9.1 Introduction
We have seen two lines mostly intersect at a point or do not intersect in a plane. In some situations they coincide with each other.
Similarly, what happens when a curve and a line is given in a plane?You know a curve may be a parabola as you have seen in a polynomials or a simple closed curve "circle" which is a collection of all those points on a plane that are at a constant distance from a fixed point.
You might have seen circular objects rolling on a plane creating a path. For example ; riding a bicycle, wheels of train on the track etc., where it seems to be a circle and a line. Does there a relation exist between them?
Let us see what happens, if a circle and a line are given in a plane.
9.1.1 A Line and a Circle
You are given a circle and a line drawn on a paper. Salman argues that there can only be 3 possible ways of presenting them.
Consider a circle 'O' and a line PQ , the three possibilities are given in figure below :
In fig (1) : the line PQ and the circle have no common point. In this case PQ is a non-intersecting line with respect to the circle.
In fig (2) : the line PQ intersects the circle at two points A and B. It forms a chord on the circle AB with two common points. In this case the line PQ is a secant of the circle.
In fig (3) : There is only one point A, common to the line PQ and the circle. This line is called a tangent to the circle.
You can see that there cannot be any other position of the line with respect to the circle.
Do you know?
The word ' tangent' comes from the latin word ' tangere', which means to touch and was introduced by Danish mathematician Thomas Fineke in 1583.
Do This :-
1) Draw a circle with any radius. Draw four tangents at different points. How many tangents can you draw to this circle?
2) How many tangents you can draw to circle from a point away from it.
3) Which of the following are tangents to the circles?
9.2 tangents of a circle
We can see that tangent can be drawn at any point lying on the circle. Can you say how many tangents can be drawn at any point on the surface of the circle.
To understand this let us consider the following activity.
Activity
Take a circular wire and attach a straight wire AB at a point P of the circular wire, so that the system rotate about the point P in a plane.The circular wire represents a circle and the straight wire AB represents a line intersects the circle at a point P.
Put the system on a table and gently rotate the wire AB about the point P to get different positions of the straight wire as shown in the figure.
The wire intersects the circular wire at P and at one more point through the points Q1, Q2 or Q3 etc.
So while it generally intersects circular wire at two points one of which is P in one particular position, it intersects the circle only at the Point P ( see position A'B' of AB ).
This is the position of a tangent at the point P of the circle. You can check that in all other positions of AB it will intersect the circle at P and at another point, A'B' is a tangent to the circle at P.
We see that there is only one tangent to the circle at Point P.
Moving wire AB in either direction from this position makes it cut the circular wire in two points. All these are therefore secants. Tangent is a special case of a secant where the two points of intersection of a line with a circle coincide.
Do This
(*) Draw a circle and a secant PQ of the circle on a paper as shown below. Draw various lines parallel to the secant on both sides of it.
(*) What happens to the length of chord coming closer and closer to the center of the circle?
(*) What is the longest chord?
(*) How many tangents can you draw to a circle, which are parallel to each other?
The common point of the tangent and the circle is called the point of contact and the tangent is said to touch the circle at the common point.
Observe the tangents to the circle in the figure given below :
How many tangents can you draw to a circle at a Point? how many tangents can you obtain to the circle in all?see the points of contact. Draw radii from the points of contact. Do you see anything special about the angle between the tangents and the radii at the points of contact. All appear to be perpendicular to the corresponding tangents. We can also prove it.
Let us see how.
Theorem-9.1 : The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Given : A circle with center ‘O’ and a tangent XY to the circle at a point P.
To prove : OP is perpendicular to XY. (i.e OP ⊥ XY)
Proof : Here, we will use the method that assumes that the statement is wrong and shows that such an assumption leads to a fallacy.
So we will suppose OP is not perpendicular to XY.
Take a point Q on XY other than P and join OQ.
The point Q must lie outside the circle (why?) (Note that if Q lies inside the circle, XY becomes a secant and not a tangent to the circle)
Therefore, OQ is longer than the radius OP of the circle [Why?]
i.e., OQ > OP.
This must happen for all points on the line XY.
It is therefore true that OP is the shortest of all the distances of the point O to the points of XY. So our assumption that OP is not perpendicular to XY is false.
Therefore , OP is perpendicular to XY
Note : The line containing the radius through the point of contact is also called the ‘normal to the circle at the point’.
Try This
How can you prove the converse of the above theorem.
“If a line in the plane of a circle is perpendicular to the radius at its endpoint on the circle, then the line is tangent to the circle”.
We can find some more results using the above theorem
(i) Since there can be only one perpendicular OP at the point P, it follows that one and only one tangent can be drawn to a circle at a given point on the circumference.
(ii) Since there can be only one perpendicular to XY at the point P, it follows that the perpendicular to a tangent at its point of contact passes through the centre.
9.2.1 Construction of Tangent to a Circle
How can we construct a line that would be tangent to a circle at a given point on it?
We use what we just found the tangent has to be perpendicular to the radius at the point of contact.
To draw a tangent through the point of contact we need to draw a line perpendicular to the radius at that point. To draw this radius we need to know the center of the circle. Let us see the steps for this construction.
Construction : Construct a tangent to a circle at a given point when the center of the circle is known.
Steps of Construction :-
We have a circle with center ‘O’ and a point P anywhere on its circumference. Then we have to construct a tangent through P.
1. Draw a circle with center ‘O’ and mark a point ‘P’ anywhere on it. Join OP.
2. Draw a perpendicular line through the point P and name it as XY, as shown in the figure.
3. XY is the required tangent to the given circle passing through P.
Can you draw one more tangent through P ? give reason.
Try This
How can you draw the tangent to a circle at a given point when the center of the circle is not known?
Hint : Draw equal angles ∠QPX and ∠PRQ . Explain the construction.
9.2.2 Finding Length of the Tangent
Can we find the length of the tangent to a circle from a given point? Is the length of tangents from a given point to the circle the same? Let us examine this.
Example : Find the length of the tangent to a circle with centre ‘O’ and radius = 6 cm.
from a point P such that OP = 10 cm.
Solution : Tangent is perpendicular to the radius at the point of contact (Theorem 9.1)
Here PA is tangent segment and OA is radius of circle
∴ OA ⊥ PA ⇒∠OAP 90 deg
Now in ∆OAP,
OP^2 = OA^2 + PA^2 (pythagoras theorem)
10^2 = 6^2 + PA^2
100 = 36 + PA^2
PA^2 = 100 - 36 = 64
∴ PA = √64 = 8 cm
(1) Fill in the blanks
1) A tangent to a circle intersects it in ............ points(s)
A) One
2) A line intersecting a circle in two points is called a .............
A) Secant of a circle.
3) A circle can have ........... parallel tangents at the most.
A) Two
4) The common point of a tangent to a circle and the circle is called.............
A) Point of contact
5) We can draw..........tangents to a given circle.
A) Infinite
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(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 OQ = 12cm. Find length of PQ.
sol)
Tangent :- It is perpendicular to the radius at point of contact.
Here PQ is tangent segment and OP is radius of circle.
OP _|_ PQ
/_OPQ = 90
In /_\ OPQ,
Using Pythagoras Theorem :-
OQ^2 = OP^2 + PQ^2
(12)^2 = (5)^2 + PQ^2
144 = 25 + PQ^2
119 = PQ^2
PQ = _/119 cm
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(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.
sol)
Steps of Construction :-
1) Draw a circle with centre 'O' and radius "r"
2) mark the point "P" anywhere on the circles Join OP
3) Draw a perpendicular line through the point "P" and name it as "AB", as shown in figure.
4) AB is the required tangent to the given circle.
5) Draw another line parallel to the tangent AB and intersect circle at two points and name it as CD
6) CD is the required secant.
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(4) Calculate the length of tangent from a point 15cm away from the centre of a circle of radius 9 cm.
sol)
Here PQ is tangent segment and OP is radius of circle
OP _|_ PQ
/_OPQ = 90
Now in /_\ OPQ
Using Pythagoras Formula :-
OQ^2 = OP^2 + PQ^2
(15)^2 = (9)^2 + PQ^2
225 = 81 + PQ^2
144 = PQ^2
PQ = 12 cm
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(5) Prove that the tangents to a circle at the end points of a diameter are parallel.
sol)
Two lines are parallel to each other only if a line that cuts through them makes the same alternate angles with both of them.
*) In the case, since the diameter is perpendicular to the tangent, the two lines are perpendicular to the diameter on the either side. Each line makes an angle of 90 degrees with diameter.
*) Hence, the alternate angles are 90 deg
Therefore, Tangents drawn at both ends of diameter are parallel to each other.
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9.3 Number of Tangent to a circle from any point
To get an idea of the number of tangents from a point on a circle, Let us perform the following
(*) Activity
(i) Draw a circle on a paper. Take a point P inside it. Can you draw a tangent to the circle through this point ?
You will find that all the lines through this point intersect the circle in two points. What are these ? These are all secants of a circle. So, it is not possible to draw any tangent to a circle through a point inside it.
(ii) Next, take a point P on the circle and draw tangents through this point. You have observed that there is only one tangent to the circle at a such a point.
(iii) Now, take a point P outside the circle and try to draw tangents to the circle from this point.
What do you observe? You will find that you can draw exactly two tangents to the circle through this point.
Now, we can summarise these facts as follows :-
Case (i) : There is no tangent to a circle passing through a point lying inside the circle.
Case (ii) : There is one and only one tangent to a circle passing through a point lying on the circle.
Case(iii) : There are exactly two tangents to a circle through a point lying outside the circle in this case, A and B are the points of contacts of the tangents PA and PB respectively.
The length of the segment from the external point P and the point of contact with the circle is called the length of the tangent from the point P to the circle.
Note that in the above figure (iii), PA and PB are the length of the tangents from P to the circle. What is the relation between lengths PA and PB
Theorem-9.2 :-
The lengths of tangents drawn from an external point to a circle are equal.
Given : A circle with centre O, P is a point lying outside the circle and PA and PB are two tangents to the circle from P. (See figure)
To prove : PA = PB
Proof : Join OA, OB and OP.
∠OAP =∠OBP = 90 deg (Now in the two right triangles)
(Angle between radii and tangents according to theorem 9.1)
∆OAP and ∆OBP,
OA = OB (radii of same circle)
OP = OP (Common)
Therfore, By R.H.S. Congruency axiom,
∆OAP ≅ ∆OBP
This gives PA = PB (CPCT)
Hence Proved.
Try This:- Use Pythagoras theorem and write proof of above theorem.
9.3.1 . Construction of Tangents to a circle from an external point.
You saw that if a point lies outside the circle, there will be exactly two tangents to the circle from this point. We shall now see how to draw these tangents.
Construction : To construct the tangents to a circle from a point outside it.
Given : We are given a circle with centre ‘O’ and a point P outside it. We have to construct two tangents from P to the circle.
Steps of construction : -
Step(i) : Join PO and draw a perpendicular bisector of it. Let M be the midpoint of PO.
Step (ii) : Taking M as centre and PM or MO as radius, draw a circle. Let it intersect the given circle at the points A and B.
Step (iii) : Join PA and PB. Then PA and PB are the required two tangents.
Proof : Now, Let us see how this construction is justified.
Join OA. Then ∠PAO is an angle in the semicircle and, therefore, ∠PAO = 90°.
Can we say that PA ⊥ OA ?
Since, OA is a radius of the given circle, PA has to be a tangent to the circle (By converse theorem of 9.1)
Similarly, PB is also a tangent to the circle.
Hence proved.
Some interesting statements about tangents and secants and their proof:
Statement-1 : The centre of a circle lies on the bisector of the angle between two tangents drawn from a point outside it. Can you think how we can prove it?
Proof : Let PQ and PR be two tangents drawn from a point P outside of the circle with centre O Join OQ and OR, triangles OQP and ORP are congruent
because we know that,
∠OQP = ∠ORP = 90 deg (Theorem 9.1)
OQ = OR (Radii)
OP is common.
This means ∠OPQ =∠ OPR (CPCT)
Therefore, OP is the angle bisector of ∠QPR .
Hence, the centre lies on the bisector of the angle between the two tangents.
Statement-2 : In two concentric circles, such that a chord of the bigger circle, that touches the smaller circle is bisected at the point of contact with the smaller circle.
Can you see how is this?
Proof : We are given two concentric circles C1 and C2 with centre O and a chord AB of the larger circle C1, touching the smaller circle C2 at the point P (See figure)
we need to prove that AP = PB.
Join OP.
Then AB is a tangent to the circle C2 at P and OP is its radius.
Therefore, by Theorem 9.1
OP ⊥ AB
Now, ∆OAP and ∆OBP are congruent. (Why?)
This means AP = PB.
Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord.
Statement-3 : If two tangents AP and AQ are drawn to a circle with centre O from an external point A
then ∠PAQ = 2 ∠ OPQ =2 ∠OQP .
Can you see?
Proof :We are given a circle with centre O, an external point A and two tangents AP and AQ to the circle, where P, Q are the points of contact (See figure).
We need to prove that
∠PAQ = 2∠ OPQ
Let ∠PAQ = θ
Now, by Theorem 9.2,
AP = AQ, So ∆APQ is an isoscecles triangle
Therefore, ∠APQ + ∠AQP + ∠PAQ = 180° (Sum of three angles)
∠ APQ = ∠AQP = 1/2 (180 -0 )
90° − 1/2 θ
Also, by Theorem 9.1,
∠OPA = 90°
So, ∠OPQ =∠OPA −∠ APQ
90°−[90− 1/2 θ ] = 1/2 θ = 1/2 ∠PAQ
This gives ∠OPQ = 1/2 ∠ PAQ.
Therefore ∠PAQ = 2∠OPQ.
Similarly ∠ PAQ = 2∠OQP
Statement-4 : If a circle touches all the four sides of a quadrilateral ABCD at points PQRS. Then AB+CD = BC + DA
Can you think how do we proceed? AB, CD, BC, DA are all chords to a circle.
For the circle to touch all the four sides of the quadrilateral at points P, Q, R, S, it has to be inside the quadrilateral. (See figure)
How do we proceed further?
Proof : The circle touched the sides AB, BC, CD and DA of Quadrilateral ABCD at the points P, Q, R and S respectively as shown
Since by theorem 9.2, the two tangents to a circle drawn from a point outside it, are equal,
AP = AS
BP = BQ
DR = DS and
CR = CQ On adding, We get
AP + BP + DR + CR = AS + BQ + DS + CQ or
(AP + PB) + (CR + DR) = (BQ + QC) + (DS + SA) or
AB + CD = BC + DA.
Let us do an example of analysing a situation and know how we would construct something.
Example-1. Draw a pair of tangents to a circle of radius 5cm which are inclined to each other at an angle 60°.
Solution : To draw the circle and the two tangents we need to see how we proceed. We only have the radius of the circle and the angle between the tangents.
We do not know the distance of the point from where the tangents are drawn to the circle and we do not know the length of the tangents either. We know only the angle between the tangents. Using this, we need to find out the distance of the point outside the circle from which we have to draw the tangents.
To begin let us consider a circle with centre ‘O’ and radius 5cm.
Let PA and PB are two tangents draw from a point ‘P’ outside the circle and the angle between them is 60 deg .
In this ∠APB = 60o . Join OP.
As we know,
OP is the bisector of ∠APB,
∠OAP= ∠OPB= 60 deg / 2 =30 deg (∵ ∆OAP ≅ ∆OBP)
Now ln ∆OAP,
sin 30 deg = Opp. side / Hyp = OA/OP
1/ 2 = 5 / OP (From trigonometric ratio)
OP = 10 cm.
Now we can draw a circle of radius 5 cm with centre ‘O’. We then mark a point at a distance of 10 cm from the centre of the circle.
Join OP and complete the construction as given in construction 9.2. Hence PA and PB are the required pair of tangents to the given circle.
You can also try this construction without using trigonometric ratio.
Try This
(*) Draw a pair of radii OA and OB such that ∠BOA = 120 deg .
(*) Draw the bisector of ∠BOA and draw lines perpendiculars to OA and OB at A and B.
(*) These lines meet on the bisector of ∠BOA at a point which is the external point and the perpendicular lines are the required tangents.
(*) Construct and Justify.
Exercise- 9.2
(1) Choose the correct answer and give justification for each.
1) The angle between a tangent to a circle and the radius drawn at the point of contact is
a) 60 deg
b) 30 deg
c) 45 deg
d) 90 deg
Answer) 90 deg
Justification :- Since the tangent at any point is perpendicular to the radius through the point of contact.
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(2) From a point "Q", the length of the tangent to a circle is 24 cm, and the distance of "Q" from the center is 25 cm. The radius of the circle is
(a) 7cm
(b) 12 cm
(c) 15 cm
(d) 24.5 cm
sol) 7 cm
justification :-
Tangent is perpendicular to the radius at point of contact here 'PQ' is tangent segment and 'OP' is radius of circle.
OP _|_ PQ
=> /_OPQ = 90
Now in /_\ OPQ,
Using Pythagoras Formula :-
OQ^2 = OP^2 + PQ^2
(25)^2 = OP^2 + (24)^2
625 = OP^2 + 576
625 - 576 = OP^2
49 = OP^2
OP = 7 cm
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(3) If AP and AQ are the two tangents a circle with center "O" so that /_POQ = 110 deg, then /_ PAQ is equal to
(a) 60 deg
(b) 70 deg
(c) 80 deg
(d) 90 deg
sol) 70 deg
Justification :-
From figure Here;
/_OPT = /_OQT = 90 deg ( Since radius is perpendicular to tangent )
Hence, /_POQ + /_PTQ = 180 deg
or, 110 + /_PTQ = 190
or, /_PTQ = 180 - 110
or, /_PTQ = 70 deg
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(4) If tangents PA and PB from a point "P" to a circle with center "O" are inclined to each other at angle of 80 deg, then /_ POA is equal to
(a) 50 deg
(b) 60 deg
(c) 70 deg
(d) 80 deg
sol) 50 deg
Justification ;-
/_AOB = 360 - 180 - 80 = 100
/_POA = /_AOB = 50 deg
2
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(5) In the figure XY and X1 Y1 are two parallel tangents to a circle with center "O" and another tangent AB with point of contact "C" intersecting XY at A and X1 Y1 at B then
/_ AOB =
(a) 80 deg
(b) 100 deg
(c) 90 deg
(d) 60 deg
sol) 90 deg
Justification:- If we join "O" and "C" we get /_OCA = 90 deg
/_OPA = 90 deg , we get OPAC is a square.
Similarly OCBQ is square
=> /_AOB = 90 deg
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2) Two concentric circles are radii 5 cm and 3 cm are drawn. Find the length of the chord of the larger circle which touches the smaller circle.
sol)
Given AO = 5 cm ( radii of larger circle)
OD = 3cm ( radii of smaller circle)
Using Pythagoras theorem :-
So, AO^2 = OD^2 + AD^2
(5)^2 = (3)^2 + AD^2
25 - 9 = AD^2
16 = AD^2
4 = AD
Since AD = DB
AB = 8cm
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3) Prove that the parallelogram circumscribing a circle is a rhombus.
sol) Construction :-
*) Draw a circle with center "O"
*) Draw parallelogram ABCD which touches the circle P,Q,R,S.
Given :- AB || DC
AD || BC
RTP :- ABCD is a rhombus
In /_\ AOB and /_\DOC
AB = DC ( Given that AD || BC )
/_AOB = /_DOC ( vertically opposite angles )
/_BAO = /_DCO ( Alternate angles )
Hence; /_\ AOB ~ /_\ DOC
Hence; AO = CO and BO = DO
Since diagonals are bisecting each other, so given parallelogram is a rhombus
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4) 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. Find the sides AB and AC.
sol)
Since the external tangent are equal
DC = CE = 3cm
BD = BF = 9cm
Let AF = AE = X cm
Using Pythagoras theorem :-
AB^2 = BC^2 + AC^2
(AF + FB )^2 = (BD + DC)^2 + ( AE + EC)^2
(x + 9)^2 = (12)^2 + (3+x)^2
x^2 + 18x + 81 = 144 + 9 + 6x + x^2
12x = 72
x = 6
So, AB = AF + FB = 15 cm
AC = EC + AE = 9cm
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5) Draw a circle of radius 6cm. From a point 10 cm away from its center, construct the pair of tangents to the circle and measure their lengths. Verify by using Pythogoras Theorem.
sol)
Using Pythagoras theorem :-
AB^2 + AO^2 = OB^2
AB^2 + (6)^2 = (10)^2
AB^2 = 100 - 36
AB = 8 cm.
************************* 6) Construct a tangent to a circle of radius 4cm from a point on the concentric circle of radius 6 cm and measure its length . Also verify the measurement by actual calculation.
sol) Steps of Construction :-
1) Draw two concentric circles with center "O" and radii 4cm and 6 cm such that OP = 6 cm, OQ = 4 cm.
2) Join OP and bisect it at "M". i.e. M is the mid-point of OP
3) Taking "M" as center with "OM " as radius draw a circle intersecting the smaller circle in two points namely "T" and "S"
4) Join PT and PS.
PT and PS are the required tangents from a point "P" to the smaller circle, whose radius is 4 cm. By measurement : PT = 4.5 cm
Verification :-
OTP is right /_\ at T
OP^2 = OT^2 + PT^2
PT^2 = OP^2 - OT^2
PT^2 = 36 - 16 = 20
PT = _/20
PT = _/4 *+/ 5
PT = 2 _/5
PT = 2 * 2.236
PT = 4.47 cm
Similarly, PS = 4.47 cm
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7) 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.
sol) Steps of construction :-
1) Draw a circle, take a point outside the circle anywhere let it be "R".
2) Join "RO" and draw a perpendicular bisector of it. Let "M" be the midpoint of "RO"
3) Make an arc across the circle at two places such as "A" and "B"
4) Join "RA" and "RB". Then "RA" and "RB" are the required two tangents.
Diagram :-
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8) 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.
sol) Let us follow a step-wise approach :-
/_ABC = 90 deg
AB is the diameter.
From "P", tangent "PQ" intersects "BC".
To show : BQ + BC
Let us Join BP.
Now, PQ and BQ are the tangents drawn from an external point "Q" and they must be equal/
So, PQ = BQ -------(1)
=> /_PBQ = /_BPQ [ angles opposite to equal sides off a triangle are equal]
AB = Diameter,
=> /_APB = 90 deg ( Angles in semi circle)
=> /_APB + /_BPC = 180 deg ( Linear Pair)
So,
/_BPC = 180 - /_APB
/_BPC = 90 deg
In /_\ BPC,
/_BPC + /_PBC + /_PCB = 180 deg (Linear Pair)
So, /_PBC + /_PCB = 180 deg - /_BPC
=> 180 deg - 90 deg
=> 90 deg ...........(2)
Now,
/_BPC = 90 deg
So, /_BPQ + /_CPQ = 90 deg -----(3)
From ,(2) and (3)
/_PBC + /_PCB = /_ BPQ + /_CPQ
=> /_PCQ = /_CPQ (as, /_BPQ = /_PBQ)
Now, in /_\ PQC
/_PCQ = /_CPQ
.^., PQ = QC ----(4)
From (1) and (4),
BQ = QC
Hence the tangent at "P" bisects "BC"
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9) Draw a tangent to a given circle with center "O" from a point " R" outside the circle. How many tangents can be drawn to the circle from that point?
[ Hint : The distance of two points to the point of contact is the same .}
Sol) Given :- We are given a circle with centre 'O' and "o" point "R" outside it. We have to construct tangent from point "R" to the circle.
Steps of construction :-
1) Join OR and draw a perpendicular bisector of it. Let "M" be the mid-point of OR
2) Make an arc across the circle at two places such as "A" and "B".
3) Join RA and RB.
.^. RA and RB are the required two tangents.
Diagram :-
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9.4 Segment of a circle form by a secant
We have seen a line and a circle. When a line meets a circle in only one point, it is a tangent. A secant is a line which intersects the circle at two distinct points represented in the chord.
Here ‘l’ is the secant and AB is the chord.
Shankar is making a picture by sticking pink and blue paper. He makes many pictures. One picture he makes is of washbasin.
How much paper does he need to make this picture? This picture can be seen in two parts. A rectangle is there, but what is the remaining part?
It is the segment of the circle. We know how to find the area of rectangle. How do we find the area of the segment? In the following discussion we will try to find this area.
Do This
Shankar made the following pictures also along with washbasin.
What shapes can they be broken into that we can find area easily? Make some more pictures and think of the shapes they can be divided into different parts.
Lets us recall how to find the area of the following geometrical figures as given in the table.
9.4.1 Finding the Area of Segment of a circle
To estimate the area of segment of a circle, Swetha made the segments by drawing secants to the circle.
As you know a segment is a region, bounded by the arc and a chord, we can see the area that is shaded (
) in
fig.(i) is a minor segment, semicircle in fig.(ii) and major segment in fig.(iii).
How do we find the area of the segment? Do the following activity.
Take a circular shaped paper and fold it along with a chord less than the diameter and shade the smaller part as shown in in the figure. What do we call this smaller part? It is a minor segment (APB).
What do we call the unshaded portion of the circle? Obviously it is a major segment (AQB).
You have already come across the sector and segment in earlier classes. The portion of some unshaded part and shaded part (minor segment) is a sector which is the combination of a triangle and a segment.
Let OAPB be a sector of a circle with centre O and radius ‘r’ as shown in the figure. Let the angle measure of ∠AOB be ‘x’
You know that area of circle when the angle measure at the centre is 360° is πr^ 2.
So, when the degree measure of the angle at the centre is 1°, then area of sector is
1 / 360 ° ×πr 2 .
Therefore, when the degree measure of the angle at the centre is x°, the area of sector
is x° / 360 ×πr 2 .
Now let us take the case of the area of the segment APB of a circle with centre ‘O’ and radius ‘r’, you can see that
Area of the segment APB
= Area of the sector OAPB - Area of ∆OAB
= x°/ 360 ° × πr^ 2 - area of ∆OAB
Try This
How can you find the area of major segment using area of minor segment?
Do This
1. Find the area of sector, whose radius is 7 cm. with the given angle:
i. 60° ii. 30° iii. 72° iv. 90° v. 120°
2. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 10 minutes
Now, we will see an example to find area of segment of a circle.
Example-1. Find the area of the segment AYB showing in the adjacent figure. If radius of the circle is 21 cm and ∠ AOB = 120 deg (Use π = 22/ 7 and _/3 = 1.732)
Solution : Area of the segment AYB = Area of sector OAYB − Area of ∆OAB
Now, area of the sector OAYB
= 120 deg / 360 × 22/7 * 21 * 21 cm2
= 462 cm2 ...(1)
For finding the area of ∆OAB, draw OM ⊥ AB as shown in the figure:-
Note OA = OB. Therefore, by RHS congruence,∆AMO ≅ ∆BMO
So, M is the midpoint of AB and ∠ AOM = ∠ BOM = 1/2 * 120 = 60 deg
Let, OM = x cm
So, from ∆OMA , OM / OA = cos 60 deg. .
or
x / 21 = 1/ 2 [because cos 60 deg = 1/2]
or
x = 21 / 2
So, OM = 21/ 2 cm
Also, AM / OA = sin 60°
AM / 21 = _/3 / 2 [because sin 60 deg = _/3 / 2 ]
So, AM = (21 _/ 3) / 2 cm.
Therefore
AB = 2AM
= ((2 * 21 _/ 3) / 2) × cm.
= 21_/ 3 cm
So, Area of /_\ OAB = 1 / 2 * AB * OM
= 1 / 2 *( 21 _/3) * (21 / 2) cm^2
= (441 / 4) * _/3) cm^2.........(2)
Therefore, area of the segment AYB = [ 462 - 441/4 * _/3] cm^2. (from (1) and (2))
= 21 / 4 ( 88 - 21 / _/3) cm^2
= 271.047 cm^2
Example-2. Find the area of the segments shaded in figure, if PQ = 24 cm., PR = 7 cm. and QR is the diameter of the circle with centre O (Take π= 22 / 7 )
Solution : Area of the segments shaded = Area of sector OQPR - Area of triangle PQR.
Since QR is diameter, ∠QPR = 90° (Angle in a semicircle)
So, using pythagoras Theorem
In ∆QPR,
QR^2 = PQ^2 + PR^2
= 24^22 + 7^2
= 576 + 49 = 625
QR = _/625 = 25 cm.
Then radius of the circle = 1/ 2 QR = 1 / 2 * (25) = 25 / 2 cm.
Now, area of semicircle OQPR = 1 / 2 πr^ 2
= (1/ 2 * 22/7 )* ( 25 /2 *25 / 2 )
= 327.38 cm 2 .... (1) .
Area of right angled triangle QPR
= 1/ 2 × PR × PQ
= 1 /2 × 7 × 24 = 84 cm 2 ..... (2)
From (1) and (2),
Area of the shaded segments = 327.38 - 84 = 243.38 cm^ 2
Example-3. A round table top has six equal designs as shown in the figure. If the radius of the table top is 14 cm., find the cost of making the designs with paint at the rate of 5 per cm 2 . (use _/3 = 1.732)
Solution : We know that the radius of circumscribing circle of a regular hexagon is equal to the length of its side.
∴ Each side of regular hexagon = 14 cm.
Therefore, Area of six design segments = Area of circle - Area of the regular hexagon.
Now, Area of circle = πr^ 2 = (22 /7) × 14 × 14 = 616 cm 2 ..... (1)
Area of regular hexagon = 6 × (_/3 /4) a ^2
= 6 × _/3 / 4 × 14 × 14
= 509.2 cm 2 ..... (2)
Hence, area of six designs
= 616 - 509.21 (from (1), (2)
= 106.79 cm 2 .
Therefore, cost of painting the design at the rate of 5 per cm 2
= 106.79 × 5
= 533.95
Exercise - 9.3
What is a chord or meaning of chord :- A lie segment connecting two points on a circle's circumference is a chord. When the chord passes through the center of a circle it is called the diameter.
1) A chord of a circle of radius 10 cm, subtends a right angle at the center. Find the area of the corresponding ( use Pi = 3.14)
(1) Minor segment (2) Major segment
sol) RTP :- Finding area of minor segment.
Given :- A chord of a circle of radius 10 cm
Radius :- OA = OB = 10 cm
o = 90 deg
Minor segment APB = Area of sector OAPB - Area of /_\AOB
(*) Finding area of OAPB :-
o = 90 deg pi = 3.14, r = 10 cm
=>o * pi * r^2
360
=> 90 * 3.14 * (10)^2
360
=> 1 * 3.14 * 100
4
=> 1 * 314
4
OAPB = 78.5 cm^2 ---------------(1)
(*) Finding Area of /_\AOB :-
It is a right angled triangle
/_ O = 90 , Base = OA and Height = OB
We have
Area of triangle = 1/2 * base * height
/_\ AOB = 1 * OA * OB
2
=> 1 *10 * 10
2
=> 5 * 10
/_\ AOB = 50 cm^2 --------------(2)
Now,
Finding Area of minor segment (APB) = OAPB - AOB
=> 78. 5 - 50
=> 28.5 cm^2
(B) Finding area of major segment :-
major sector = Area of circle - Area of sector OAPB
Area of circle :- Pi * r^2
=> 3.14 * (10)^2
=> 3.14 * 100
=> 314 cm^2
Now
Area of major sector = Area of circle - OAPB
=> 314 - 78.5
=> 235.5 cm^2
************************************************
2) A chord of a circle of radius 12 cm, subtends an angle of 120 deg at the center. Find the area of the corresponding minor segment of the circle ( use Pi = 3.14 and _/3 = 1.732)
sol) Given :- Radius(r) = 12 cm and Angle(0) = 120 deg
Area of minor segment (APB) =Area of sector OAPB - /_\ OAB
(1) Finding Area of sector OAPB :-
=> 0 * Pi * r^2
360
=> 120 * 3.14 * (12)^2
360
=> 1 * 3.14 * 12 * 12
3
=> 1 * 3.14 * 4 * 12
=> 150. 72 cm^2
(2) Finding area of /_\ AOB :-
Area of /_\ AOB = 1/2 * Base * Height
We draw OM _|_ AB
.^. /_OMB = /_OMA = 90 deg
In /_\ OMA & /_\ OMB
/_OMA = /_OMB ( Both 90 deg)
OA = OB ( Both radius)
OM = OM ( Common)
.^. /_\ OMA ~ /_\ OMB ( By R.H.S congruency)
=> /_AOM ~ /_ BOM ( CPCT )
.^. /_AOM = /_ BOM = 1/2 /_BOA
.^. /_AOM = /_BOM = 1/2 * 120 = 60 deg
Also, since /_\ OMB ~ /_\ OMA
.^. BM = AM ( CPCT )
=> BM = AM = 1/2 AB ----------(1)
In right triangle /_\ OMA
Sin 0 = side opposite to angle 0
Hypotenuse
Sin 60 = AM
AO
_/3 = AM
2 12
-/3 * 12 = AM
2
6_/3 = AM
In right triangle /_\ OMA :-
Cos 0 = Side adjacent to angle 0
Hypotenuse
Cos 60 = OM
AO
1 = OM
2 12
12 = OM
2
6 = OM
OM = 6
Now,
AM = 1/2 * AB ( From 1 )
2AM = AB
AB = 2 (6_/3) [ Putting AM = 6 _/3 ]
AB = 12_/3
Now,
Area of /_\ AOB = 1/2 * Base * height
=> 1 / 2 * AB * OM
=> 1/2 * 12_/3 * 6
=> 36 _/3
=> 36 * 1.73
Area of /_\ AOB = 62.28 cm^2
Area of minor segment APB = OAPB - /_\ OAB
=> 150.72 - 62.28
=> 88.44 cm^2
******************************************
3) A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm. Sweeping through an angle of 115 deg . Find the total area cleaned at each sweep of the blades ( use Pi = 22/7).
sol) Given :- Radius (r) = 25 cm and sweeping angle(0) = 115 deg
Total area cleaned by the two wipers = 2 * Area cleaned by 1 wiper
=> 2 * Area of sector with angle 115 deg
=> 2 * 0 * Pi * r^2
360
=> 2 * 115 * 22 * ( 25)^2
360 7
= 115 * 22 * 25 * 25
180 7
+> 115 * 11 * 25 * 25
90 7
=> 115 * 11 * 5 * 25
18 7
=> 158125 cm^2
126
Hence , Area cleaned by both wiper = 1,254.96 cm^2
**********************************************
4) Find the area of the shaded region in the figure, where ABCD is a square of side 10 cm, and semicircles are drawn with each side of the square as diameter ( use Pi = 3.14)
sol) Given :- Side(s) of square ABCD = 10 cm
Formula :- Area of Square = s^2
=> (10)^2 = 100 cm^2
Given :- semicircle is drawn with side of square as diameter,
So, Diameter of semi-circle = side of a square = 10 cm
radius(r) of semicircle = side = 10 = 5 cm
2 2
Now,
Area of semi-circle = 1/2 * Pi * r^2
=> 1/2 * 22/7 * (5)^2
=> 1/2 * 22/7 * 25
=> 1 * 11 /7 * 25
=> 39.28
Area of 4 semi-circle = 4 * 39.28 = 157.12 cm^2
Area of shaded region :- area of 4 semi-circle - area of square
=> 157.12 - 100
=> 57.12 cm^2
******************************************************
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 Pi = 22/7)
sol) Given :- Side(s) of square ABCD = 7 cm
Formula :- Area of Square = s^2
=> (7)^2 = 49 cm^2
radius(r) of semicircle = side = 7 = 3.5 cm
2 2
Area of 2 semi-circle = 2 * ( 1/2 * Pi * r^2)
=> Pi * r^2
=> 3.14 * (3.5)^2
=> 3.14 * 3.5 * 3.5
=> 3.14 * 12. 25
=> 38.465 cm^2
area of shaded region = area of square - area of 2 semicircle
=> 49 - 38.465
Area of shaded region = 10.5 cm^2
****************************************************
6) In figure, OACB is a quadrant of a circle with center 'O" and radius 3.5 cm. If OD = 2cm, find the area of the shaded region. ( use Pi = 22/7)
sol) RTP :- Finding area of shaded region
Area of shaded region = Area of quadrant OACB - Area of /_\ BOD
(1) Finding area of quadrant ( OACB) :-
Given :- radius (r) = 3.5 cm
We know that
Area of Quadrant OACB = 1/4 * Area of circle
=> 1/4 * Pi * r^2
=> 1/4 * 22/7 * (3.5)^2
=> 1/4 * 22/7 * 3.5 * 3.5
Area of Quadrant (OACB) = 9.625 cm^2
(2) Area of /_\ BOD :-
Area of /_\ BOD = 1/2 * Base * Height
=> 1/2 * OB * OD
=> 1/2 * 3.5 * 2
=> 3.5 cm^2
Now
Area of shaded region = Area of quadrant OACB - Area of /_\ BOD
=> 9.625 - 3.5
Area of shaded region = 6.125 cm^2
********************************************
7) AB and CD are respectively arcs of two concentric circles of radii 21 cm, and 7 cm , with center "O" ( see figure) . IF /_AOB = 30 deg, find the area of the shaded region ( use Pi = 22/7)
sol) Given :- radii(R1 = 21 cm and radii(r2) = 7 cm
0 = 30 deg
RTP :- Area of shaded region ABCD
Area of shaded region is given by :- 0 * Pi [ R1^2 - r2^2 ]
360
=> 30 * 22 [ (21)^2 - (7)^2 ]
360 7
=> 1 * 22 [ 441 - 49 ]
12 7
=> 22 [ 392 ]
7 * 12
=> 11 * 392
7 * 6
=> 102. 67 cm^2
*********************************************
8) calculate the area of the designed region if figure, common between the two quadrants of the circles of radius 10 cm, each ( use Pi = 3. 14)
sol) Given :- radius(r) = 10 cm
RTP :- Area of designed region
Area of designed region = area of two quadrant - Area of square
(1) Area of two quadrant :-
=> 2 * 1/4 * Pi *r^2
=> 2 * ( 1/4 * 3.14 * (10)^2 )
=> 2 * ( 1/4 * 3.14 * 100 )
=> 2 * ( 3.14 * 25)
=> 157 cm^2
(2) Area of square :-
=> s ^2 => (10)^2 = 100 cm^2
Now.
Area of designed region = area of two quadrant - Area of square
=> 157 - 100
=> 57 cm^2
**************************************
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Optional exercise :-
1. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line - segment joining the points of contact at the center.
sol) Given : A circle with center 'O'
Tangents PA and PB drawn from external point P.
R.T.P :- /_APB + /_ AOB = 180 deg
Proof : Since PA is tangent,
OA _|_ PA
(Tangent at any point of circle is perpendicular to the radius through point of contact)
Therefore, /_OAP = 90 deg ------(1)
similarly /_OBP = 90 deg .........(2)
In quadrilateral OAPB
( According,Angle sum property of Quadrilateral)
/_OAP + /_APB + /_OBP +/_AOB = 360 deg
90 deg + /_APB + 90 deg + /_AOB = 360 deg
180 deg + /_APB + /_ AOB = 360 deg
/_APB + /_AOB = 360 deg - 180 deg =180 deg
Hence proved
**********************************************
2. PQ is a chord of length 8cm of a circle of radius 5cm. The tangents at P and Q intersect at a point T (See figure). Find the length of TP.
Solution :- Join OT
Let OT intersect PQ at R
From theorem 10.2,
Length of tangents from external points are equal
So, TP = TQ
In /_\ TPQ
TP = TQ (i.e. two sides are equal)
So, /_\ TPQ is an isosceles triangle.
Here , OT bisector of /_PTQ
So, OT _|_ PQ ( Angle bisector and altitude of isosceles triangle are same)
Since OT _|_ PQ
PR = RQ ( _|_ from center to a chord, bisects the chord) ----(1)
since PQ = 8 cm
QR = 4 cm
PR = QR = 1 / 2 PQ = 8 / 2 = 4 cm.
In right triangle ORP
By Pythagoras Theorem
OP^2 = PR^2 + OR ^2
5^2 = 4^2 + OR^2
25 = 16 + OR^2
9 = OR^2
3 = OR
Let TP = x
In right triangle PRT
(TP)^2 = (PR)^2 + (RT)^2
x^2 = 4^2 + RT^2
x^2 = 16 + RT^2.........(1)
Now,
Since TP is a tangent
OP _|_ TP
(Tangent at any point of a circle is perpendicular to the radius through point of contact)
Therefore, /_OPT = 90 deg
In Right angled triangle /_\ OPT
(OT)^2 = (OP)^2 + (TP)^2
(OT)^2 = 5^2 + x^2
(OR + RT )^2 = 25 + x^2
( 3 + RT )^2 = 25 + x^2
(3 + RT)^2 is in ( A+ B)^2 form
3^2 + RT^2 + 2 (3)(RT) = 25 + x^2
9 + RT^2 + 6RT = 25 +x^2
But : x^2 = 16 + RT^2 from (1) substitute
we get,
9 + RT^2 + 6RT = 25 + (16 + RT^2)
6RT = 32
RT = 32 / 6
RT = 16 / 3
Now putting RT value in (1) we get
x^2 = 16 + RT^2
x^2 = 16 + ( 16/3) ^2
x^2 = 16 + ( 256 / 9)
x^2 = (16 * 9) + 256) / 9
x^2 = 400 / 9
x = 20 / 3
Hence proved, length of tangent TP = x = 20/ 3 cm
********************************************
3) Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
sol) A circle with center 'O' touches the sides AB,BC,CD and DA of a quadrilateral ABCD at the points P,Q,R, and S respectively.
R.T.P :- Opposite sides of a quadrilateral circumscribing a circle , i.e
/_ AOB + /_ COD = 180 deg
/_ AOD + /_ BOC = 180 deg
Construction :- Join OP, OQ, OR and OS.
Proof :- In /_\ AOP and /_\ AOS
(Lengths of tangents drawn from external point are equal )
AP = AS
AO = AO ( common)
OP = OS ( Radius)
Therefore, /_\ AOP = /_\ AOS (SSS congruence rule )
/_AOP = /_AOS (CPCT)
i.e, /_1 = /_8 ......(1)
similarly ,we can prove
/_2 = /_3 ....(2)
/_5 = /_4.....(3)
/_6 = /_7.......(4)
Now,
/_1 + /_2 + /_3 + /_4 + /_ 5 + /_6 +/_7 +/_8 = 360 deg
from (1) (2) (3) (4)
/_1 + /_2 + /_2 + /_5 + /_5 + /_6 + /_6 +/_1 = 360 deg
2 ( /_1 + /_2 + /_5 + /_6) = 360
/_1 + /_2 + /_5 + /_6 = 360/2
(/_1 + /_2) +( /_5 + /_6 )= 180
/_ AOB + /_ COD = 180 deg
Hence both angle are supplementary
similarly we can prove /_AOD + /_ BOC = 180 deg
Hence proved.
**********************************************
6). Find the area of the shaded region in the figure, given in which two circles with centers A and B touch each other at the point C. If AC = 8 cm. and AB = 3 cm.
sol) Given AB = 3 cm, AC = 8 cm,, BC = 8-3 = 5cm
Now area of bigger circle = Pi * r^2
= Pi * (8)^2
= 64Pi cm^2
Now area of smaller circle = Pi * r^2
= Pi * ( 5)^2
=25Pi cm^2
Now,
Area of shaded region = Area of bigger circle - smaller circle
= 64Pi - 25Pi
= 39Pi cm^2
*********************************************
7. ABCD is a rectangle with AB = 14 cm. and BC = 7 cm. Taking DC, BC and AD as diameters, three semicircles are drawn as shown in the figure. Find the area of the shaded region
sol) Given:
Length of a rectangle (AB) = (DC) = 14 cm
Breadth of a rectangle ( BC) = (AD)= 7 cm
Area Of Semicircle with Diameter DC:
Diameter (D) = 14
radius (r) = 14/2 =7
Area of semicircle = 1/2πr²
= 1 /2 * (22/7) * (7)²
= 11/7 × 49
= 11 * 7
= 77 cm² -----(1)
(*) AREA OF RECTANGLE (ABCD)
= Length × Breadth
= AB × BC
= 14* 7
= 98 cm ----(2)
Area of 2 semicircle with diameter BC and AD
Here, Diameter (D) = 7 cm
radius(r) = (D/2) = 7 / 2
Area of 2 semicircle =2× (1/2 πr² )
= Pi * (r)^2
=(22/7) × (7/2)²
= (22/7) * (49/4)
= (11/7) * ( 49/2)
= 11 ×7 / 2
= 77 /2 cm² -----------(3)
AREA OF SHADED REGION :- (2) - (1) + (3)
= Area of rectangle ABCD - area of semicircle with diameter DC + Area of 2 semicircle with diameter BC and AD
Area of shaded region = 98 - 77 + 77/2
= 21 + 77/2
= (42 +77)/2
= 119/2
= 59.5 cm²
Area of shaded region = 59.5 cm²
Hence, the Area of shaded region is 59.5 cm².
*****************************************************
What we have discussed :-
In this chapter, we have studied the following points.
1. The meaning of a tangent to a circle and a secant. We have used the idea of the chord of a circle.
2. We used the ideas of different kinds of triangles particularly right angled triangles and isosceles triangles.
3. We learn the following:
a) The tangents to a circle is perpendicular to the radius through the point of contact.
b) The lengths of the two tangents from an external point to a circle are equal.
4. We learnt to do the following:
a) To construct a tangent to a circle at a given point when the center of the circle is known.
b) To construct the pair of tangents from an external point to a circle.
5. We learnt to understand how to prove some statements about circles and tangents. In this process learnt to use previous results and build on them logically to from new results.
6. We have learnt Area of segment of a circle =
Area of the corresponding sector - Area of the corresponding triangle
We have seen two lines mostly intersect at a point or do not intersect in a plane. In some situations they coincide with each other.
Similarly, what happens when a curve and a line is given in a plane?You know a curve may be a parabola as you have seen in a polynomials or a simple closed curve "circle" which is a collection of all those points on a plane that are at a constant distance from a fixed point.
You might have seen circular objects rolling on a plane creating a path. For example ; riding a bicycle, wheels of train on the track etc., where it seems to be a circle and a line. Does there a relation exist between them?
Let us see what happens, if a circle and a line are given in a plane.
9.1.1 A Line and a Circle
You are given a circle and a line drawn on a paper. Salman argues that there can only be 3 possible ways of presenting them.
Consider a circle 'O' and a line PQ , the three possibilities are given in figure below :
In fig (1) : the line PQ and the circle have no common point. In this case PQ is a non-intersecting line with respect to the circle.
In fig (2) : the line PQ intersects the circle at two points A and B. It forms a chord on the circle AB with two common points. In this case the line PQ is a secant of the circle.
In fig (3) : There is only one point A, common to the line PQ and the circle. This line is called a tangent to the circle.
You can see that there cannot be any other position of the line with respect to the circle.
Do you know?
The word ' tangent' comes from the latin word ' tangere', which means to touch and was introduced by Danish mathematician Thomas Fineke in 1583.
Do This :-
1) Draw a circle with any radius. Draw four tangents at different points. How many tangents can you draw to this circle?
2) How many tangents you can draw to circle from a point away from it.
3) Which of the following are tangents to the circles?
9.2 tangents of a circle
We can see that tangent can be drawn at any point lying on the circle. Can you say how many tangents can be drawn at any point on the surface of the circle.
To understand this let us consider the following activity.
Activity
Take a circular wire and attach a straight wire AB at a point P of the circular wire, so that the system rotate about the point P in a plane.The circular wire represents a circle and the straight wire AB represents a line intersects the circle at a point P.
Put the system on a table and gently rotate the wire AB about the point P to get different positions of the straight wire as shown in the figure.
The wire intersects the circular wire at P and at one more point through the points Q1, Q2 or Q3 etc.
So while it generally intersects circular wire at two points one of which is P in one particular position, it intersects the circle only at the Point P ( see position A'B' of AB ).
This is the position of a tangent at the point P of the circle. You can check that in all other positions of AB it will intersect the circle at P and at another point, A'B' is a tangent to the circle at P.
We see that there is only one tangent to the circle at Point P.
Moving wire AB in either direction from this position makes it cut the circular wire in two points. All these are therefore secants. Tangent is a special case of a secant where the two points of intersection of a line with a circle coincide.
Do This
(*) Draw a circle and a secant PQ of the circle on a paper as shown below. Draw various lines parallel to the secant on both sides of it.
(*) What happens to the length of chord coming closer and closer to the center of the circle?
(*) What is the longest chord?
(*) How many tangents can you draw to a circle, which are parallel to each other?
The common point of the tangent and the circle is called the point of contact and the tangent is said to touch the circle at the common point.
Observe the tangents to the circle in the figure given below :
How many tangents can you draw to a circle at a Point? how many tangents can you obtain to the circle in all?see the points of contact. Draw radii from the points of contact. Do you see anything special about the angle between the tangents and the radii at the points of contact. All appear to be perpendicular to the corresponding tangents. We can also prove it.
Let us see how.
Theorem-9.1 : The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Given : A circle with center ‘O’ and a tangent XY to the circle at a point P.
To prove : OP is perpendicular to XY. (i.e OP ⊥ XY)
Proof : Here, we will use the method that assumes that the statement is wrong and shows that such an assumption leads to a fallacy.
So we will suppose OP is not perpendicular to XY.
Take a point Q on XY other than P and join OQ.
The point Q must lie outside the circle (why?) (Note that if Q lies inside the circle, XY becomes a secant and not a tangent to the circle)
Therefore, OQ is longer than the radius OP of the circle [Why?]
i.e., OQ > OP.
This must happen for all points on the line XY.
It is therefore true that OP is the shortest of all the distances of the point O to the points of XY. So our assumption that OP is not perpendicular to XY is false.
Therefore , OP is perpendicular to XY
Note : The line containing the radius through the point of contact is also called the ‘normal to the circle at the point’.
Try This
How can you prove the converse of the above theorem.
“If a line in the plane of a circle is perpendicular to the radius at its endpoint on the circle, then the line is tangent to the circle”.
We can find some more results using the above theorem
(i) Since there can be only one perpendicular OP at the point P, it follows that one and only one tangent can be drawn to a circle at a given point on the circumference.
(ii) Since there can be only one perpendicular to XY at the point P, it follows that the perpendicular to a tangent at its point of contact passes through the centre.
9.2.1 Construction of Tangent to a Circle
How can we construct a line that would be tangent to a circle at a given point on it?
We use what we just found the tangent has to be perpendicular to the radius at the point of contact.
To draw a tangent through the point of contact we need to draw a line perpendicular to the radius at that point. To draw this radius we need to know the center of the circle. Let us see the steps for this construction.
Construction : Construct a tangent to a circle at a given point when the center of the circle is known.
Steps of Construction :-
We have a circle with center ‘O’ and a point P anywhere on its circumference. Then we have to construct a tangent through P.
1. Draw a circle with center ‘O’ and mark a point ‘P’ anywhere on it. Join OP.
2. Draw a perpendicular line through the point P and name it as XY, as shown in the figure.
3. XY is the required tangent to the given circle passing through P.
Can you draw one more tangent through P ? give reason.
Try This
How can you draw the tangent to a circle at a given point when the center of the circle is not known?
Hint : Draw equal angles ∠QPX and ∠PRQ . Explain the construction.
9.2.2 Finding Length of the Tangent
Can we find the length of the tangent to a circle from a given point? Is the length of tangents from a given point to the circle the same? Let us examine this.
Example : Find the length of the tangent to a circle with centre ‘O’ and radius = 6 cm.
from a point P such that OP = 10 cm.
Solution : Tangent is perpendicular to the radius at the point of contact (Theorem 9.1)
Here PA is tangent segment and OA is radius of circle
∴ OA ⊥ PA ⇒∠OAP 90 deg
Now in ∆OAP,
OP^2 = OA^2 + PA^2 (pythagoras theorem)
10^2 = 6^2 + PA^2
100 = 36 + PA^2
PA^2 = 100 - 36 = 64
∴ PA = √64 = 8 cm
(1) Fill in the blanks
1) A tangent to a circle intersects it in ............ points(s)
A) One
2) A line intersecting a circle in two points is called a .............
A) Secant of a circle.
3) A circle can have ........... parallel tangents at the most.
A) Two
4) The common point of a tangent to a circle and the circle is called.............
A) Point of contact
5) We can draw..........tangents to a given circle.
A) Infinite
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(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 OQ = 12cm. Find length of PQ.
sol)
Tangent :- It is perpendicular to the radius at point of contact.
Here PQ is tangent segment and OP is radius of circle.
OP _|_ PQ
/_OPQ = 90
In /_\ OPQ,
Using Pythagoras Theorem :-
OQ^2 = OP^2 + PQ^2
(12)^2 = (5)^2 + PQ^2
144 = 25 + PQ^2
119 = PQ^2
PQ = _/119 cm
**************************
(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.
sol)
Steps of Construction :-
1) Draw a circle with centre 'O' and radius "r"
2) mark the point "P" anywhere on the circles Join OP
3) Draw a perpendicular line through the point "P" and name it as "AB", as shown in figure.
4) AB is the required tangent to the given circle.
5) Draw another line parallel to the tangent AB and intersect circle at two points and name it as CD
6) CD is the required secant.
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(4) Calculate the length of tangent from a point 15cm away from the centre of a circle of radius 9 cm.
sol)
Here PQ is tangent segment and OP is radius of circle
OP _|_ PQ
/_OPQ = 90
Now in /_\ OPQ
Using Pythagoras Formula :-
OQ^2 = OP^2 + PQ^2
(15)^2 = (9)^2 + PQ^2
225 = 81 + PQ^2
144 = PQ^2
PQ = 12 cm
**************************
(5) Prove that the tangents to a circle at the end points of a diameter are parallel.
sol)
Two lines are parallel to each other only if a line that cuts through them makes the same alternate angles with both of them.
*) In the case, since the diameter is perpendicular to the tangent, the two lines are perpendicular to the diameter on the either side. Each line makes an angle of 90 degrees with diameter.
*) Hence, the alternate angles are 90 deg
Therefore, Tangents drawn at both ends of diameter are parallel to each other.
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9.3 Number of Tangent to a circle from any point
To get an idea of the number of tangents from a point on a circle, Let us perform the following
(*) Activity
(i) Draw a circle on a paper. Take a point P inside it. Can you draw a tangent to the circle through this point ?
You will find that all the lines through this point intersect the circle in two points. What are these ? These are all secants of a circle. So, it is not possible to draw any tangent to a circle through a point inside it.
(ii) Next, take a point P on the circle and draw tangents through this point. You have observed that there is only one tangent to the circle at a such a point.
(iii) Now, take a point P outside the circle and try to draw tangents to the circle from this point.
What do you observe? You will find that you can draw exactly two tangents to the circle through this point.
Now, we can summarise these facts as follows :-
Case (i) : There is no tangent to a circle passing through a point lying inside the circle.
Case (ii) : There is one and only one tangent to a circle passing through a point lying on the circle.
Case(iii) : There are exactly two tangents to a circle through a point lying outside the circle in this case, A and B are the points of contacts of the tangents PA and PB respectively.
The length of the segment from the external point P and the point of contact with the circle is called the length of the tangent from the point P to the circle.
Note that in the above figure (iii), PA and PB are the length of the tangents from P to the circle. What is the relation between lengths PA and PB
Theorem-9.2 :-
The lengths of tangents drawn from an external point to a circle are equal.
Given : A circle with centre O, P is a point lying outside the circle and PA and PB are two tangents to the circle from P. (See figure)
To prove : PA = PB
Proof : Join OA, OB and OP.
∠OAP =∠OBP = 90 deg (Now in the two right triangles)
(Angle between radii and tangents according to theorem 9.1)
∆OAP and ∆OBP,
OA = OB (radii of same circle)
OP = OP (Common)
Therfore, By R.H.S. Congruency axiom,
∆OAP ≅ ∆OBP
This gives PA = PB (CPCT)
Hence Proved.
Try This:- Use Pythagoras theorem and write proof of above theorem.
9.3.1 . Construction of Tangents to a circle from an external point.
You saw that if a point lies outside the circle, there will be exactly two tangents to the circle from this point. We shall now see how to draw these tangents.
Construction : To construct the tangents to a circle from a point outside it.
Given : We are given a circle with centre ‘O’ and a point P outside it. We have to construct two tangents from P to the circle.
Steps of construction : -
Step(i) : Join PO and draw a perpendicular bisector of it. Let M be the midpoint of PO.
Step (ii) : Taking M as centre and PM or MO as radius, draw a circle. Let it intersect the given circle at the points A and B.
Step (iii) : Join PA and PB. Then PA and PB are the required two tangents.
Proof : Now, Let us see how this construction is justified.
Join OA. Then ∠PAO is an angle in the semicircle and, therefore, ∠PAO = 90°.
Can we say that PA ⊥ OA ?
Since, OA is a radius of the given circle, PA has to be a tangent to the circle (By converse theorem of 9.1)
Similarly, PB is also a tangent to the circle.
Hence proved.
Some interesting statements about tangents and secants and their proof:
Statement-1 : The centre of a circle lies on the bisector of the angle between two tangents drawn from a point outside it. Can you think how we can prove it?
Proof : Let PQ and PR be two tangents drawn from a point P outside of the circle with centre O Join OQ and OR, triangles OQP and ORP are congruent
because we know that,
∠OQP = ∠ORP = 90 deg (Theorem 9.1)
OQ = OR (Radii)
OP is common.
This means ∠OPQ =∠ OPR (CPCT)
Therefore, OP is the angle bisector of ∠QPR .
Hence, the centre lies on the bisector of the angle between the two tangents.
Statement-2 : In two concentric circles, such that a chord of the bigger circle, that touches the smaller circle is bisected at the point of contact with the smaller circle.
Can you see how is this?
Proof : We are given two concentric circles C1 and C2 with centre O and a chord AB of the larger circle C1, touching the smaller circle C2 at the point P (See figure)
we need to prove that AP = PB.
Join OP.
Then AB is a tangent to the circle C2 at P and OP is its radius.
Therefore, by Theorem 9.1
OP ⊥ AB
Now, ∆OAP and ∆OBP are congruent. (Why?)
This means AP = PB.
Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord.
Statement-3 : If two tangents AP and AQ are drawn to a circle with centre O from an external point A
then ∠PAQ = 2 ∠ OPQ =2 ∠OQP .
Can you see?
Proof :We are given a circle with centre O, an external point A and two tangents AP and AQ to the circle, where P, Q are the points of contact (See figure).
We need to prove that
∠PAQ = 2∠ OPQ
Let ∠PAQ = θ
Now, by Theorem 9.2,
AP = AQ, So ∆APQ is an isoscecles triangle
Therefore, ∠APQ + ∠AQP + ∠PAQ = 180° (Sum of three angles)
∠ APQ = ∠AQP = 1/2 (180 -
90° − 1/2 θ
Also, by Theorem 9.1,
∠OPA = 90°
So, ∠OPQ =∠OPA −∠ APQ
90°−[90− 1/2 θ ] = 1/2 θ = 1/2 ∠PAQ
This gives ∠OPQ = 1/2 ∠ PAQ.
Therefore ∠PAQ = 2∠OPQ.
Similarly ∠ PAQ = 2∠OQP
Statement-4 : If a circle touches all the four sides of a quadrilateral ABCD at points PQRS. Then AB+CD = BC + DA
Can you think how do we proceed? AB, CD, BC, DA are all chords to a circle.
For the circle to touch all the four sides of the quadrilateral at points P, Q, R, S, it has to be inside the quadrilateral. (See figure)
How do we proceed further?
Proof : The circle touched the sides AB, BC, CD and DA of Quadrilateral ABCD at the points P, Q, R and S respectively as shown
Since by theorem 9.2, the two tangents to a circle drawn from a point outside it, are equal,
AP = AS
BP = BQ
DR = DS and
CR = CQ On adding, We get
AP + BP + DR + CR = AS + BQ + DS + CQ or
(AP + PB) + (CR + DR) = (BQ + QC) + (DS + SA) or
AB + CD = BC + DA.
Let us do an example of analysing a situation and know how we would construct something.
Example-1. Draw a pair of tangents to a circle of radius 5cm which are inclined to each other at an angle 60°.
Solution : To draw the circle and the two tangents we need to see how we proceed. We only have the radius of the circle and the angle between the tangents.
We do not know the distance of the point from where the tangents are drawn to the circle and we do not know the length of the tangents either. We know only the angle between the tangents. Using this, we need to find out the distance of the point outside the circle from which we have to draw the tangents.
To begin let us consider a circle with centre ‘O’ and radius 5cm.
Let PA and PB are two tangents draw from a point ‘P’ outside the circle and the angle between them is 60 deg .
In this ∠APB = 60o . Join OP.
As we know,
OP is the bisector of ∠APB,
∠OAP= ∠OPB= 60 deg / 2 =30 deg (∵ ∆OAP ≅ ∆OBP)
Now ln ∆OAP,
sin 30 deg = Opp. side / Hyp = OA/OP
1/ 2 = 5 / OP (From trigonometric ratio)
OP = 10 cm.
Now we can draw a circle of radius 5 cm with centre ‘O’. We then mark a point at a distance of 10 cm from the centre of the circle.
Join OP and complete the construction as given in construction 9.2. Hence PA and PB are the required pair of tangents to the given circle.
You can also try this construction without using trigonometric ratio.
Try This
(*) Draw a pair of radii OA and OB such that ∠BOA = 120 deg .
(*) Draw the bisector of ∠BOA and draw lines perpendiculars to OA and OB at A and B.
(*) These lines meet on the bisector of ∠BOA at a point which is the external point and the perpendicular lines are the required tangents.
(*) Construct and Justify.
Exercise- 9.2
(1) Choose the correct answer and give justification for each.
1) The angle between a tangent to a circle and the radius drawn at the point of contact is
a) 60 deg
b) 30 deg
c) 45 deg
d) 90 deg
Answer) 90 deg
Justification :- Since the tangent at any point is perpendicular to the radius through the point of contact.
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(2) From a point "Q", the length of the tangent to a circle is 24 cm, and the distance of "Q" from the center is 25 cm. The radius of the circle is
(a) 7cm
(b) 12 cm
(c) 15 cm
(d) 24.5 cm
sol) 7 cm
justification :-
Tangent is perpendicular to the radius at point of contact here 'PQ' is tangent segment and 'OP' is radius of circle.
OP _|_ PQ
=> /_OPQ = 90
Now in /_\ OPQ,
Using Pythagoras Formula :-
OQ^2 = OP^2 + PQ^2
(25)^2 = OP^2 + (24)^2
625 = OP^2 + 576
625 - 576 = OP^2
49 = OP^2
OP = 7 cm
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(3) If AP and AQ are the two tangents a circle with center "O" so that /_POQ = 110 deg, then /_ PAQ is equal to
(a) 60 deg
(b) 70 deg
(c) 80 deg
(d) 90 deg
sol) 70 deg
Justification :-
From figure Here;
/_OPT = /_OQT = 90 deg ( Since radius is perpendicular to tangent )
Hence, /_POQ + /_PTQ = 180 deg
or, 110 + /_PTQ = 190
or, /_PTQ = 180 - 110
or, /_PTQ = 70 deg
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(4) If tangents PA and PB from a point "P" to a circle with center "O" are inclined to each other at angle of 80 deg, then /_ POA is equal to
(a) 50 deg
(b) 60 deg
(c) 70 deg
(d) 80 deg
sol) 50 deg
Justification ;-
/_AOB = 360 - 180 - 80 = 100
/_POA = /_AOB = 50 deg
2
*********************************
(5) In the figure XY and X1 Y1 are two parallel tangents to a circle with center "O" and another tangent AB with point of contact "C" intersecting XY at A and X1 Y1 at B then
/_ AOB =
(a) 80 deg
(b) 100 deg
(c) 90 deg
(d) 60 deg
sol) 90 deg
Justification:- If we join "O" and "C" we get /_OCA = 90 deg
/_OPA = 90 deg , we get OPAC is a square.
Similarly OCBQ is square
=> /_AOB = 90 deg
******************************
2) Two concentric circles are radii 5 cm and 3 cm are drawn. Find the length of the chord of the larger circle which touches the smaller circle.
sol)
Given AO = 5 cm ( radii of larger circle)
OD = 3cm ( radii of smaller circle)
Using Pythagoras theorem :-
So, AO^2 = OD^2 + AD^2
(5)^2 = (3)^2 + AD^2
25 - 9 = AD^2
16 = AD^2
4 = AD
Since AD = DB
AB = 8cm
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3) Prove that the parallelogram circumscribing a circle is a rhombus.
sol) Construction :-
*) Draw a circle with center "O"
*) Draw parallelogram ABCD which touches the circle P,Q,R,S.
Given :- AB || DC
AD || BC
RTP :- ABCD is a rhombus
In /_\ AOB and /_\DOC
AB = DC ( Given that AD || BC )
/_AOB = /_DOC ( vertically opposite angles )
/_BAO = /_DCO ( Alternate angles )
Hence; /_\ AOB ~ /_\ DOC
Hence; AO = CO and BO = DO
Since diagonals are bisecting each other, so given parallelogram is a rhombus
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4) 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. Find the sides AB and AC.
sol)
Since the external tangent are equal
DC = CE = 3cm
BD = BF = 9cm
Let AF = AE = X cm
Using Pythagoras theorem :-
AB^2 = BC^2 + AC^2
(AF + FB )^2 = (BD + DC)^2 + ( AE + EC)^2
(x + 9)^2 = (12)^2 + (3+x)^2
x^2 + 18x + 81 = 144 + 9 + 6x + x^2
12x = 72
x = 6
So, AB = AF + FB = 15 cm
AC = EC + AE = 9cm
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5) Draw a circle of radius 6cm. From a point 10 cm away from its center, construct the pair of tangents to the circle and measure their lengths. Verify by using Pythogoras Theorem.
sol)
Using Pythagoras theorem :-
AB^2 + AO^2 = OB^2
AB^2 + (6)^2 = (10)^2
AB^2 = 100 - 36
AB = 8 cm.
************************* 6) Construct a tangent to a circle of radius 4cm from a point on the concentric circle of radius 6 cm and measure its length . Also verify the measurement by actual calculation.
sol) Steps of Construction :-
1) Draw two concentric circles with center "O" and radii 4cm and 6 cm such that OP = 6 cm, OQ = 4 cm.
2) Join OP and bisect it at "M". i.e. M is the mid-point of OP
3) Taking "M" as center with "OM " as radius draw a circle intersecting the smaller circle in two points namely "T" and "S"
4) Join PT and PS.
PT and PS are the required tangents from a point "P" to the smaller circle, whose radius is 4 cm. By measurement : PT = 4.5 cm
Verification :-
OTP is right /_\ at T
OP^2 = OT^2 + PT^2
PT^2 = OP^2 - OT^2
PT^2 = 36 - 16 = 20
PT = _/20
PT = _/4 *+/ 5
PT = 2 _/5
PT = 2 * 2.236
PT = 4.47 cm
Similarly, PS = 4.47 cm
************************
7) 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.
sol) Steps of construction :-
1) Draw a circle, take a point outside the circle anywhere let it be "R".
2) Join "RO" and draw a perpendicular bisector of it. Let "M" be the midpoint of "RO"
3) Make an arc across the circle at two places such as "A" and "B"
4) Join "RA" and "RB". Then "RA" and "RB" are the required two tangents.
Diagram :-
************************************************
8) 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.
sol) Let us follow a step-wise approach :-
/_ABC = 90 deg
AB is the diameter.
From "P", tangent "PQ" intersects "BC".
To show : BQ + BC
Let us Join BP.
Now, PQ and BQ are the tangents drawn from an external point "Q" and they must be equal/
So, PQ = BQ -------(1)
=> /_PBQ = /_BPQ [ angles opposite to equal sides off a triangle are equal]
AB = Diameter,
=> /_APB = 90 deg ( Angles in semi circle)
=> /_APB + /_BPC = 180 deg ( Linear Pair)
So,
/_BPC = 180 - /_APB
/_BPC = 90 deg
In /_\ BPC,
/_BPC + /_PBC + /_PCB = 180 deg (Linear Pair)
So, /_PBC + /_PCB = 180 deg - /_BPC
=> 180 deg - 90 deg
=> 90 deg ...........(2)
Now,
/_BPC = 90 deg
So, /_BPQ + /_CPQ = 90 deg -----(3)
From ,(2) and (3)
/_PBC + /_PCB = /_ BPQ + /_CPQ
=> /_PCQ = /_CPQ (as, /_BPQ = /_PBQ)
Now, in /_\ PQC
/_PCQ = /_CPQ
.^., PQ = QC ----(4)
From (1) and (4),
BQ = QC
Hence the tangent at "P" bisects "BC"
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9) Draw a tangent to a given circle with center "O" from a point " R" outside the circle. How many tangents can be drawn to the circle from that point?
[ Hint : The distance of two points to the point of contact is the same .}
Sol) Given :- We are given a circle with centre 'O' and "o" point "R" outside it. We have to construct tangent from point "R" to the circle.
Steps of construction :-
1) Join OR and draw a perpendicular bisector of it. Let "M" be the mid-point of OR
2) Make an arc across the circle at two places such as "A" and "B".
3) Join RA and RB.
.^. RA and RB are the required two tangents.
Diagram :-
*************************************************************
9.4 Segment of a circle form by a secant
We have seen a line and a circle. When a line meets a circle in only one point, it is a tangent. A secant is a line which intersects the circle at two distinct points represented in the chord.
Here ‘l’ is the secant and AB is the chord.
Shankar is making a picture by sticking pink and blue paper. He makes many pictures. One picture he makes is of washbasin.
How much paper does he need to make this picture? This picture can be seen in two parts. A rectangle is there, but what is the remaining part?
It is the segment of the circle. We know how to find the area of rectangle. How do we find the area of the segment? In the following discussion we will try to find this area.
Do This
Shankar made the following pictures also along with washbasin.
What shapes can they be broken into that we can find area easily? Make some more pictures and think of the shapes they can be divided into different parts.
Lets us recall how to find the area of the following geometrical figures as given in the table.
9.4.1 Finding the Area of Segment of a circle
To estimate the area of segment of a circle, Swetha made the segments by drawing secants to the circle.
As you know a segment is a region, bounded by the arc and a chord, we can see the area that is shaded (
fig.(i) is a minor segment, semicircle in fig.(ii) and major segment in fig.(iii).
How do we find the area of the segment? Do the following activity.
Take a circular shaped paper and fold it along with a chord less than the diameter and shade the smaller part as shown in in the figure. What do we call this smaller part? It is a minor segment (APB).
What do we call the unshaded portion of the circle? Obviously it is a major segment (AQB).
You have already come across the sector and segment in earlier classes. The portion of some unshaded part and shaded part (minor segment) is a sector which is the combination of a triangle and a segment.
Let OAPB be a sector of a circle with centre O and radius ‘r’ as shown in the figure. Let the angle measure of ∠AOB be ‘x’
You know that area of circle when the angle measure at the centre is 360° is πr^ 2.
So, when the degree measure of the angle at the centre is 1°, then area of sector is
1 / 360 ° ×πr 2 .
Therefore, when the degree measure of the angle at the centre is x°, the area of sector
is x° / 360 ×πr 2 .
Now let us take the case of the area of the segment APB of a circle with centre ‘O’ and radius ‘r’, you can see that
Area of the segment APB
= Area of the sector OAPB - Area of ∆OAB
= x°/ 360 ° × πr^ 2 - area of ∆OAB
Try This
How can you find the area of major segment using area of minor segment?
Do This
1. Find the area of sector, whose radius is 7 cm. with the given angle:
i. 60° ii. 30° iii. 72° iv. 90° v. 120°
2. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 10 minutes
Now, we will see an example to find area of segment of a circle.
Example-1. Find the area of the segment AYB showing in the adjacent figure. If radius of the circle is 21 cm and ∠ AOB = 120 deg (Use π = 22/ 7 and _/3 = 1.732)
Solution : Area of the segment AYB = Area of sector OAYB − Area of ∆OAB
Now, area of the sector OAYB
= 120 deg / 360 × 22/7 * 21 * 21 cm2
= 462 cm2 ...(1)
For finding the area of ∆OAB, draw OM ⊥ AB as shown in the figure:-
Note OA = OB. Therefore, by RHS congruence,∆AMO ≅ ∆BMO
So, M is the midpoint of AB and ∠ AOM = ∠ BOM = 1/2 * 120 = 60 deg
Let, OM = x cm
So, from ∆OMA , OM / OA = cos 60 deg. .
or
x / 21 = 1/ 2 [because cos 60 deg = 1/2]
or
x = 21 / 2
So, OM = 21/ 2 cm
Also, AM / OA = sin 60°
AM / 21 = _/3 / 2 [because sin 60 deg = _/3 / 2 ]
So, AM = (21 _/ 3) / 2 cm.
Therefore
AB = 2AM
= ((2 * 21 _/ 3) / 2) × cm.
= 21_/ 3 cm
So, Area of /_\ OAB = 1 / 2 * AB * OM
= 1 / 2 *( 21 _/3) * (21 / 2) cm^2
= (441 / 4) * _/3) cm^2.........(2)
Therefore, area of the segment AYB = [ 462 - 441/4 * _/3] cm^2. (from (1) and (2))
= 21 / 4 ( 88 - 21 / _/3) cm^2
= 271.047 cm^2
Example-2. Find the area of the segments shaded in figure, if PQ = 24 cm., PR = 7 cm. and QR is the diameter of the circle with centre O (Take π= 22 / 7 )
Solution : Area of the segments shaded = Area of sector OQPR - Area of triangle PQR.
Since QR is diameter, ∠QPR = 90° (Angle in a semicircle)
So, using pythagoras Theorem
In ∆QPR,
QR^2 = PQ^2 + PR^2
= 24^22 + 7^2
= 576 + 49 = 625
QR = _/625 = 25 cm.
Then radius of the circle = 1/ 2 QR = 1 / 2 * (25) = 25 / 2 cm.
Now, area of semicircle OQPR = 1 / 2 πr^ 2
= (1/ 2 * 22/7 )* ( 25 /2 *25 / 2 )
= 327.38 cm 2 .... (1) .
Area of right angled triangle QPR
= 1/ 2 × PR × PQ
= 1 /2 × 7 × 24 = 84 cm 2 ..... (2)
From (1) and (2),
Area of the shaded segments = 327.38 - 84 = 243.38 cm^ 2
Example-3. A round table top has six equal designs as shown in the figure. If the radius of the table top is 14 cm., find the cost of making the designs with paint at the rate of 5 per cm 2 . (use _/3 = 1.732)
Solution : We know that the radius of circumscribing circle of a regular hexagon is equal to the length of its side.
∴ Each side of regular hexagon = 14 cm.
Therefore, Area of six design segments = Area of circle - Area of the regular hexagon.
Now, Area of circle = πr^ 2 = (22 /7) × 14 × 14 = 616 cm 2 ..... (1)
Area of regular hexagon = 6 × (_/3 /4) a ^2
= 6 × _/3 / 4 × 14 × 14
= 509.2 cm 2 ..... (2)
Hence, area of six designs
= 616 - 509.21 (from (1), (2)
= 106.79 cm 2 .
Therefore, cost of painting the design at the rate of 5 per cm 2
= 106.79 × 5
= 533.95
Exercise - 9.3
What is a chord or meaning of chord :- A lie segment connecting two points on a circle's circumference is a chord. When the chord passes through the center of a circle it is called the diameter.
1) A chord of a circle of radius 10 cm, subtends a right angle at the center. Find the area of the corresponding ( use Pi = 3.14)
(1) Minor segment (2) Major segment
sol) RTP :- Finding area of minor segment.
Given :- A chord of a circle of radius 10 cm
Radius :- OA = OB = 10 cm
Minor segment APB = Area of sector OAPB - Area of /_\AOB
(*) Finding area of OAPB :-
=>
360
=> 90 * 3.14 * (10)^2
36
4
=> 1 * 314
4
OAPB = 78.5 cm^2 ---------------(1)
(*) Finding Area of /_\AOB :-
It is a right angled triangle
/_ O = 90 , Base = OA and Height = OB
We have
Area of triangle = 1/2 * base * height
/_\ AOB = 1 * OA * OB
2
=> 1 *
=> 5 * 10
/_\ AOB = 50 cm^2 --------------(2)
Now,
Finding Area of minor segment (APB) = OAPB - AOB
=> 78. 5 - 50
=> 28.5 cm^2
(B) Finding area of major segment :-
major sector = Area of circle - Area of sector OAPB
Area of circle :- Pi * r^2
=> 3.14 * (10)^2
=> 3.14 * 100
=> 314 cm^2
Now
Area of major sector = Area of circle - OAPB
=> 314 - 78.5
=> 235.5 cm^2
************************************************
2) A chord of a circle of radius 12 cm, subtends an angle of 120 deg at the center. Find the area of the corresponding minor segment of the circle ( use Pi = 3.14 and _/3 = 1.732)
sol) Given :- Radius(r) = 12 cm and Angle(
Area of minor segment (APB) =Area of sector OAPB - /_\ OAB
(1) Finding Area of sector OAPB :-
=> 0 * Pi * r^2
360
=> 120 * 3.14 * (12)^2
360
=> 1 * 3.14 * 12 * 12
3
=> 1 * 3.14 * 4 * 12
=> 150. 72 cm^2
(2) Finding area of /_\ AOB :-
Area of /_\ AOB = 1/2 * Base * Height
We draw OM _|_ AB
.^. /_OMB = /_OMA = 90 deg
In /_\ OMA & /_\ OMB
/_OMA = /_OMB ( Both 90 deg)
OA = OB ( Both radius)
OM = OM ( Common)
.^. /_\ OMA ~ /_\ OMB ( By R.H.S congruency)
=> /_AOM ~ /_ BOM ( CPCT )
.^. /_AOM = /_ BOM = 1/2 /_BOA
.^. /_AOM = /_BOM = 1/
Also, since /_\ OMB ~ /_\ OMA
.^. BM = AM ( CPCT )
=> BM = AM = 1/2 AB ----------(1)
In right triangle /_\ OMA
Sin
Sin 60 = AM
AO
_/3 = AM
2 12
-/3 * 12 = AM
2
6_/3 = AM
In right triangle /_\ OMA :-
Cos
Hypotenuse
Cos 60 = OM
AO
1 = OM
2 12
12 = OM
2
6 = OM
OM = 6
Now,
AM = 1/2 * AB ( From 1 )
2AM = AB
AB = 2 (6_/3) [ Putting AM = 6 _/3 ]
AB = 12_/3
Now,
Area of /_\ AOB = 1/2 * Base * height
=> 1 / 2 * AB * OM
=> 1/2 * 12_/3 * 6
=> 36 _/3
=> 36 * 1.73
Area of /_\ AOB = 62.28 cm^2
Area of minor segment APB = OAPB - /_\ OAB
=> 150.72 - 62.28
=> 88.44 cm^2
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3) A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm. Sweeping through an angle of 115 deg . Find the total area cleaned at each sweep of the blades ( use Pi = 22/7).
sol) Given :- Radius (r) = 25 cm and sweeping angle
Total area cleaned by the two wipers = 2 * Area cleaned by 1 wiper
=> 2 * Area of sector with angle 115 deg
=> 2 *
360
=>
= 115 *
+> 115 * 11 *
=> 115 * 11 * 5 * 25
18 7
=> 158125 cm^2
126
Hence , Area cleaned by both wiper = 1,254.96 cm^2
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4) Find the area of the shaded region in the figure, where ABCD is a square of side 10 cm, and semicircles are drawn with each side of the square as diameter ( use Pi = 3.14)
sol) Given :- Side(s) of square ABCD = 10 cm
Formula :- Area of Square = s^2
=> (10)^2 = 100 cm^2
Given :- semicircle is drawn with side of square as diameter,
So, Diameter of semi-circle = side of a square = 10 cm
radius(r) of semicircle = side =
2 2
Now,
Area of semi-circle = 1/2 * Pi * r^2
=> 1/2 * 22/7 * (5)^2
=> 1/
=> 1 * 11 /7 * 25
=> 39.28
Area of 4 semi-circle = 4 * 39.28 = 157.12 cm^2
Area of shaded region :- area of 4 semi-circle - area of square
=> 157.12 - 100
=> 57.12 cm^2
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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 Pi = 22/7)
sol) Given :- Side(s) of square ABCD = 7 cm
Formula :- Area of Square = s^2
=> (7)^2 = 49 cm^2
radius(r) of semicircle = side =
2 2
Area of 2 semi-circle = 2 * ( 1/2 * Pi * r^2)
=> Pi * r^2
=> 3.14 * (3.5)^2
=> 3.14 * 3.5 * 3.5
=> 3.14 * 12. 25
=> 38.465 cm^2
area of shaded region = area of square - area of 2 semicircle
=> 49 - 38.465
Area of shaded region = 10.5 cm^2
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6) In figure, OACB is a quadrant of a circle with center 'O" and radius 3.5 cm. If OD = 2cm, find the area of the shaded region. ( use Pi = 22/7)
sol) RTP :- Finding area of shaded region
Area of shaded region = Area of quadrant OACB - Area of /_\ BOD
(1) Finding area of quadrant ( OACB) :-
Given :- radius (r) = 3.5 cm
We know that
Area of Quadrant OACB = 1/4 * Area of circle
=> 1/4 * Pi * r^2
=> 1/4 * 22/7 * (3.5)^2
=> 1/4 * 22/7 * 3.5 * 3.5
Area of Quadrant (OACB) = 9.625 cm^2
(2) Area of /_\ BOD :-
Area of /_\ BOD = 1/2 * Base * Height
=> 1/2 * OB * OD
=> 1/2 * 3.5 * 2
=> 3.5 cm^2
Now
Area of shaded region = Area of quadrant OACB - Area of /_\ BOD
=> 9.625 - 3.5
Area of shaded region = 6.125 cm^2
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7) AB and CD are respectively arcs of two concentric circles of radii 21 cm, and 7 cm , with center "O" ( see figure) . IF /_AOB = 30 deg, find the area of the shaded region ( use Pi = 22/7)
sol) Given :- radii(R1 = 21 cm and radii(r2) = 7 cm
RTP :- Area of shaded region ABCD
Area of shaded region is given by :-
360
=> 3
36
=> 1 * 22 [ 441 - 49 ]
12 7
=> 22 [ 392 ]
7 * 12
=> 11 * 392
7 * 6
=> 102. 67 cm^2
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8) calculate the area of the designed region if figure, common between the two quadrants of the circles of radius 10 cm, each ( use Pi = 3. 14)
sol) Given :- radius(r) = 10 cm
RTP :- Area of designed region
Area of designed region = area of two quadrant - Area of square
(1) Area of two quadrant :-
=> 2 * 1/4 * Pi *r^2
=> 2 * ( 1/4 * 3.14 * (10)^2 )
=> 2 * ( 1/4 * 3.14 * 100 )
=> 2 * ( 3.14 * 25)
=> 157 cm^2
(2) Area of square :-
=> s ^2 => (10)^2 = 100 cm^2
Now.
Area of designed region = area of two quadrant - Area of square
=> 157 - 100
=> 57 cm^2
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Optional exercise :-
1. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line - segment joining the points of contact at the center.
sol) Given : A circle with center 'O'
Tangents PA and PB drawn from external point P.
R.T.P :- /_APB + /_ AOB = 180 deg
Proof : Since PA is tangent,
OA _|_ PA
(Tangent at any point of circle is perpendicular to the radius through point of contact)
Therefore, /_OAP = 90 deg ------(1)
similarly /_OBP = 90 deg .........(2)
In quadrilateral OAPB
( According,Angle sum property of Quadrilateral)
/_OAP + /_APB + /_OBP +/_AOB = 360 deg
90 deg + /_APB + 90 deg + /_AOB = 360 deg
180 deg + /_APB + /_ AOB = 360 deg
/_APB + /_AOB = 360 deg - 180 deg =180 deg
Hence proved
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2. PQ is a chord of length 8cm of a circle of radius 5cm. The tangents at P and Q intersect at a point T (See figure). Find the length of TP.
Solution :- Join OT
Let OT intersect PQ at R
From theorem 10.2,
Length of tangents from external points are equal
So, TP = TQ
In /_\ TPQ
TP = TQ (i.e. two sides are equal)
So, /_\ TPQ is an isosceles triangle.
Here , OT bisector of /_PTQ
So, OT _|_ PQ ( Angle bisector and altitude of isosceles triangle are same)
Since OT _|_ PQ
PR = RQ ( _|_ from center to a chord, bisects the chord) ----(1)
since PQ = 8 cm
QR = 4 cm
PR = QR = 1 / 2 PQ = 8 / 2 = 4 cm.
In right triangle ORP
By Pythagoras Theorem
OP^2 = PR^2 + OR ^2
5^2 = 4^2 + OR^2
25 = 16 + OR^2
9 = OR^2
3 = OR
Let TP = x
In right triangle PRT
(TP)^2 = (PR)^2 + (RT)^2
x^2 = 4^2 + RT^2
x^2 = 16 + RT^2.........(1)
Now,
Since TP is a tangent
OP _|_ TP
(Tangent at any point of a circle is perpendicular to the radius through point of contact)
Therefore, /_OPT = 90 deg
In Right angled triangle /_\ OPT
(OT)^2 = (OP)^2 + (TP)^2
(OT)^2 = 5^2 + x^2
(OR + RT )^2 = 25 + x^2
( 3 + RT )^2 = 25 + x^2
(3 + RT)^2 is in ( A+ B)^2 form
3^2 + RT^2 + 2 (3)(RT) = 25 + x^2
9 + RT^2 + 6RT = 25 +x^2
But : x^2 = 16 + RT^2 from (1) substitute
we get,
9 + RT^2 + 6RT = 25 + (16 + RT^2)
6RT = 32
RT = 32 / 6
RT = 16 / 3
Now putting RT value in (1) we get
x^2 = 16 + RT^2
x^2 = 16 + ( 16/3) ^2
x^2 = 16 + ( 256 / 9)
x^2 = (16 * 9) + 256) / 9
x^2 = 400 / 9
x = 20 / 3
Hence proved, length of tangent TP = x = 20/ 3 cm
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3) Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
sol) A circle with center 'O' touches the sides AB,BC,CD and DA of a quadrilateral ABCD at the points P,Q,R, and S respectively.
R.T.P :- Opposite sides of a quadrilateral circumscribing a circle , i.e
/_ AOB + /_ COD = 180 deg
/_ AOD + /_ BOC = 180 deg
Construction :- Join OP, OQ, OR and OS.
Proof :- In /_\ AOP and /_\ AOS
(Lengths of tangents drawn from external point are equal )
AP = AS
AO = AO ( common)
OP = OS ( Radius)
Therefore, /_\ AOP = /_\ AOS (SSS congruence rule )
/_AOP = /_AOS (CPCT)
i.e, /_1 = /_8 ......(1)
similarly ,we can prove
/_2 = /_3 ....(2)
/_5 = /_4.....(3)
/_6 = /_7.......(4)
Now,
/_1 + /_2 + /_3 + /_4 + /_ 5 + /_6 +/_7 +/_8 = 360 deg
from (1) (2) (3) (4)
/_1 + /_2 + /_2 + /_5 + /_5 + /_6 + /_6 +/_1 = 360 deg
2 ( /_1 + /_2 + /_5 + /_6) = 360
/_1 + /_2 + /_5 + /_6 = 360/2
(/_1 + /_2) +( /_5 + /_6 )= 180
/_ AOB + /_ COD = 180 deg
Hence both angle are supplementary
similarly we can prove /_AOD + /_ BOC = 180 deg
Hence proved.
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4). Draw a line segment AB of length 8cm. Taking A as centre,
draw a circle of radius 4cm and taking B as centre, draw
another circle of radius 3cm. Construct tangents to each circle
from the center of the other circle.
sol) Steps of construction :-
1) Draw a line segment AB of length 8cm
A------------8cm-----------------------B
2) Taking 'A' as center, draw a circle of radius 4 cm
3) Taking 'B' as center, draw a circle of radius 3 cm.
Now, we need to draw tangent from point 'A' to the right circle, and from point 'B' to the left circle.
To draw tangents to the circle, we need to draw perpendicular bisector of line 'AB'
4) Make perpendicular bisector of 'AB'
Let M be the midpoint of 'AB'
5) Taking 'M' as center and 'MA' as radius, draw a circle.
6) Let green circle intersect left circle at P, Q
7) Let green circle intersect right circle at R,S
8) Join BP,BQ,AR and AS
Therefore AR, AS and BP, BQ are the required tangents
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5). Let ABC be a right triangle in which AB = 6cm, BC = 8 cm and ∠B = 90deg . BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
sol) Steps of construction :-
(*) Steps to draw /_\ABC :-
1) Draw Base BC of side 8 cm
B-----------8cm---------C
2) Draw /_B = 90 deg
3) Taking 'B' as center, 6 cm as radius we draw an arc .Let the point where arc intersects the ray be point A
4) Join AC
Therefore ,/_\ ABC is the required triangle
Steps to draw circle and tangents:-
1) Draw _|_ bisector of line BC. Let the line intersect BC at point 'E'
2) Now, E is the mid-point of BC. Taking 'E' as center, and 'BE' as radius,draw a circle.
Now, we need to construct tangents from point 'A' to the circle
3 Join point 'A' to center of circle 'E'. Make perpendicular bisector of 'AE'. Let 'M' be the mid-point of 'AE'.
4) Taking 'M' as center and 'MA' as radius, draw a circle.
5) Let blue circle intersect the other circle at 'B' and 'Q'
6) Join 'AQ'
Thus, AB and AQ are the required tangents.
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sol) Steps of construction :-
1) Draw a line segment AB of length 8cm
A------------8cm-----------------------B
2) Taking 'A' as center, draw a circle of radius 4 cm
3) Taking 'B' as center, draw a circle of radius 3 cm.
Now, we need to draw tangent from point 'A' to the right circle, and from point 'B' to the left circle.
To draw tangents to the circle, we need to draw perpendicular bisector of line 'AB'
4) Make perpendicular bisector of 'AB'
Let M be the midpoint of 'AB'
5) Taking 'M' as center and 'MA' as radius, draw a circle.
6) Let green circle intersect left circle at P, Q
7) Let green circle intersect right circle at R,S
8) Join BP,BQ,AR and AS
Therefore AR, AS and BP, BQ are the required tangents
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5). Let ABC be a right triangle in which AB = 6cm, BC = 8 cm and ∠B = 90deg . BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
sol) Steps of construction :-
(*) Steps to draw /_\ABC :-
1) Draw Base BC of side 8 cm
B-----------8cm---------C
2) Draw /_B = 90 deg
3) Taking 'B' as center, 6 cm as radius we draw an arc .Let the point where arc intersects the ray be point A
4) Join AC
Therefore ,/_\ ABC is the required triangle
Steps to draw circle and tangents:-
1) Draw _|_ bisector of line BC. Let the line intersect BC at point 'E'
2) Now, E is the mid-point of BC. Taking 'E' as center, and 'BE' as radius,draw a circle.
Now, we need to construct tangents from point 'A' to the circle
3 Join point 'A' to center of circle 'E'. Make perpendicular bisector of 'AE'. Let 'M' be the mid-point of 'AE'.
4) Taking 'M' as center and 'MA' as radius, draw a circle.
5) Let blue circle intersect the other circle at 'B' and 'Q'
6) Join 'AQ'
Thus, AB and AQ are the required tangents.
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6). Find the area of the shaded region in the figure, given in which two circles with centers A and B touch each other at the point C. If AC = 8 cm. and AB = 3 cm.
sol) Given AB = 3 cm, AC = 8 cm,, BC = 8-3 = 5cm
Now area of bigger circle = Pi * r^2
= Pi * (8)^2
= 64Pi cm^2
Now area of smaller circle = Pi * r^2
= Pi * ( 5)^2
=25Pi cm^2
Now,
Area of shaded region = Area of bigger circle - smaller circle
= 64Pi - 25Pi
= 39Pi cm^2
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7. ABCD is a rectangle with AB = 14 cm. and BC = 7 cm. Taking DC, BC and AD as diameters, three semicircles are drawn as shown in the figure. Find the area of the shaded region
sol) Given:
Length of a rectangle (AB) = (DC) = 14 cm
Breadth of a rectangle ( BC) = (AD)= 7 cm
Area Of Semicircle with Diameter DC:
Diameter (D) = 14
radius (r) = 14/2 =7
Area of semicircle = 1/2πr²
= 1 /2 * (22/7) * (7)²
= 11/
= 11 * 7
= 77 cm² -----(1)
(*) AREA OF RECTANGLE (ABCD)
= Length × Breadth
= AB × BC
= 14* 7
= 98 cm ----(2)
Area of 2 semicircle with diameter BC and AD
Here, Diameter (D) = 7 cm
radius(r) = (D/2) = 7 / 2
Area of 2 semicircle =
= Pi * (r)^2
=(22/7) × (7/2)²
= (
= (11/
= 11 ×7 / 2
= 77 /2 cm² -----------(3)
AREA OF SHADED REGION :- (2) - (1) + (3)
= Area of rectangle ABCD - area of semicircle with diameter DC + Area of 2 semicircle with diameter BC and AD
Area of shaded region = 98 - 77 + 77/2
= 21 + 77/2
= (42 +77)/2
= 119/2
= 59.5 cm²
Area of shaded region = 59.5 cm²
Hence, the Area of shaded region is 59.5 cm².
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What we have discussed :-
In this chapter, we have studied the following points.
1. The meaning of a tangent to a circle and a secant. We have used the idea of the chord of a circle.
2. We used the ideas of different kinds of triangles particularly right angled triangles and isosceles triangles.
3. We learn the following:
a) The tangents to a circle is perpendicular to the radius through the point of contact.
b) The lengths of the two tangents from an external point to a circle are equal.
4. We learnt to do the following:
a) To construct a tangent to a circle at a given point when the center of the circle is known.
b) To construct the pair of tangents from an external point to a circle.
5. We learnt to understand how to prove some statements about circles and tangents. In this process learnt to use previous results and build on them logically to from new results.
6. We have learnt Area of segment of a circle =
Area of the corresponding sector - Area of the corresponding triangle
Optional exercise
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