Problem Set 3 Geometry 10th Std Maths Part 2 Answers Chapter 3 Circle
Problem Set 3 Geometry Class 10 Question 1.
Four alternative answers for each of the following questions are given. Choose the correct alternative.
i. Two circles of radii 5.5 cm and 3.3 cm respectively touch each other. What is the distance between their centres?
(A) 4.4 cm
(B) 8.8 cm
(C) 2.2 cm
(D) 8.8 or 2.2 cm
Answer: (D)
Two circles can touch each other internally or externally.
∴ Distance between centres = 5.5 + 3.3 or 5.5 – 3.3 = 8.8 or 2.2
ii. 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
Answer: (B)
PQ is the radius = 12 cm
iii. A circle touches all sides of a parallelogram. So the parallelogram must be a __________.
(A) rectangle
(B) rhombus
(C) square
(D) trapezium
Answer: (B)
꠸ABCD is a rhombus.
Note: It cannot be square as the angles are not mentioned as 90°.
iv. 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
Answer: (C)
In ∆OAB, ∠B = 90° [Tangent theorem]
∴ OA2= OB2+ AB2[Pythagoras theorem]
∴ 12.52= OB2+ 122
∴ OB2= 156.25- 144
∴ OB = = 3.5 cm
∴ Diameter = 2 × OB = 2 × 3.5 = 7 cm
v. If two circles are touching externally, how many common tangents of them can be drawn?
(A) One
(B) Two
(C) Three
(D) Four
Answer: (C)
line l, line m and line n are the tangents.
vi. ∠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°
Answer: (D)
m∠ACB = m(arc AB) [Inscribed angle theorem]
∴ m(arc AB) = 2 m∠ACB =
= 2 × 65
= 130°
m(arc ACB) = 360° – m(arc AB) [Measure of a circle is 360°]
= 360° – 130°
= 230°
vii. 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
Answer: (A)
Chords AB and CD intersect
internally at E.
AE × EB = CE × ED [Theorem of internal division of chords]
∴ 5.6 × 10 = 8 × ED
∴ ED = 7 units
viii. 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°
Answer: (B)
∠A + ∠C = 180° [Theorem of cyclic quadrilateral]
∴ 2∠A + 2∠C = 2 × 180° [Multiplying both sides by 2]
∴ 3∠C + 2∠C = 360° [∵ 2∠A = 3∠C]
∴ 5∠C = 360°
∴ ∠C = 72°
ix. 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
Answer: (A)
m(arc AB) + m(arc BC) + m(arc AC) = 360° [Measure of a circle is 360°]
∴ 120° + 120° + m (arc AC) = 360°
∴ m(arc AC) = 120°
∴ arc AB = arc BC = arc AC
∴ seg AB ≅ seg BC ≅ seg AC [Corresponding chords of congruents arcs of a circle are congruent]
∴ ∆ABC is an equilateral triangle.
x. Seg XZ is a diameter of a circle. Point Y lies in 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
Answer: (C)
x. seg XZ is the diameter.
∴ ∠XWZ is a right angle. [Angle inscribed in a semicircle]
Since, Y lies in the interior of ∆XWZ,
∴ ∠XYZ > 90°
i.e., ∠XYZ is an obtuse angle.
Problem Set 3 Question 2.
Line l touches a circle with centre O at point P. If radius of the circle is 9 cm, answer the following.
i. What is d(O, P) = ? Why?
ii. If d(O, Q) = 8 cm, where does the point Q lie?
iii. If d(O, R) = 15 cm, how many locations of point R are on line l ?
At what distance will each of them be from point P?
Solution:
i. seg OP is the radius of the circle.
∴ d(0, P) = 9 cm
ii. Here, 8 cm < 9 cm
∴ d(0, Q) < d(0, P)
∴ d(0, Q) < radius
Point Q lies in the interior of the circle.
iii. There can be two locations of point R on line l.
d(0, R) = 15 cm
Now, in ∆OPR, ∠OPR = 90° [Tangent theorem]
∴ OR2= OP2+ PR2[Pythagoras theorem]
∴ 152= 92+ PR2
∴ 225 = 81 + PR2
∴ PR2= 225 – 81 = 144 [Taking square root of both sides]
∴ PR =
= 12 cm
Standard 10th Geometry Problem Set 3 Question 3.
In the adjoining figure, M is the centre of the circle and seg KL is a tangent segment. If MK = 12, KL = 6,then find
i. Radius of the circle.
ii. Measures of ∠K and ∠M.
Solution:
i. Line KL is the tangent to the circle at point L and seg ML is the radius. [Given]
∴ ∠MLK = 90°…………. (i) [Tangent theorem]
In ∆MLK, ∠MLK = 90°
∴ MK2= ML2+ KL2[Pythagoras theorem]
∴ 122= ML2+ (6)2
∴ 144 = ML2+ 108
∴ ML2= 144 – 108
∴ ML2= 36
∴ ML = = 6 units. [Taking square root of both sides]
∴ Radius of the circle is 6 units.
ii. We know that,
ML = MK
∴ ∠K = 30° …………… (ii) [Converse of 30° – 60° – 90° theorem]
In ∆MLK,
∠L = 90° [From (i)]
∠K = 30° [From (ii)]
∴ ∠M = 60° [Remaining angle of ∆MLK]
10th Class Geometry Problem Set 3 Question 4.
In the adjoining figure, 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.
Given: O is the centre of circle.
seg AB and seg AC are the tangents, radius = r, /(AB) = r.
To prove: □ABOC is a square.
Construction: Draw seg OB and seg OC.
Proof:
seg AB and seg AC are the tangents to the circle. [Given]
∴ AB = AC [Tangent segment theorem]
But, AB = r [Given]
∴ AB = AC = r ……….. (i)
Also, OB = OC = r ……….. (ii) [Radii of the same circle]
∴ AB = AC = OB = OC [From (i) and (ii)]
∴ □ABOC is a rhombus.
∠OBA = 90° [Tangent theorem]
∴ □ABOC is a square [A rhombus is a square, if one of its angles is a right angle]
10th Std Geometry Circle Problem Set Question 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.
Given: X and Y are the centres of circle.
To prove: radius XA || radius YB
Construction: Draw segments XZ and YZ.
Proof:
By theorem of touching circles, points X, Z,
Y are collinear.
∴ ∠XZA ≅ ∠BZY [Vertically opposite angles]
Let ∠XZA = ∠BZY = a …………….. (i)
Now, seg XA seg XZ [Radii of the same circIe]
∴ ∠XAZ ≅∠XZA = a …………….. (ii) [Isosceles triangle theorem]
Similarly, seg YB ≅ seg YZ [Radii of the same circie]
∴ ∠BZY = ∠ZBY = a …………….. (iii) [Isosceles triangle theorem]
∴ ∠XAZ = ∠ZBY [From (i), (ii) and (iii)]
∴ radius XA || radius YB [Alternate angles test]
Circle Problem Set 3 Question 8.
In the adjoining figure, 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.
Given: X and Y are the centres of the circle.
To prove: seg AX || seg BY
Proof:
In ∆XAZ,
seg XA ≅ seg XZ [Radii of the same circle]
∴ ∠XZA ≅ ∠XAZ ………… (i) [Isosceles triangle theorem]
Also, in ∆YBZ,
seg YB ≅ seg YZ [Radii of the sanie circle]
∴ ∠YZB ≅∠YBZ [Isosceles triangle theorem]
∴ ∠XZA ≅ ∠YBZ ………….. (ii) [Y – X – Z,B – A – Z]
∴ ∠XAZ ≅ ∠YBZ [From (i) and (ii)]
∴ seg AX || seg BY [Corresponding angles test]
Question 11
Maharashtra Board Solution
Draw circles with centres A, B and C each of radius 3 cm, such that each circle touches the other two circles. Analysis: Let the circles with centres A, B, C touch each other at points P, Q, R. [∴ A – P – B A – Q – C B – R – C ] [Theorem of touching circles] ∴ AB = AP + BP [A – P – B] ∴ AB = 3 + 3 = 6 cm Similarly, BC = 6 cm, AC = 6 cm So, if we construct triangle ∆ABC of side 6 cm each, then with A, B, C as the centres and radius 3 cm, the touching circles can be drawn. |
Question 12. Prove that any three points on a circle cannot be collinear Given: A circle with centre O. Points A, B and C lie on the circle. To prove: Points A, B and C are not collinear. Proof: OA = OB [Radii of the same circle] ∴ Point O is equidistant from the endpoints A and B of seg AB. ∴ Point O lies on the perpendicular bisector of AB. [Perpendicular bisector theorem] Similarly, we can prove that, Point O lies on the perpendicular bisector of BC. ∴ Point O is the point of intersection of perpendicular bisectors of AB and BC (i.e., circumcentre of ∆ABC) ……… (i) Now, suppose that the points A, B, C are collinear. Then, the perpendicular bisector of AB and BC will be parallel. [Perpendiculars to the same line are parallel] ∴ The perpendicular bisector do not intersect at O. This contradicts statement (i) that the perpendicular bisectors intersect each other at O. ∴ Our supposition that A, B, C are collinear is false. ∴ Points A, B and C are non collinear points. |
Question 13. In the adjoining figure, line PR touches the circle at point Q.
Solution & Step-by-Step Answer:
the following questions with the help of the figure. i. What is the sum of ∠TAQ and ∠TSQ? ii. Find the angles which are congruent to ∠AQP. iii. Which angles are congruent to ∠QTS? iv. ∠TAS = 65°, find the measures of ∠TQS and arc TS. v. If ∠AQP = 42° and ∠SQR = 58° find measure of ∠ATS.
Solution & Step-by-Step Answer:
i. ꠸AQST is a cyclic quadrilateral. [Given] ∴ ∠TAQ + ∠TSQ = 180° [Opposite angles of a cyclic quadrilateral are supplementary] ii. line PR is the tangent and seg AQ is the secant. [Given] ∴∠AQP = m(arc AQ) [Theorem of angle between tangent and secant] But,∠ASQ = m(arc AQ) [Inscribed angle theorem] ∴∠AQP ≅ ∠ZASQ Similarly, we can prove that, ∠AQP ≅ ∠ATQ iii. ∠QTS = m(arc QS) [Inscribed angle theorem] But, ∠SQR = m(arc QS) [Theorem of angle between tangent and secant] ∴ ∠QTS ≅ ∠SQR Also, ∠QTS = ∠QAS [Angles inscribed in the same arc] iv. ∠TQS = ∠TAS [Angles inscribed in the same arc] ∴ ∠TQS = 65° Now, ∠TQS = m(arc TS) [Inscribed angle theorem] ∴ 65°= m(arcTS) ∴ m(arc TS) = 65° × 2 ∴ m(arc TS) = 130° v. ∠AQP + ∠AQS + ∠SQR = 180° [Angles in a linear pair] ∴ 42° + ∠AQS + 58° = 180° ∴∠AQS + 100° = 180° ………… (i) But, ꠸AQST is a cyclic quadrilateral. ∴ ∠AQS + ∠ATS = 180° ………… (ii) [Theorem of cyclic quadrilateral] ∴ ∠ATS = 100° [From (i) and (ii)]
Question 14
Maharashtra Board Solution
In the adjoining figure, O is the centre of a circle, chord PQ ≅ chord RS. If ∠POR = 70° and (arc RS) = 80°, find i. m (arc PR) ii. m (arc QS) iii. m (arc QSR).
Solution & Step-by-Step Answer:
i. m(arc PR) = m∠POR [Definition of measure of arc] ∴ m(arc PR) = 70° ii. chord PQ chord RS [Given] ∴ m(arc PQ) = m(arc RS) = 80° [Corresponding arcs of congruents chords of a circle are congruent] Now, m(arc QS) + m(arc PQ) + m(arc PR) + m(arcRS) = 360° ∴ m(arc QS) + 80° + 70° + 80° = 360° [Measure of a circle is 360°] ∴ m(arc QS) + 230° = 360° ∴ m(arc QS) = 130° iii. m(arc QSR) = m(arc QS) + m(arc SR) [Arc addition property] = 130° + 80° ∴ m(arc QSR) = 210°
Question 15
Maharashtra Board Solution
In the adjoining figure, m(arc WY) = 44°, m(arc ZX) = 68°, then i. Find the measure of ∠ZTX. ii. If WT = 4.8, TX = 8.0, YT = 6.4, find TZ. iii. If WX = 25, YT = 8, YZ = 26, find WT.
Solution & Step-by-Step Answer:
i. Chords WX and YZ intersect internally at point T. ∴ ∠ZTX = m(arc WY) + m(arc ZX)] = (44° + 68°) = × 112° ∴ m ∠ZTX = 56° ii. WT × TX = YT × TZ [Theorem of internal division of chords] ∴ 4.8 × 8.0 = 6.4 × TZ ∴ TZ = ∴ l(TZ) = 6.0 units iii. Let the value of WT be x. [W – T – X] WT + TX = WX ∴ x + TX = 25 ∴ TX = 25 – x Also, YT + TZ = YZ [Y – T – Z] ∴ 8 + TZ = 26 ∴ TZ = 26 – 8 = 18 units But, WT × TX = YT × TZ [Theorem of internal division of chords] ∴ x × (25 – x) = 8 × 18 ∴ 25x – x2 = 144 ∴ x2 – 25x + 144 = 0 ∴ (x – 16)(x – 9) = 0 ∴ x = 16 or x = 9 ∴ WT = 16 units or WT = 9units
Question 16
Maharashtra Board Solution
In the adjoining figure, i. Mn (arc CE) = 54°, m (arc BD) = 23°, find measure of ∠CAE. ii. If AB = 4.2,BC = 5.4, AE = 12.0, find AD. iii. If AB = 3.6, AC = 9.0, AD = 5.4, find AE.
Solution & Step-by-Step Answer:
i. Chords BC and ED intersect each other externally at point A. ∴ ∠CAE = [m(arc CE) – m(arc BD)] = (54° – 23°) = × 31° ∴ m∠CAE = 15.5° ii. AC = AB + BC [A – B – C] = 4.2 + 5.4 = 9.6 units Now, AB × AC = AD × AE [Theorem of external division of chords] ∴ 4.2 × 9.6 = AD × 12.0 ∴ AD = ∴ AD = 3.36 units iii. AB × AC = AD × AE [Theorem of external division of chords] ∴ 3.6 × 9.0 = 5.4 × AE ∴ AE = ∴ AE = 6 units
Geometry Problem Set 3 Question 17. SSC Geometry Circle Chapter Solutions Pdf Question 18.
Question 19
Maharashtra Board Solution
In the adjoining figure, 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. Given: Circles with centres C and D touch each other internally. To prove: seg EA ≅ seg AB Construction: Join seg ED and seg DA. Proof: E – C – D [Theorem of touching circles] seg ED is the diameter of smaller circle. ∴∠EAD = 90° [Angle inscribed in a semicircle] ∴ seg AD ⊥ chord EB ∴ seg EA ≅ seg AB [Perpendicular drawn from the centre of the circle to the chord bisects the chord] |
Question 20
Maharashtra Board Solution
In the adjoining figure, seg AB is a diameter of a circle with centre O. The bisector of ∠ACB intersects the circle at point D. Prove that, seg AD ≅ seg BD. Complete the following proof by filling the blanks. Given: seg AB is a diameter, seg CD bisects ∠ACB. To prove: seg AD ≅ seg BD Construction: Draw seg OD. Proof: ∠ACB = 90° [Angle inscribed in a semicircle] ∠DCB = ∠DCA = 45° [CD is the bisector of ∠C] m(arcDB) = 2∠DCA = 90° [Inscribed angle theorem] ∠DOB = m(arc DB) = 90° ………… (i) [Definition of measure of arc] segOA ≅ segOB …………. (ii) [[Radii of the same circle] ∴ line OD is the perpendicular biscctor of [From (i) and (ii)] seg AB. ∴ seg AD ≅ seg BD |
10th Geometry Circle Question 21.
In the adjoining figure, 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(0, L) = 5. Find the radius of the circle.
Construction: Draw seg OK ⊥ chord MN. Join OM.
seg OK ⊥chord MN [Construction]
∴ MK = MN [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
= × 25
= 12.5 units
MK = ML + LK [M – L – K]
∴ 12.5 = 9 + LK
∴ LK= 12.5 – 9 = 3.5 units
In ∆OKL, ∠OKL = 90°
∴ OL2= KL2+ OK2[Pythagoras theorem]
∴ 52= 3.52+ OK2
∴ OK2= 25 – 12.25 = 12.75
Now, in ∆OKM, ∠OKM = 90°
∴ OM2= OK2+ MK2
= 12.75 + 12.52
= 12.75 + 156.25
= 169
∴ OM =
= 13 units [Taking square root of both sides]
∴ The radius of the given circle is 13 units.
Question 22. In the adjoining figure, 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°. Given: Two circles intersect each other at points S and R. line PQ is a common tangent. To prove: ∠PRQ + ∠PSQ = 180° Proof: Line PQ is the tangent at point P and seg PR is a secant. ∴ [∠RPQ = ∠PSR …………. (i) and ∠PQR = ∠QSR] ………… (ii) [Tangent secant theorem] In ∆ PQR, ∠PQR + ∠PRQ + ∠RPQ = 180° [Sum of the measures of angles of a triangle is 180°] ∴ ∠QSR + ∠PRQ + ∠PSR = 180° [From (i) and (ii)] ∴ ∠PRQ + ∠QSR + ∠PSR = 180° ∴ ∠PRQ + ∠PSQ = 180° [Angle addition property]
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