Question 9.

Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Verify by using Pythogoras Theorem.


Answer:

Step1: Draw circle of radius 6cm with center A, mark point C at 10 cm from center



Step 2: find perpendicular bisector of AC



Step3: Take this point as center and draw a circle through A and C



Step4:Mark the point where this circle intersects our circle and draw tangents through C



Length of tangents = 8cm


AE is perpendicular to CE (tangent and radius relation)


In ΔACE


AC becomes hypotenuse


AC2 = CE2 + AE2


102 = CE2 + 62


CE2 = 100-36


CE2 = 64


CE = 8cm



Question 10.

Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.


Answer:

Step1:Draw circles of radius 4 and 6 cm



Step 3: Draw tangent to inner circle from C



AD is perpendicular to DC- tangent and radius


In Δ ADC


AC is radius


AC2 = AD2 + DC2


62 = 42 + DC2


36 = 16 + DC2


DC2 = 20


DC = 2 cm



Question 11.

Draw a circle with the help of a bangle, Take a point outside the circle. Construct the pair of tangents from this point to the circle measure them. Write conclusion.


Answer:

Step1: Draw a circle with bangle, its center is not known



Step2: to find the center draw 2 random chords C and FE



Step 3:Find the perpendicular bisectors of these chords, the points where they intersect is the center of circle



Step 4:


Draw a line from center to a point outside it.


Step 5: draw its perpendicular bisector



Step 6: Cut arcs over the circle with this point as center and point outside it as radius


Step 7: Draw tangent to the circle where the arcs cut the circle



Question 12.

In a right triangle ABC, a circle with a side AB as diameter is drawn to intersect the hypotenuse AC in P. Prove that the tangent to the circle at P bisects the side BC.



Answer:

ΔABC is right angled triangle


∠ ABC = 90°


Drawn with AB as diameter that intersects AC at P, PQ is the tangent to the circle which intersects BC at Q.


Join BP.


PQ and BQ are tangents from an external point Q.


∴ PQ = BQ ---1 tangent from external point


⇒ ∠PBQ = ∠BPQ = 45° (isosceles triangle)


Given that, AB is the diameter of the circle.


∴ ∠APB = 90° (Angle subtended by diameter)


∠APB + ∠BPC = 180° (Linear pair)


∴ ∠BPC = 180° – ∠APB = 180° – 90° = 90°


Consider ΔBPC,


∠BPC + ∠PBC + ∠PCB = 180° (Angle sum property of a triangle)


∴ ∠PBC + ∠PCB = 180° – ∠BPC = 180° – 90° = 90° ---2


∠BPC = 90°


∴ ∠BPQ + ∠CPQ = 90° ...3


From equations 2 and 3, we get


∠PBC + ∠PCB = ∠BPQ + ∠CPQ


⇒ ∠PCQ = ∠CPQ (Since, ∠BPQ = ∠PBQ)


Consider ΔPQC,


∠PCQ = ∠CPQ


∴ PQ = QC ---4


From equations 1 and 4, we get


BQ = QC


Therefore, tangent at P bisects the side BC.



Question 13.

Draw a tangent to a given circle with center O from a point ‘R’ outside the circle. How many tangent can be drawn to the circle from that point?

Hint : The distance of two points to the point of contact is the same.


Answer:

Let O be center of circle and R be a point outside it.


Draw tangent to circle at A and B through R





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


Answer:


i. Let Major segment A1 minor segment be A2


∠A = 90°







area of sector ACD = 78.5cm2


Pythagoras


AC2 = CD2 + AD2


AC2 = 102 + 102


AC2 = 200


AC = 10


Height of triangle =


Cos 45° = 



AE = 


Area of minor segment = Area of sector ACD – Area ΔACD


= 78.5 –10


= 78.5-50


= 28.5 cm2


ii. Major segment = Area of circle – minor segment


= π × r2 –minor segment


= π × 102 –28.5


= 314-28.5


= 285.5cm2



Question 2.

A chord of a circle of radius 12 cm. subtends an angle of 120o at the centre. Find the area of the corresponding minor segment of the circle

(use π = 3.14 and √3 = 1.732)


Answer:


Let Major segment A1 minor segment be A2







area of sector ACD = 150.72 sq.cm


Sin 30° = 



AE = 6cm


Cos 30° = 



DE = 1.732 × 6


DE = 10.392cm


CD = 2 × 10.392 = 20.784cm


Area of minor segment = Area of sector ACD – Area ΔACD


= 150.72 –20.784


= 1550.72-62.352


= 88.44 cm2



Question 3.

A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm. sweeping through an angle of 115o. Find the total area cleaned at each sweep of the blades.

(use π = 22/7)


Answer:

Radius = 25cm


Angle = 115°






Area of sector ACD = 627.22cm2


For 2 such wipers: 2 × 627.22


Area = 1254.45 cm2



Question 4.

Find the area of the shaded region in figure, where ABCD is a square of side 10 cm. and semicircles are drawn with each side of the square as diameter

(use π = 3.14)



Answer:

Consider midpoint side AB, it forms Smaller square of size 5cm


For this smaller square


Area of shaded region


= Area of 1st quadrant + area of 2nd quadrant –area of square


Area of both quadrants is same


∴ Area of shaded region


= 2 × Area of quadrant - area of square


= 2 ×  – r × r


= 2 ×  – 5 × 5


= 2 ×  – 5 × 5


= 14.25 cm2


Area of total shaded region = 4 × 14.25


= 57 cm2



Question 5.

Find the area of the shaded region in figure, if ABCD is a square of side 7 cm. and APD and BPC are semicircles.
(use π = 22/7)



Answer:

Area of shaded region = Area of square – area of 2 semicircles


Area of square = 7 × 7 = 49 sq.cm


Area of semicircles = 


= π × r2


= π × 3.52


= 38.5 sq.cm


Area of shaded region = 49-38.5 = 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 = 22/7)



Answer:

Area of Shaded region = Area of sector OABC-Area of ΔDOB


OB = OA = 3.5cm






area of sector OABC = 9.625 cm2


Area of ΔDOB = 


= 3.5 cm2


Area of Shaded region = 9.625-3.5


= 6.125 cm2



Question 7.

AB and CD are respectively arcs of two concentric circles of radii 21 cm. and 7 cm. with centre O (See figure). If ∠AOB = 30°, find the area of the shaded region.

(use π = 22/7)



Answer:

Area of shaded region = Area(Sector OAB)-Area(Sector OCD)




= 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)



Answer:

Area of shaded region


= Area of 1st quadrant + area of 2nd quadrant –area of square


Area of both quadrants is same


∴ Area of shaded region


= 2 × Area of quadrant - area of square


= 2 ×  – r × r


= 2 ×  – 10 × 10


= 2 ×  – 10 × 10


= 57.07 cm2


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