Maharashtra Board Class 10 Math Part – 2 Solution Chapter 3 Problem Set 3 – Circle
Balbharati Maharashtra Board Class 10 Math Part – 2 Solution Chapter 3: Circle. Marathi or English Medium Students of Class 10 get here Circle full Exercise Solution.
Std |
Maharashtra Class 10 |
Subject |
Math Part 2 Solution |
Chapter |
Circle |
Problem Set |
3 |
Problem Set – 3
(1) Four alternative answers for each of the following questions are given. Choose the correct alternative.
(1) Two circles of radii 5.5 cm and 3.3 cm respectively touch each other. What is the distance between their centers?
(A) 4.4 cm
(B) 8.8 cm
(C) 2.2 cm
(D) 8.8 or 2.2 cm
Solution:
Sum of radius of two circle = distance between their centers when they touch each other externally.
∴ Distance = 5.5 + 3.3 = 8.8 cm
Also, difference in their radius = distance between them when they touch each other internally.
∴ Distance = 5.5 – 3.3 = 2.2m.
The question doesn’t mention whether the circle touch each other externally or internally.
Ans: distance between their center = 8.8 or 2.2 cm Option – (D).
2) Two circles intersect each other such that each circle passes through the centre of the other. If the distance between their centres is 12, what is the radius of each circle?
(A) 6 cm
(B) 12 cm
(C) 24 cm
(D) Can’t say
Solution:
(B) 12 cm
3) A circle touches all sides of a parallelogram. So the parallelogram must be a, ……………….
(A) rectangle
(B) Rhombus
(C) Square
(D) Trapezium
Solution:
Option (B) rhombus or option (c) square
4) Length of a tangent segment drawn from a point which is at a distance 12.5 cm from the centre of a circle is 12 cm, find the diameter of the circle.
(A) 25 cm
(B) 24 cm
(C) 7 cm
(D) 14 cm
Solution:
(A) 12.5 + 12.5 = 25cm
5) If two circles are touching externally, how many common tangents of them can be drawn? (A) One
(B) Two
(C) Three
(D) Four
Solution:
(c) Three.
6) ∠ACB is inscribed in arc ACB of a circle with centre O. If ∠ACB = 65°, find m (arc ACB).
(A) 65°
(B) 130°
(C) 295°
(D) 230°
Solution:
m (arc ACB) = 2 × ∠ACB [Inscribed angle theorem]
= 2 × 65° = 130°
Ans- (B) = 130°
7) Chords AB and CD of a circle intersect inside the circle at point E. If AE = 5.6, EB = 10, CE = 8, find ED.
(A) 7
(B) 8
(C) 11.2
(D) 9
Solution:
AE×EB = CE×ED
Or, 5.6 × 10 = 8 × ED
Or, ED = 56/8 = 7
∴ Ans => (A) 7
8) In a cyclic ABCD, twice the measure of ∠A is thrice the measure of ∠c. Find the measure of ∠c?
(A) 36
(B) 72
(C) 90
(D) 108
Solution:
2∠A = 3∠C
∴ ∠A + ∠C = 180°
Or, ∠A = 3∠C/2
Or, 3∠C/2 + ∠C = 180°
Or, 5∠C = 180×2
Or, ∠c = 180×2/8 = 72°
(B) 72°
9) Points A, B, C are on a circle, such that m(arc AB) = m (arc BC) = 120°. No point, except point B, is common to the arcs. Which is the type of ∆ABC?
(A) Equilateral triangle
(B) Scalene triangle
(C) Right angled triangle
(D) Isosceles triangle
Solution:
m (arc AB) + m (arc BC) + m (arc AC) = 360° [Definition of measured arc]
Or, 120° + 120° + m (arc AC) = 360°
Or, m (arc AC) = 360° – 240°
= 120°
∴ Corresponding chords are equal for congruent arcs.
∴ AB = BC = AC
∴∆ABC is an equilateral △
10) Seg XZ is a diameter of a circle. Point Y lies in the its interior. How many of the following statements are true?
(i) It is not possible that ∠XYZ is an acute angle.
(ii) ∠XYZ can’t be a right angle
(iii) ∠XYZ is an obtuse angle.
(iv) Can’t make a definite statement for measure of ∠XYZ
(A) Only one
(B) Only two
(C) Only three
(D) All
Solution:
True statements.
(ii) ∠XYZ can’t be a right angle.
(iii) ∠XYZ is an obtuse angle
∴ Ans => (B) Only are true.
(2) Line l touches a circle with centre O at point P. If radius of the circle is 9 cm, answer the following.
(1) What is d (O, P) =? Why?
(2) If d (O, Q) = 8 cm, where does point Q lie?
(3) If d(PQ) = 15 cm, How many locations of point R are line on line l? At what distance will each of them be from point P?
Solution:
In the adjacent fig it is given that, O is Centre of circle.
l is tangent touching the circle at point P. radius of circle = 9 cm.
(1) ∴ d (O, P) = 9 cm
[∵ P is the point where tangent l touches the circle]
(2) If d (O, Q) = 8cm
Point Q li in the interior of the circle.
(3)In figure 3.83, M is the centre of the circle and seg KL is a tangent segment
If MK = 12, KL = 6√3 then find –
(1) Radius of the circle
(2) Measures of ∠K and ∠M
Solution:
In the adjacent fig. it is given that, M is the centre of the circle.
KL is the tangent circle
MK = 12, KL = 6√3
∴ In ∆MLK
∠MLK = 90° [∵ ML is a radius of circle and KL is a tangent touching at L]
(1) ∴ By Pythagoras theorem,
ML^{2} + KL^{2} = MK^{2}
Or, ML^{2} + (6√3)^{2} = 12^{2}
Or, ML^{2} = 1444 – 108
Or, ML = √36 = 6
∴ Radius of circle = ML = 6
(2) In ∆MLK
ML/MK = 6/12 = 1/2 [Base/hypotenuse]
Also, KL/MK = 6√3/12 = √3/2 [height/hypotenuse]
∴ ∠M = 60° [Converse of 30 – 60 – 90 theorem]
∠K = 30° [Converse of 30 – 60 – 90 theorem]
(4) In figure 3.84, O is the centre of the circle. Seg AB, seg AC are tangent segments. Radius of the circle is r and l(AB) = r, Prove that,ABOC is a square.
Solution
In the adjacent fig O is the centre of circle,
AB & AC are tangents.
Radius of circle = r
l (AB) = r
Construction, joint OB & OC
Now, in ABOC
OB = OC = radius of circle [∵ B and C are part of points on tangents of the circle where they touch the circle]
∴ OB = OC = r
Also AB = r [Given]
Also, ∠ABO = ∠ACO = 90° [Tangents are ⊥ to radius of circle]
∴ Opposite angle are 90°
∴ All the angles of ABOC are 90° [∵ two opposite angles are 90°]
Therefore ABOC is a parallelogram.
Now, 3 sides OB, OC, AB of ABOC are equal.
OB = OC = AB = r
∴ Side AC is also equal to r because ABOC is a parallelogram.
∴ All sides of ABOC are equal and all their interior angles are 90°.
Therefore, ABOC is a square (Proved)
(5) In figure 3.85, ABCD is a parallelogram. It circumscribes the circle with centre T. Point E, F, G, H are touching points. If AE = 4.5, EB = 5.5, find AD.
Solution:
In the adjacent fig. It is given that, ABCD is a parallelogram.
ABCD circumscribes the circle with centre T and touching points E, F, G, H.
AE = 4.5, EB = 5.5
∴ AB = DC & AD = BC —- (i) [Opposite sides of Parallelogram are equal]
Also, AH = AE —– (ii)
DH = DG —- (iii)
FC = GC (iv)
BF = EB — (v)
[Tangents drawn from some point outside the circle are equal]
∴ Adding (ii), (iii), (iv), (v) we get,
AH + DH + FC + BF = AE + DG + GC + EB
Or, AD + BC = AE + EB + DG + GC
Or, AD + BC = AB + DC
Also, AD = BC & AB = DC [From equation (ii)]
AD + AD = AB + AB
Or, 2AD = 2AB
Or, AD = AB
∴ AB = AE + EB
= 4.5 + 5.5
= 10
∴ AD = AB = 10
(6) In figure 3.86, circle with centre M touches the circle with centre N at point T. Radius RM touches the smaller circle at S. Radii of circles are 9 cm and 2.5 cm. Find the answers to the following questions hence find the ratio MS:SR.
(1) Find the length of segment MT
(2) Find the length of seg MN
(3) Find the measure of ∠NSM
Solution:
In the adjacent fig. it is given that,
Two circle touch each other internally.
Radius RN touches smaller circle at S.
Radius of bigger circle = 9cm
Radius of smaller circle = 2.5 cm
(1) Length of seg MT = radius of bigger circle
= 9 cm [∵ T is where both circle touches each other]
(2) Length of seg MN = distance between circle’s centres
= 9 – 2.5 [∵ they touch each other internally]
= 6.5 cm
(3) ∠NSM = 90° [∵ MS is the tangent of smaller circle touching at point S and NS is the radius of smaller circle]
Construction join NS & NM.
∴ In ∆NMS
∠NSM = 90°
∴ By Pythagoras theorem,
NS^{2} + MS^{2} = NM^{2}
Or, 2.5^{2} + MS^{2} = 6.5^{2} [∵ NS is the radius of smaller circle & NN is the distance between the centres of two circle]
Or, MS = √42.25 – 6.25
= √36
= 6
∴ RS = MR – MS where R is a cyclic point on the bigger circle
∴ RS = 9 – 6 [∵ MR is radius of bigger circle]
Or, SR = 3
∴ MS:SR = 6:3 = 2:1
(7) In the adjoining figure circles with centres X and Y touch each other at point Z. A secant passing through Z intersects the circles at points A and B respectively. Prove that, radius XA||radius YB.
Fill in the blanks and complete the proof.
Construction: Draw segments XZ and ……….
Proof: By theorem of touching circles, points X, Z, Y are ……….
∴∠XZA ≅ …….. opposite angles
Let ∠XZA = ∠BZY = a ………. (I)
Now, seg XA ≅seg XZ ……… (………..)
∴∠XAZ = ……. = a …… (isosceles triangle theorem) (II)
Similarly, seg YB ≅ …… ……. (……)
∴∠BZY ….. = a …… (……) (III)
∴From (I), (II), (III),
∠XAZ = …..
∴Radius XA||radius YB ….. (…..)
Solution:-
Construction: Draw segments XZ and ZY
Proof: By theorem of touching circles, points X, Z, Y are collinear.
∴ ∠XZA ≅ ∠YZB – opposite angles
Let ∠XZA = ∠BZY = a —- (I)
Now, seg XA ≅ seg XZ — (Radius of circle with centre x)
∴ ∠XAZ = ∠XZA = a (isosceles triangle theorem) —- (II)
Similarly, seg YB ≅ YZ (Radius of circle with centre y)
∴ ∠BZY = ∠YBZ = a — (Isosceles triangle theorem) — (III)
∴From (I), (II), (III),
∠XAZ = ∠YBZ = a
∴ Radius XA || radius YB [∵ their alternate angle along seg AB are equal]
(8) In figure 3.88, circles with centres X and Y touch internally at point Z. Seg BZ is a chord of bigger circle and it intersects smaller circle at point A. Prove that, seg AX || seg BY.
Solution:
In the adjacent fig. it is given that X & Y are centres of the circles. BZ is a chord of bigger circle and it intersects the smaller circle at A.
Construction : draw a line connecting Y, X, Z
Now, In ∆AXZ,
XZ ≅ XA [∵ Z and A are radius of smaller circle]
∴ ∠XZA ≅ ∠XAZ [Isosceles triangle theorem]
Now, In ∆BYZ.
BY≅ YZ [Radius of bigger circle]
∴ ∠YZB ≅ ∠YBZ [Isosceles triangle theorem]
Or, ∠XAZ ≅ ∠YBZ [∵ Z – A – B is a chord]
∴ Radius XA||radius YB [∵ their corresponding angles along seg BZ are equal]
(9) In figure 3.89, line l touches the circle with centre O at point P. Q is the mid point of radius OP. RS is a chord through Q such that chords RS||line l. If RS = 12 find the radius of the circle.
Solution:
In the adjacent fig. it is given that O is the centre of circle.
Line l is a tangent that touches the circle at point P.
RS is a chord through Q.
RS || line l, RS = 12
Q is the mid-point of OP.
Let, the radius of circle be r.
Now, OP⊥lin l [∵ line l is the tangent meeting the circle at point P]
∴ RS ⊥ OP [∵ RS || line l]
∴ RQ = QS = 1/2 RS [∵ Perpendicular draw from the centre of a circle to a chord bisects the chord]
= ½ × 12
= 6
OQ = 1/2 × OP [∵ Q is the midpoint of OP]
= ½ ×r [∵ OP radius]
= r/2
Now, construction : join OS
In ∆OQS
∠OQS = 90° [∵ OP ⊥ RS]
∴ By Pythagoras theorem,
OQ^{2} + QS^{2} = OS^{2}
Or, (r/2)^{2} + 6^{2} = r^{2} [∵ OS is a radius]
Or, r^{2}/4 + 36 = r^{2}
Or, r^{2} – r^{2}/4 = 36
Or, 3r/4 = 36
Or, r^{2} = 36×4/3
Or, r = √48 = 4√3
∴ Radius 1 circle is 4√3.
(10) In figure 3.90, seg AB is a diameter of a circle with centre C. Line PQ is a tangent, which touches the circle at point T.
seg AP ⊥ line PQ and seg BQ ⊥ line PQ. Prove that, seg CP≅seg CQ.
Solution:
In the adjacent fig. it is given that, AB is the diameter of the circle with centre C.
PQ is a tangent that touches the circle at point T. Let, radius = r
AP⊥PQ, BQ⊥PQ
CT⊥PQ [Tangent theorem]
∴ AP||CT||BQ [∵ all are ⊥ to PQ]
∴ PT/TQ = AC/BC [∵ Theorem of three parallel lines and their transversal]
Or, PT/TQ = r/r [∵ AB is the diameter and C is the centre ∴ AC and BC are radius ]
Or, PT/TQ = 1
Or, PT = TQ
∴ In∆CTP
By Pythagoras theorem, CP^{2} = PT^{2} + CT^{2}
Or, CP^{2} = TQ^{2} + CT^{2} [∵ PT = TQ] —– (i)
Now, In∆CTQ,
By Pythagoras theorem,
CQ^{2} = TQ^{2} + CT^{2} —– (ii)
Comparing equation (i) and (ii) we get.
CP^{2} = CQ^{2}
Or, CP = CQ
Or, CP ≅ CQ Proved
(11) Draw circles with centres A, B and C each of radius 3 cm, such that each circle touches the other two circles.
Solution:
To draw three circle of radius 3cm we have to find the distance their centre first.
The below fig show the three circle.
Distance between their centre = 3+3 = 6cm
Now, first draw an equilateral ∆ with 6cm as each side and each vertex representing the centre of each circle.
The below fig has 3 circles with centre A, B, C of radius 3cm touching each other at point D, E, F.
(12) Prove that any three points on a circle cannot be collinear.
Solution:
In the adjacent fig. A, B, C are random points. Line K & L are perpendicular bisector of chord AB & BC respectively meeting the centre O of circle by perpendicular bisector of chords theorem.
Now, If AB, c are collinear then perpendicular bisector of AB & BC must be parallel.
∴ Line k & line L must not meet at any point on the 2D surface.
But we can see that the perpendicular bisects are meeting at point O.
∴ It contradicts the statement that point A, B, C, are collinear.
∴ A, B, C are non-collinear and any point on the circumference of a circle are non-collinear (Proved).
(13) In figure 3.91, line PR touches the circle at point Q. Answer the following questions with the help of the figure.
(1) What is the sum of ∠TAQ and ∠TSQ?
(2) Fid the angles which are congruent to ∠AQP.
(3) Which angles are congruent to ∠QTS?
(4) ∠TAS = 65°, find the measure of ∠TQS and arc TS.
(5) If ∠AQP = 42° and ∠SQR = 58° find measure of ∠ATS.
Solution:
In the adjacent fig. it is given that, tangent PR touches the circle at point Q.
(1) From the given, information in the adjacent fig. We can say that AQST is a cyclic quadrilateral size all points A, Q, S, T lies on the circumference of the circle.
∴ ∠TAQ + ∠TSQ = 180° [∵ Sum of opposite of a cyclic quadrilateral is 180°]
(2) The angle between the chords touching the point of contact of a —- and the arc opposite to it are congruent.
∴ Here the chords are AQ & QS and PR is the tangent.
∴ ∠AQP ≅ ∠ATQ & ∠AQP ≅ ∠ASQ
∴ ∠AQP ≅ ∠ATQ ≅ ∠ASQ
(3)∠QTS ≅ ∠SAQ [∵ angles inscribed on the same chord as base are congruent]
Also, from the property mentioned in the previous question we can say that,
∠QTS ≅ ∠SQR
∴ ∠QTS ≅ ∠SAQ ≅ ∠SQR
(4) Given, ∠TAS = 65°
∠TAS = ∠TQS = 65° [∵ angle inscribed in the same are arc congruent]
∴ m (arc TS) = 2× ∠TAS [inscribed angle theorem]
Or, m (arc TS) = 2 × 65 = 130°
(5) Given, ∠AQP = 42°, ∠SQR = 58°
∴ ∠AQS = 180° – (∠AQP + ∠SQR) [∵ all three angles are supplementary angles]
= 180° – (42 + 58)
= 180° – 110
= 70°
∴ ∠ATS + ∠AQS = 180° [Opposite angles of cyclic quadrilateral are supplementary]
Or, ∠ATS + 70° = 180°
Or, ∠ATS = 180° – 70°
Or, ∠ATS = 110°
(14) In figure 3.92, O is the centre of a circle, chord PQ ≅ chord RS
If ∠POR = 70°and (arc RS) = 80°, find –
(1) m(arc PR)
(2) m(arc QS)
(3) m(arc QSR)
Solution:
In the adjacent fig. it is given that, O is the centre of the circle.
PR ≅ RS, ∠POR = 70°
m (arc RS) = 80°
(1) m (arc PR) = ∠POR
Here, PO and OR are radius of the circle.
∴ m (arc PR) = 70°
[Angle between two radius is the measure of the chord between them]
(2) m (arc PQ) = m (arc RS) [∵ arcs subtended by equal chords are congruent]
Or, m (arc PQ) = 80°
∴ m (arc QS) = 360° – [m (arc PQ) + m (arc RS) + m (arc PR)]
Or, m (arc QS) = 360° – [80° + 70° + 80°] [Definition of measure of circle]
Or, m (arc 30) = 360° – 230 = 130°
(3) m (arc QSR) = m (arc QS) + m (arc RS)
Or, m (arc QSR) = 130° + 30°
Or, m (arc QSR) = 210°
(15)In figure 3.93, m (arc WY) = 44°, m (arc ZX) = 68°, then
(1) Find the measure of ∠ZTX
(2) If WT = 4.8, TX = 8.0, YT = 6.4, find TZ.
(3) If WX = 25, YT = 8, YZ = 26, find WT.
Solution:
m (arc NY) = 44°
m (arc ZX) = 68° given,
(1) Construction : join YX
∴ ∠ZYX = 1/2 XM (arc ZX) [inscribed angle theorem]
Or, ∠ZYX = 1/2 × 68 = 34°
Also, ∠WXY = 1/2 × m (arc WY) [Inscribed angle theorem]
Or, ∠WXY = ½ × 44 = 22°
∴ In ∆TYX,
∠TYX = ∠ZYX [∵ Z – T – Y]
= 34°
∠TXY = ∠WXY [∵ W – T – X]
∴ ∠YTX = 180° – (∠TYX + ∠TXY) [sum of interior angles of a triangle is 180°]
Or, ∠YTX = 180° – (22° + 34°)
= 180° – 56
= 124°
∴ ∠ZTX = 180° – ∠YTX [Collinear angles]
= 180°- 124°
= 56°
(2) Given, WT = 4.8, TX = 8.0, YT = 6.4
∴ By theorem of intersecting chords,
WT × TX = YT × TZ
Or, 4.8 × 8 = YT × 6.4
Or, YT = 4.8×8/6.4
Or, YT = 6
(3) Given, WY = 25, YT = 8, YZ = 26
∴ By theorem of intersecting chords,
WT × TX = YT × TZ
Or, WT × (WX – WT) = 8 × (YZ – YT)
Or, WT × (25 – WT) = 8 × (26 – 8)
Or, WT × 25 – WT^{2} = 8 × 18
Or, WT^{2} – 25 × WT + 144 = 0
Or, WT^{2} – 16 × WT – 9 × WT + 144 = 0
Or, WT (WT – 16) – 9 (WT – 16) = 0
Or, (WT – 16) (WT – 9) = 0
Either, WT – 16 = 0
Or, WT = 16
Or, WT – 9 = 0
Or, WT = 9
∴ WT = 16, 9
(16) In figure 3.94,
(1) m (arc CE) = 54°,
m (arc BD) = 23°, find measure of ∠CAE
(2) If AB = 4.2, BC = 5.4
AE = 12.0, find AD
(3) If AB = 3.6, AC = 9.0,
AD = 5.4, find AE
Solution:
In the adjacent fig. It is given that,
(1) m (arc CE) = 54°, m (arc BD) = 23°
Construction : join CD .
∴ ∠CDE = 1/2 × m (arc CE) [Inscribed angle theorem]
Or, ∠CDE = 1/2 × 54° = 27°
∠DCB = ½ ×m (arc BD) = [Inscribed angle theorem]
= ½ × 23 = 11.5°
∴ In ∆ADC,
∠ADC = 180° – ∠CDE [Collinear angles]
Or, ∠ADC = 180° – 27°
= 153°
So, ∠DCB = ∠DCA [∵ C – B – A]
∴ ∠DCA = 11.5°
∴ ∠CAD = 180° – (∠DCA + ∠ADC) [Sum of interior angles of a triangle is 180°]
= 180° – 164.5
= 15.5°
∴ ∠CAE = ∠CAD = 15.5° [∵ A – B – E]
(2) Given AB = 4.2, BC = 5.4, AE = 12.0
∴ By theorem of intersecting chords outside the circle.
AB × AC = AD × AE
Or, AB × (AB + BC) = AD × AE
Or, 4.2 × (4.2 + 5.4) = AD × 12
Or, 4.2 × 9.6 = AD × 12
Or, AD = 4.2×9.6/12 = 3.36
Or, AD = 3.36
(3) Given AB = 3.36, AC = 9.0, AD = 5.4
AB × AC = AD × AE
Or, 3.6 × 9 = 5.4 × AE
Or, AE = 3.6×9/5.4
Or, AE = 6
(17) In figure 3.95, chord EF||chord GH. Prove that, chord EG ≅ chord FH.
Fill in the blanks and write the proof.
Proof: Draw seg GF.
∠EFG = ∠FGH …… —– (I)
∠EFG = —– ……. inscribed angle theorem (II)
∠FGH = —– ……. inscribed angle theorem (III)
∴m (arc EG) = —– from (I), (II), (III).
Chord EG ≅ chord FH …… ——
Solution:
In the adjacent fig. it is given that,
EF||GH
Construction : draw GF
∠EFG = ∠FGH [alternate angles] —- (I)
∠EFG = 1/2 m (arc EG) [inscribed angle theorem] —– (II)
∠FGH = 1/2 m (arc FH) [inscribed angle theorem] —– (III)
∴ m (arc EG) = m (arc FH) from (I), (II), (III).
Chord EG ≅ chord FH [∵ congruent arcs substands congruent chords]
(18) In figure 3.96 is the point of contact
(1) If m (arc PR) = 140°,
∠POR = 36°,
Find m (arc) PQ
(2) If OP = 7.2, OQ = 3.2,
Find OR and QR
(3) If OP = 7.2, OR = 16.2,
Find OR.
Solution:
(1) Given, m (arc PR) = 140°
∠POR = 36°
Construction : join RP & QP
∴ ∠PQR = 1/2 × m (arc PR) [Inscribed angle theorem]
Or, ∠PQR = 1/2 × 140 = 70°
∴ In ∆POQ,
∠PQR is exterior angle to ∠PQO.
∴ ∠PQR = ∠POQ + ∠OPQ [Exterior angle theorem of o triangle]
Or, 70° = 36° + ∠OPQ
Or, ∠OPQ = 70° – 36° = 34°
Now, ∠PRQ = ∠OPQ [∵ angle between tangent of a circle and a chord drawn from the point of contact P as are PQ is congruent to the angle inscribed in the arc opposite to the arc intercepted by that angle]
Or, ∠PRQ = 34°
∴ ∠PRQ = 1/2 × m (arc PQ) [Inscribed angle theorem]
Or, m (arc PQ) = 2 × 34 = 68°
(2) Given, OP = 7.2, OQ = 3.2
∴ By tangent secant segment theorem,
OP^{2} = OQ × OR
Or, 7.2^{2 }= 3.2 × OR
Or, OR = 7.2 × 7.2/3.2 = 16.2
Or, OR = 16.2
Also, OQ + QR = OR
Or, 3.2 + QR = 16.2
Or, QR = 16.2 – 3.2 = 13
Or, QR = 13
(3) Give, OP = 7.2, OR = 16.2
∴ By tangent secant segment theorem,
OP^{2} = OQ×OR
Or, 7.2^{2} = OQ×16.2
Or, OQ = 7.2×7.2/16.2 = 3.2
Or, OQ = 3.2
∴ Also, OQ + QR = OR
Or, 3.2 + QR = 16.2
Or, QR = 16.2 – 3.2 = 13
Or, QR = 13
(19) In figure 3.97, circles with centres C and D touch internally at point E. D lies on the inner circle. Chord EB of the outer circle intersects inner circle at point A. Prove that, seg EA ≅seg AB.
Solution:
In the adjacent fig it is given that, circle with centre C touches circle with centre D internally.
D lies on inner circle.
Construction : join AD & B
∴ ED is the diameter of small circle.
∴ ∠EAD = 90° [angle inscribed in a semi-circle is 90°]
∴ ∠DAB = 180° – ∠EAD [Collinear angles]
= 90°
∴ Let, radius of small circle be x
∴ Radius of bigger circle = 2x [∵ DE is the radius bigger circle]
∴ In ∆EAD, ∠EAD = 90°
By Pythagoras theorem
DE^{2} = AE^{2} + AD^{2}
Or, (2x)^{2} = AE^{2} + AD^{2}
Or, AE^{2} = 4x^{2} – AD^{2} —- (i)
In ∆ADB, ∠DAB = 90°
∴ By Pythagoras theorem,
DB^{2} = AB^{2} + AD^{2}
Or, (2x)^{2} = AB^{2 }+ AD^{2} [∵ DB is radius of bigger or circle]
Or, AB^{2} = 4x^{2} – AD^{2} — (ii)
Comparing equation & (ii) we get,
AB^{2} = AE^{2}
Or, AB = AE
Or, AB≅AE
Or, EA≅AB (Proved)
(20) In figure 3.98, seg AB is a diameter of a circle with centre O. The bisector of ∠ACD intersects the circle at point D. Prove that, segAD≅seg BD.
Complete the following proof by filling in the blanks.
Proof: Draw seg OD.
∠ACB = —- … angle inscribed in semicircle
∠DCB = —- …… CD is the bisector of ∠C
m (arc DB) = —- ……. inscribed angle theorem
∠DOB = —- ….. definition of measure of an arc (I)
seg OA ≅seg OB ….. —— (II)
∴line OD is —– of seg AB …… from (I) and (II)
∴segAD ≅ seg BD
Solution:
In the adjacent fig it is given that, O is centre of circle.
Bisector of ∠ACB in intersects the circle at point O.
Construction : draw OD.
∠ACB = 90° [Angle inscribed in semi circle]
∠DCB = 90/2 = 45° ( CD bisects ∠C )
m (arc DB) = 2×∠DCB = 2×45 = 90° [Inscribed angle theorem]
∠DOB = 90° [Definition of measure of x] —– (I)
segOA ≅ seg OB — [Radius of circle] —- (II)
∴ line OD is perpendicular bisector of seg AB —– from (I) and (II)
∴ seg AD ≅ seg BD
(21) In figure 3.99, seg MN is a chord of a circle with centre O. MN = 25, L is a point on chord MN such that ML = 9 and d(O,L) = 5. Find the radius of the circle.
Solution:
In the adjacent fig it is given that, O is the centre of circle, Now is chord.
MN = 25,
ML = 9, d (O, L) = 5
Construction : draw OP⊥MN
∴ MP = MN/2 = 25/2 [∵ perpendicular drawn from centre of circle to chord bisects the chord]
∴ LP = MP – ML
= 25/2 – 9
= 7/2
∴ In ∆OPL,
By Pythagoras theorem,
OL^{2} = LP^{2} + OP^{2}
Or, 5^{2} = (7/2)^{2} + OP^{2}
Or, OP^{2} = 25 – 49/4
Or, OP^{2} = 100-49/4 = 51/4
Now, In ∆OMP
OM^{2} = MP^{2} + OP^{2} [By Pythagoras theorem]
Or, OM^{2} = (25/2)^{2} + 51/4
Or, OM^{2} = 625/4 + 51/4
Or, OM = √169
Or, OM = 13
∴ The radius of circle OM = 13
(22) In figure 3.100, two circles intersect each other at points S and R. Their common tangent PQ touches the circle at points P, Q. Prove that, ∠PRQ + ∠PSQ = 180°
Solution:
In the adjacent fig it is given that, tangent PQ touches the two circle at point P and Q.
Construction : join RS
∴ ∠RQP = ∠RSQ
[Angle between targets of a circle PQ and a chord drawn from the point of contact Q is congruent to the angle inscribed in the arc opposite to the arc intercepted by that angle] —- (i)
Similarly,
∠RPQ = ∠PSR [Angle between tangent of a circle PQ and a chord drawn PR from the point of contact P is congruent to the angle inscribed in the arc opposite to the arc intercepted by that angle] — (ii)
Now, ∠PSR + ∠RSQ = ∠PSQ
Or, ∠RPQ + ∠RQP = ∠PSQ [From (i) & (ii)] —- (iii)
∴ In ∆PRQ
∠PRQ + ∠RPQ + ∠RQP = 180° [Sum of interior angles of a triangle is 180°]
Or, ∠PRQ + ∠PSQ = 180° [From equation (iii)] (Proved)
(23) In figure 3.101, two circles intersect at points M and N. Secants drawn through M and N intersect the circles at points R, S and P, Q respectively.
Prove that: seg SQ||seg RP
Solution:
In the adjacent fig it is given that,
Construction join NN.
∴ I MRPN, is a cyclic quadrilateral,
∠MRF = ∠MNQ [exterior angle opposite to interior angle of a cyclic quadrilateral is equal]
Also MSQN is a cyclic quadrilateral.
∴ ∠MNQ + ∠NSQ = 180° [Opposite interior angles of a cyclic quadrilateral is supplementary]
∴ ∠MRF + ∠NSQ = 180°
Now, ∠MRF & ∠NSQ are a pair of interior angle on the same side of transversal lines SQ and RF.
∴ SQ||MN||QF (Proved) [By interior angle theorem of three transversal line]
(24) In figure 3.102, two circles intersect each other at points A and E. Their common secant through E intersects the circles at points B and D. The tangents of the circles at points B and D intersect each other at point C. Prove thatABCD is cyclic.
Solution:
Construction : join AB, AD & AE
Now, In the smaller circle.
BC is the tangent which touches the circle at point B.
BE & BA and AE are chords.
∴ ∠EBC = ∠EAB —- (i)
[Angle between the chord AE and the tangent BC of a circle is congruent to the angle inscribe in the circle in the opposite arc intercepted by the angle]
Similarly, In the bigger circle,
ED, EA & AD are chords,
DC is the tangent which touches the circle a point D.
∴ ∠EDC = ∠EAD —- (ii)
[The angle between the tangent DC and the chord DE drawn at the point D that touch as the circle is congruent to the angle inscribed in the opposite arc intercepted by the angle]
∴ ∠BAD = ∠EAB + ∠EAD — (iii)
Now, In ∆BDC,
∠EBC + ∠EDC + ∠BCD = 180° [Sum of interior angles of a triangle is 180°]
Or, ∠EAB + ∠EAD + ∠BCD = 180° [From (i) & (ii)]
Or, ∠BAD + ∠BCD = 180° [From (iii)] — (iv)
∴ In ABCD,
∠BAD + ∠BCD = 180° [From (iv)]
∴ Opposite angles of ABCD are supplementary.
Therefore ABCD is a cyclic quadrilateral (Proved)
(25) In figure 3.103, segAD ⊥ side BC, seg BE ⊥ side AC, seg CF ⊥ side AB. Point O is orthocentre. Prove that, point O is the incentre of ∆DEF.
Solution:
In the adjacent fig it is given that, AD⊥BC, BE ⊥ AC, CF⊥AB.
O is the orthocentre point of ∆ABC.
∠AFO + ∠AEO = 90° + 90° = 180°
[∵ CF⊥AB & BE ⊥ AC]
Therefore, AFOE is a cyclic quadrilateral]
Similarity, BFOD is a cyclic quadrilateral &CEOD is a cyclic quadrilateral.
∴ In AFOE,
∠OAE = ∠OFE — (i) [Angle inscribed by the same arc — congruent]
In BFOD,
∠OBD = ∠OFD — (ii) [Angle inscribed by the same arc — congruent]
In ∆ACD,
∠DAC + ∠ACD = 180° – 90°
= 90° [Sum of angle of a triangle is 180°] —– (iii)
In ∆BCE,
∠BCE + ∠CBE = 90° — (iv) [Sum of angles of a triangles is 180°]
Comparing equation (iii) & (iv) we get,
∠DAC + ∠ACD = ∠BCE + ∠CBE
Or, ∠DAC = ∠CBE — (v)
Comparing equation (ii) & (iv) we get,
∠OFE = ∠OFD
∴ OF is bisector of ∠EFD
Similarly, OE & OD bisects of ∠DEF & ∠EDF respectively
Hence, O is the incentre of ∆DEF (Proved)
Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 3 Circle Problem Set 3
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