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NCERT Class 9 Mathematics MCQ Circles with Answers
Question :In the figure, triangle ABC is an isosceles triangle with AB = AC and measure of angle ABC = 50°. Then the measure of angle BDC and angle BEC will be
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(a) 60°, 100°
(b) 80°, 100°
(c) 50°, 100°
(d) 40°, 120°
Answer : (b) 80°, 100°Show Answer :
Question :The region between chord and either of the arc is called
(a) a sector
(b) a semicircle
(c) a segment
(d) a quarter circle
Answer : (c) a segmentShow Answer :
Question : An equilateral triangle of side 9 cm is inscribed a circle. The radius of the circle is
(a) 3 cm
(b) 3√2 cm
(c) 3√3 cm
(d) 6 cm
Answer :(c) 3√3 cmShow Answer :
Question : AB is a chord of a circle with radius ‘r’. If P is any point on the circle such that ∠APB is a right angle , then AB is equal to
(a) 3r
(b) r
(c) 2r
(d) r2
Answer :(c) 2rShow Answer :
Question : A line that intersects a circle in two distinct points is a
(a) Secant
(b) Chord
(c) Radius
(d) Diameter
Answer :(a) SecantShow Answer :
Question : The region between chord and either of the arc is called
(a) a sector
(b) a semicircle
(c) a segment
(d) a quarter circle
Answer :(c) a segmentShow Answer :
Question : The angle subtended by the diameter of a semi-circle is:
(a) 90
(b) 45
(c) 180
(d) 60
Answer :(c) 180Show Answer :
Question :The region between an arc and the two radii joining the centre of the end points of the arc is called a:
(a) Segment
(b) Semi circle
(c) Minor arc
(d) Sector
Answer : (d) SectorShow Answer :
Question :If a line intersects two concentric circles with centre O at A, B, C and D, then:
(a) AB = CD
(b) AB > CD
(c) AB < CD
(d) None of the above
Answer : (a) AB = CDShow Answer :
Question :A chord of a circle which is twice as long as its radius is a ____ of the circle
(a) Diameter
(b) perpendicular
(c) arc
(d) secant
Answer : (a) DiameterShow Answer :
Question :A regular octagon is inscribed in a circle. The angle that each side of the octagon subtends at the centre is
(a) 45°
(b) 75°
(c) 90°
(d) 60°
Answer : (a) 45°Show Answer :
Question :Equal _____ of the congruent circles subtend equal angles at the centers.
(a) Segments
(b) Radii
(c) Arcs
(d) Chords
Answer : (d) ChordsShow Answer :
Question :The angle subtended by the diameter of a semi-circle is:
(a) 90
(b) 45
(c) 180
(d) 60
Answer : (c) 180Show Answer :
Question :The degree measure of a semicircle is
(a) 0°
(b) 90°
(c) 360°
(d) 180°
Answer : (d) 180°Show Answer :
Question :If there are two separate circles drawn apart from each other, then the maximum number of common points they have:
(a) 0
(b) 1
(c) 2
(d) 3
Answer : (a) 0Show Answer :
Question :D is diameter of a circle and AB is a chord. If AD = 50 cm, AB = 48 cm, then the distance of AB from the centre of the circle is
(a) 6 cm
(b) 8 cm
(c) 5 cm
(d) 7 cm
Answer : (d) 7 cmShow Answer :
Question :In a circle with center O and a chord BC, points D and E lie on the same side of BC. Then, if∠BDC=80°, then ∠BEC =
(a) 80°
(b) 20°
(c) 160°
(d) 40°
Answer : (a) 80°Show Answer :
Question : In the given figure, ∠AOB = 90º and ∠ABC = 30º, then ∠CAO is equal to:
(a) 30º
(b) 45º
(c) 60º
(d) 90º
Answer :c Show Answer :
Explanation:
Given that ∠AOB = 90º and ∠ABC = 30º
OA = OB (Radii of the circle)
Let x= ∠OAB = ∠OBA = x
In the triangle OAB,
∠OAB + ∠OBA + ∠AOB = 180° (By using the angle sum property of triangle)
⇒ x + x + 90° = 180°
⇒ 2x = 180° – 90°
⇒ x = 90°/ 2 = 45°
Therefore, ∠OAB = 45° and ∠OBA = 45°
By using the theorem, “ the angles subtended by arcs at the centre of the circle double the angle subtended at the remaining part of the circle”, we can write
∠AOB = 2∠ACB
This can also be written as,
∠ACB = ½ ∠AOB = (½) × 90° = 45°
Now, apply the angle sum property of triangle on the triangle ABC,
∠ACB + ∠BAC + ∠CBA = 180°
∠ACB + [∠BAO + ∠CAO] + ∠CBA = 180° (As,∠BAC = ∠BAO + ∠CAO)
Now, substitute the known values, we get
45° + (45° + ∠CAO) + 30° = 180°
∠CAO = 180°- (30° + 45° + 45°)
∠CAO = 180°-120°
∠CAO = 60°
Hence, ∠CAO is equal to 60°.
Question : ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to:
(a) 30º
(b) 40º
(c) 50º
(d) 80º
Answer :cShow Answer :
Explanation:
Given that ABCD is a cyclic quadrilateral and ∠ADC = 140º.
We know that the sum of opposite angles of a cyclic quadrilateral is 180°.
Hence, ∠ADC + ∠ABC = 180°
Now, substitute ∠ADC = 140º in the above equation, we get
140° + ∠ABC = 180°
∠ABC = 180° – 140° = 40°
since the angle subtended by a diameter at the circumference of the circle, is 90°
Hence, ∠ACB = 90°
By using the angle property of triangle in the triangle, ABC,
∠CAB + ∠ABC + ∠ACB = 180°
∠CAB + 40° + 90° = 180°
∠CAB = 180° – 90° – 40°
∠CAB = 50°
Therefore, ∠CAB or ∠BAC =50°.
Question : In the given figure, if ∠OAB = 40º, then ∠ACB is equal to
(a) 40º
(b) 50º
(c) 60º
(d) 70º
Answer :bShow Answer :
Explanation:
Given that ∠OAB = 40º,
In the triangle OAB,
OA = OB (radii)
Since, the angles opposite to equal sides are equal,
∠OAB = ∠OBA (i.e.) ∠OBA = 40°
Now, by using the angle sum property of triangle, we can write
∠AOB + ∠OBA + ∠BAO = 180°
Now, substitute the known values,
∠AOB + 40° + 40° = 180°
∠AOB = 180 – 80° = 100°
∠AOB = 2 ∠ACB (Since, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle)
∠ACB = ½ ∠AOB
Hence, ∠ACB = 100°/2 = 50°.
Question : If chords AB and CD of congruent circles subtend equal angles at their centres, then:
(a) AB = CD
(b) AB > CD
(c) AB < AD
(d) None of the above
Answer : aShow Answer :
Explanation: Take the reference of the figure from above question.
In triangles AOB and COD,
∠AOB = ∠COD (given)
OA = OC and OB = OD (radii of the circle)
So, ΔAOB ≅ ΔCOD. (SAS congruency)
∴ AB = CD (By CPCT)
Question : If there are two separate circles drawn apart from each other, then the maximum number of common points they have:
(a) 0
(b) 1
(c) 2
(d) 3
Answer : aShow Answer :
Question : The angle subtended by the diameter of a semi-circle is:
(a) 90
(b) 45
(c) 180
(d) 60
Answer : cShow Answer :
Explanation: The semicircle is half of the circle, hence the diameter of the semicircle will be a straight line subtending 180 degrees.
Question : If AB and CD are two chords of a circle intersecting at point E, as per the given figure. Then:
(a) ∠BEQ > ∠CEQ
(b) ∠BEQ = ∠CEQ
(c) ∠BEQ < ∠CEQ
(d) None of the above
Answer : bShow Answer :
Explanation:
OM = ON (Equal chords are always equidistant from the centre)
OE = OE (Common)
∠OME = ∠ONE (perpendiculars)
So, ΔOEM ≅ ΔOEN (by RHS similarity criterion)
Hence, ∠MEO = ∠NEO (by CPCT rule)
∴ ∠BEQ = ∠CEQ
Question : If a line intersects two concentric circles with centre O at A, B, C and D, then:
(a) AB = CD
(b) AB > CD
(c) AB < CD
(d) None of the above
Answer : aShow Answer :
Explanation: See the figure below:
From the above fig., OM ⊥ AD.
Therefore, AM = MD — 1
Also, since OM ⊥ BC, OM bisects BC.
Therefore, BM = MC — 2
From equation 1 and equation 2.
AM – BM = MD – MC
∴ AB = CD
Question : If a line intersects two concentric circles with centre O at A, B, C and D, then:
(a) AB = CD
(b) AB > CD
(c) AB < CD
(d) None of the above
Answer :(a) AB = CDShow Answer :
Question : The angle subtended by the diameter of a semi-circle is:
(a) 90
(b) 45
(c) 180
(d) 60
Answer :(c) 180Show Answer :
Question : An equilateral triangle of side 9 cm is inscribed a circle. The radius of the circle is
(a) 3 cm
(b) 3√2 cm
(c) 3√3 cm
(d) 6 cm
Answer :(c) 3√3 cmShow Answer :
Question : A regular octagon is inscribed in a circle. The angle that each side of the octagon subtends at the centre is
(a) 45°
(b) 75°
(c) 90°
(d) 60°
Answer :(a) 45°Show Answer :
Question : In the below figure, the value of ∠ADC is:
(a) 60°
(b) 30°
(c) 45°
(d) 55°
Answer : cShow Answer :
Explanation: ∠AOC = ∠AOB + ∠BOC
So, ∠AOC = 60° + 30°
∴ ∠AOC = 90°
An angle subtended by an arc at the centre of the circle is twice the angle subtended by that arc at any point on the rest part of the circle.
So,
∠ADC = 1/2∠AOC
= 1/2 × 90° = 45°
Question : In the given figure, find angle OPR.
(a) 20°
(b) 15°
(c) 12°
(d) 10°
Answer : dShow Answer :
Explanation: The angle subtended by major arc PR at the centre of the circle is twice the angle subtended by that arc at point, Q, on the circle.
So, ∠POR = 2 × ∠PQR, here ∠POR is the exterior angle
We know the values of angle PQR as 100°
So, ∠POR = 2 × 100° = 200°
∴ ∠ROP = 360° – 200° = 160° [Full rotation: 360°]
Now, in ΔOPR,
OP and OR are the radii of the circle
So, OP = OR
Also, ∠OPR = ∠ORP
By angle sum property of triangle, we know:
∠ROP + ∠OPR + ∠ORP = 180°
∠OPR + ∠OPR = 180° – 160°
As, ∠OPR = ∠ORP
2∠OPR = 20°
Thus, ∠OPR = 10°
Question : In the given figure, BC is the diameter of the circle and ∠BAO = 60º. Then ∠ADC is equal to
(a) 30º
(b) 45º
(c) 60º
(d) 120º
Answer :cShow Answer :
Explanation:
Given that ∠BAO = 60°
Since OA = OB,∠OBA = 60°
Then ∠ADC = 60° (As, the angles in the same segment are equal).
Question : In the given figure, if ∠DAB = 60º, ∠ABD = 50º, then ∠ACB is equal to:
(a) 50º
(b) 60º
(c) 70º
(d) 80º
Answer : cShow Answer :
Explanation:
Given that, ∠DAB = 60º, ∠ABD = 50º
By using the angle sum property in the triangle ABD
∠DAB+∠ABD+∠ADB=180º
60º+50º+∠ADB=180º
∠ADB=180º-110º
∠ADB=70º
∠ADB=∠ACB (Since the angles subtended at the circumference by the same arc are equal)
We know that the angles subtended at the circumference by the same arc are equal.
∠ACB =70º
Question : In the given figure, if AOB is a diameter of the circle and AC = BC, then ∠CAB is equal to:
(a) 30º
(b) 45º
(c) 60º
(d) 90º
Answer :bShow Answer :
Explanation:
We know that the angle at circumference subtended by the diameter of the circle is the right angle.
Hence, ∠ACB = 90°
Also, given that AC = BC
Therefore, ∠CAB = ∠CBA (As, the angles opposite to equal sides are also equal)
Now, by using the angle sum property of triangle in ∆ACB, we can write
∠CAB + ∠ABC + ∠BCA = 180°
∠CAB + ∠CAB + 90° = 180°
2∠CAB = 180° – 90°
∠CAB = 45°
Therefore, ∠CAB is equal to 45°.
Question : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is:
(a) 6 cm
(b) 8 cm
(c) 10 cm
(d) 12 cm
Answer : cShow Answer :
Explanation:
Given that AB = 12 cm, BC = 16 cm and AB is perpendicular to BC.
Hence, AC is the diameter of the circle passing through points A, B and C.
Hence, ABC is a right-angled triangle.
Thus by using the Pythagoras theorem:
AC2 = (CB)2 + (AB)2
⇒ AC2 = (16)2 + (12)2
⇒ AC2 = 256 + 144
⇒ AC2 = 400
Hence, the diameter of the circle, AC = 20 cm.
Thus, the radius of the circle is 10 cm.
Question : AD is the diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is
(a) 4 cm
(b) 8 cm
(c) 15 cm
(d) 17 cm
Answer : bShow Answer :
Explanation:
Given that, Diameter, AD = 34 cm.
Chord, AB = 30 cm.
Hence, the radius of the circle, OA = 17 cm
Now, consider the figure.
From the centre “O”. OM is perpendicular to the chord AB.
(i.e) OM ⊥ AM
AM = ½ AB
AM = ½ (30) = 15 cm
Now by using the Pythagoras theorem in the right triangle AOM,
AO2 = OM2 + AM2
OM2 = AO2– AM2
OM2= 172 – 152
OM2 = 64
OM = √64
OM = 8 cm
Question : In the given figure, PT touches the circle at R whose centre is O. Diameter SQ when produced meets PT at P. Given ∠SPR = x° and ∠QRP = y°. Then,
(A) x° + 2y° = 90°
(b) 2x° + y° = 90°
(c) x° + y° = 120°
(d) 3x° + 2y° = 120°
Answer :AShow Answer :
Question : In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm, CD = 4 cm, then AD equals ________.
(a) 10 cm
(b) 13 cm
(c) 11 cm
(d) 3 cm
Answer 😀Show Answer :
Question : In the given figure, a circle touches the side BC of DABC at P and touches AB and AC produced at Q and R r e s p e c t i v e l y. I f AQ = 5 cm, find the perimeter of DABC.
(a) 11 cm
(b) 10 cm
(c) 6 cm
(d) 7 cm
Answer :BShow Answer :
Question : In the given figure, AB and PQ intersect at M. If A and B are centres of circles then ________.
(a) PM = MQ
(b) PQ ⊥ AB
(c) Both (A) and (B)
(d) PQ = AB
Answer :CShow Answer :
Question : Two circles of radii 10 cm and 8 cm intersect each other and the length of common chord is 12 cm. The distance between their centres is ________.
(a) √7 cm
(b) 3 √7 cm
(c) 4 √7 cm
(d) (8 + 2 √7) cm
Answer 😀Show Answer :
Question : If O is the centre of a circle, AOC is its diameter and B is a point on the circle such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT =
(a) 40°
(b) 50°
(c) 60°
(d) 65°
Answer :BShow Answer :
Question : In the given figure, A and B are the centres of two circles that intersect at X and Y. PXQ is a straight line. If reflex angle QBY = 210°, find obtuse angle PAY.
(a) 210°
(b) 150°
(c) 160°
(d) 120°
Answer :BShow Answer :
Question : AB is a chord of length 24 cm of a circle of radius 13 cm. The tangents at A and B intersect at a point C. Find the length AC.
(a) 31.2 cm
(b) 12 cm
(c) 28.8 cm
(d) 25 cm
Answer :AShow Answer :
Question : Match the columns. Column-I Column-II
(a) Radius of the given circle is (p) 20 cm
(b) In the given figure, value of x is (q) 6 cm
(c) Perimeter of DPST with PQ = 10 cm is (r) 5 cm
(A) (a) → (p), (b) → (r), (c) → (q)
(b) (a) → (r), (b) → (q), (c) → (p)
(c) (a) → (q), (b) → (r), (c) → (p)
(d) (a) → (r), (b) → (p), (c) → (q)
Answer :BShow Answer :
Question :The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the the bigger circle. BD is a tangent to the smaller circle touching it at D. Find the length AD.
(a) 19 cm
(b) 20 cm
(c) 16 cm
(d) √105 cm
Answer :AShow Answer :
Question : In the given figure, O is the centre of a circle; PQL and PRM are the tangents at the points Q and R respectively and S is a point on the circle such that ∠SQL = 50° and ∠SRM = 60°. Then, find ∠QSR and ∠RPQ.
(a) 40°, 140°
(b) 50°, 140°
(c) 60°, 120°
(d) 70°, 40°
Answer 😀Show Answer :
Question : In the following figure, PT is of length 8 cm. OP is 10 cm. Then the radius of the circle is ________
(a) 2 cm
(b) 18 cm
(c) (5/4) cm
(d) 6 cm
Answer 😀Show Answer :
Question : A circle inscribed in DABC having AB = 10 cm, BC = 12 cm, CA = 28 cm touching sides at D, E, F (respectively). Then AD + BE + CF is ________.
(a) 25 cm
(b) 20 cm
(c) 22 cm
(d) 18 cm.
Answer :AShow Answer :
Question : In the given figure, O is the centre of the circle, then ∠XOZ is ________.
(a) 2 ∠XZY
(b) 2 ∠Y
(c) 2 ∠Z
(d) 2(∠XZY + ∠YXZ)
Answer 😀Show Answer :
Question :In the given figure, O is the centre and SAT is a tangent to the circle at A. If ∠BAT = 30°, find ∠AOB and ∠AQB.
(a) 60°, 150°
(b) 30°, 150°
(c) 60°, 60°
(d) None of these
Answer :AShow Answer :
Question : In the given figure, O is the centre of the circle. If PA and PB are tangents, then the value of ∠AQB is
(a) 100°
(b) 80° 80° Q
(c) 60°
(d) 50°
Answer 😀Show Answer :
Question : Two concentric circles of radii a and b, where a > b, are given. The length of a chord of the larger circle which touches the other circle is
(a) √a2 − b2
(b) 2 √a2 − b2
(c) √a2 + b2
(d) 2 √a2 + b2
Answer :BShow Answer :
Question : Let s denote the semiperimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle touches the sides BC, CA, AB at D, E, F, respectively ,find BD.
(a) s – b
(b) 2s + h
(c) b + s
(d) 3b – s
Answer :AShow Answer :
Question : In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30°, then BA : AT is
(a) 3 : 1 P
(b) 4 : 1
(c) 2 : 1
(d) 3 : 2
Answer :CShow Answer :
Question : How many tangents can a circle have?
(a) 1
(b) 2
(c) 4
(d) Infinite
Answer 😀Show Answer :
Question : In the given figure, if ∠ABC = 20º, then ∠AOC is equal to:
(a) 10º
(b) 20º
(c) 40º
(d) 60º
Answer :cShow Answer :
Explanation:
Given that, ∠ABC = 20º.
∠AOC = 2∠ABC (since the angle subtended by an arc at the centre of the circle is double the angle subtended at the remaining part.)
Now, substitute the values, we get
∠AOC = 2 × 20°
Therefore, ∠AOC = 40°.
Question : In the given figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to:
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 5 cm
Answer :aShow Answer :
Explanation:
From the given diagram, we can observe that OC is perpendicular to chord AB. Therefore, OC bisects the chord AB Hence. AC=CB
Also,
AC+CB=AB
AC+CB=8
2AC = 8 (Since, AC = CB)
AC = 8/2 = 4 cm
As, the triangle OCA is a right-angled triangle, by using Pythagoras theorem, we can write
AO2=AC2+OC2
52=42+OC2
52−42=OC2
OC2=9
OC=3 cm
As, OD is the radius of the circle, OA=OD=5cm
CD=OD-OC
CD = 5-3 = 2 cm
Hence, the value of CD is equal to 2cm.
Question :The center of the circle lies in______ of the circle.
(a) Interior
(b) Exterior
(c) Circumference
(d) None of the above
Answer : (a) InteriorShow Answer :
Question :If chords AB and CD of congruent circles subtend equal angles at their centres, then:
(a) AB = CD
(b) AB > CD
(c) AB < AD
(d) None of the above
Answer : (a) AB = CDShow Answer :
Question :Segment of a circle is the region between an arc and ………..of the circle.
(a) perpendicular
(b) radius
(c) chord
(d) secant
Answer : (c) chordShow Answer :
Question :In the given figure if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to
(a) 4 cm
(b) 3 cm
(c) 5 cm
(d) 2 cm
Answer :Show Answer :
Question :AB is a chord of a circle with radius ‘r’. If P is any point on the circle such that ∠APB is a right angle , then AB is equal to
(a) 3r
(b) r
(c) 2r
(d) r2
Answer : (c) 2rShow Answer :
Question :In a circle with center O and a chord BC, the point D lies on the same side BC as O. If ∠ BOC = 50°,then ∠ BDC =
(a) 25°
(b) 100°
(c) 75°
(d) 150°
Answer : (a) 25°Show Answer :
Question : The longest chord of the circle is:
(a) Radius
(b) Arc
(c) Diameter
(d) Segment
Answer :(c) DiameterShow Answer :
Question : A chord is at a distance of 8 cm from the centre of a circle of radius 17 cm. The length of the chord is
(a) 25 cm
(b) 12.5 cm
(c) 30 cm
(d) 9 cm
Answer :(c) 30 cmShow Answer :
Question : In a circle with center O and a chord BC, the point D lies on the same side BC as O. If ∠ BOC = 50°,then ∠ BDC =
(a) 25°
(b) 100°
(c) 75°
(d) 150°
Answer :(a) 25°Show Answer :
Question : The perpendicular distance of a chord 8 cm long from the centre of a circle of radius 5 cm is
(a) 2 cm
(b) 9 cm
(c) 4 cm
(d) 3 cm
Answer :(d) 3 cmShow Answer :
Question : The angle subtended by the diameter of a semi-circle is:
(a) 90
(b) 45
(c) 180
(d) 60
Answer :(c) 180Show Answer :
Question : If there are two separate circles drawn apart from each other, then the maximum number of common points they have:
(a) 0
(b) 1
(c) 2
(d) 3
Answer :(a) 0Show Answer :
Question : Number of circles that can be drawn through three non-collinear points is
(a) 3
(b) 1
(c) 2
(d) 0
Answer :(b) 1Show Answer :
Question : AB is a chord of a circle with radius ‘r’. If P is any point on the circle such that ∠APB is a right angle , then AB is equal to
(a) 3r
(b) r
(c) 2r
(d) r2
Answer :(c) 2rShow Answer :
Question : The radius of a circle which has a 6 cm long chord, 4 cm away from the centre of the circle is
(a) 9 cm
(b) 8 cm
(c) 10 cm
(d) 5 cm
Answer :(d) 5 cmShow Answer :
Question : The center of the circle lies in______ of the circle.
(a) Interior
(b) Exterior
(c) Circumference
(d) None of the above
Answer : aShow Answer :
Question : If chords AB and CD of congruent circles subtend equal angles at their centres, then:
(a) AB = CD
(b) AB > CD
(c) AB < AD
(d) None of the above
Answer :(a) AB = CDShow Answer :
Question : Segment of a circle is the region between an arc and …..of the circle.
(a) perpendicular
(b) radius
(c) chord
(d) secant
Answer :(c) chordShow Answer :
Question : AD is diameter of a circle and AB is a chord. If AD = 50 cm, AB = 48 cm, then the distance of AB from the centre of the circle is
(a) 6 cm
(b) 8 cm
(c) 5 cm
(d) 7 cm
Answer :(d) 7 cmShow Answer :
Question : The angle subtended by the diameter of a semi-circle is:
(a) 90
(b) 45
(c) 180
(d) 60
Answer :(c) 180Show Answer :
Question : Greatest chord of a circle is called its
(a) chord
(b) secant
(c) radius
(d) diameter
Answer :(d) diameterShow Answer :
Question : The longest chord of the circle is:
(a) Radius
(b) Arc
(c) Diameter
(d) Segment
Answer : cShow Answer :
Question : Equal _____ of the congruent circles subtend equal angles at the centers.
(a) Segments
(b) Radii
(c) Arcs
(d) Chords
Answer : dShow Answer :
Explanation: Let ΔAOB and ΔCOD are two triangles inside the circle.
OA = OC and OB = OD (radii of the circle)
AB = CD (Given)
So, ΔAOB ≅ ΔCOD (SSS congruency)
∴ By CPCT rule, ∠AOB = ∠COD.
Hence, this prove the statement.
