\nLet C1<\/sub>; C2<\/sub> be two coni oniric circles with their centre C.
\n
![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-1.png?resize=146%2C133&ssl=1\")
\nChord AB of circle C2<\/sup> touches C1<\/sub> at P AB is tangent at P and PC is radius
\nCP \u22a5 AB
\nGiven, \u2220P = 90\u00b0, CP = 4 cm and CA = 5 cm
\n\u2234 IN is tangent angle \u2206PAC,
\nAP2<\/sup> = AC2<\/sup> – PC2<\/sup>
\n= 52<\/sup> – 42<\/sup>
\n= 25 – 16
\n= 9
\nAP = 3 cm
\n\u2234 Perpendicular from centre to chord bisects the chord.
\n\u2234 AB = 2AP
\n= 2 \u00d7 3
\n= 6 cm.<\/p>\n
![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions.png?resize=159%2C13&ssl=1\")
<\/p>\n
Question 2.<\/p>\n
In the given figure, if \u2220AOB = 125\u00b0, then \u2220COD is equal to:<\/h2>\n
(A) 62.5\u00b0
\n(B) 45\u00b0
\n(C) 35\u00b0
\n(D) 55\u00b0
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-2.png?resize=397%2C201&ssl=1\")
\nAnswer:
\n(D) 55\u00b0<\/p>\n
Explanation:
\nSince, quadrilateral circumscribing a circle subtends supplementary angles at the centre of the circle.
\n\u2234\u2220AOB + \u2220COD = 180\u00b0
\n125\u00b0 + \u2220COD = 180\u00b0
\n\u2220COD = 180\u00b0 – 125\u00b0 = 55\u00b0<\/p>\n
Question 3.<\/p>\n
In the given figure, AB is a chord of the circle and AOC is its diameter, such that \u2220ACB = 50\u00b0. If AT is the tangent to the circle at the point A, then \u2220BAT is equal to:<\/h2>\n
(A) 65\u00b0
\n(B) 60\u00b0
\n(C) 50\u00b0
\n(D) 40\u00b0
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-3.png?resize=295%2C175&ssl=1\")
\nAnswer:
\n(C) 50\u00b0<\/p>\n
Explanation:
\nSince, the angle between chord and tangent is equal to the angle subtended by the same chord in alternate segment of circle. \u21d2 \u2220BAT = 50\u00b0.<\/p>\n
Question 4.<\/p>\n
From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is:<\/h2>\n
(A) 60 cm2<\/sup> Explanation: Question 5.<\/p>\n (A) 4cm Explanation: Question 6.<\/p>\n (A) 4 cm Explanation: Question 7.<\/p>\n (A) 100\u00b0 Explanation: Question 8.<\/p>\n (A) 25\u00b0 Explanation: Assertion and Reason Based MCQs<\/span><\/p>\n Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. Question 1.<\/p>\n Answer: Explanation: Question 2.<\/p>\n Answer: Explanation: Question 3.<\/p>\n Answer: Explanation: Question 4.<\/p>\n Answer: Explanation: \u2234 Assertion is correct. Case – Based MCQs<\/span><\/p>\n Attempt any four sub-parts from each question. Each sub-part carries 1 mark. Question 1.<\/p>\n (A) SK \u2260 SC Explanation: Question 2.<\/p>\n (A) 3 m Explanation: Question 3.<\/p>\n (A) 45\u00b0 Question 4.<\/p>\n (A) \\(\\frac{10}{3} \\mathrm{~m}\\) Explanation: Question 5.<\/p>\n (A) Constructions Explanation: II. Read the following text and answer the following question on the basis of the same: Question 1.<\/p>\n (A) 6 m Explanation: Question 2.<\/p>\n (A) 18 m Explanation: Question 3.<\/p>\n (A) 12 m Explanation: Question 4.<\/p>\n (A) 60\u00b0 Explanation: Question 5.<\/p>\n (A) 22 m Explanation: III. Read the following text and answer the following question on the basis of the same: Question 1.<\/p>\n (A) 60\u00b0 Explanation: Question 2.<\/p>\n (A) 75\u00b0 Explanation: Question 3.<\/p>\n (A) 60\u00b0 Question 4.<\/p>\n (A) 90\u00b0 Explanation: IV Read the following text and answer the following question on the basis of the same: Question 1.<\/p>\n (A) 7 Explanation: Question 2.<\/p>\n (A) 8 Explanation: Question 3.<\/p>\n (A) 9 Explanation: Question 4.<\/p>\n (A) 20 Explanation: Circles Class 10 MCQ Questions with Answers Question 1 If the radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is the tangent to the other circle is: (A) 3 cm (B) 6 cm (C) 9 cm (D) 1cm Answer: (B) 6 cm …<\/p>\n
\n(B) 65 cm2<\/sup>
\n(C) 30 cm2<\/sup>
\n(D) 32.5 cm2<\/sup>
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-4.png?resize=289%2C138&ssl=1\")
\nAnswer:
\n(A) 60 cm2<\/sup><\/p>\n
\nPQ is tangent and QO is radius at contact point Question
\n\u2234 PQO = OP2<\/sup> – OQ2<\/sup>
\n= 132 – 52
\n= 169 – 25 = 144
\nPQ = 12 cm
\n\u2206OPQ = \u2206OPR [SSS congruence]
\n\u2234 Area of \u2206OPQ = area of \u2206OPR
\n[Since, congruent figures are equal in areas]
\nArea of quadrilateral QORP = 2 area of \u2206OPR
\n= 2 \u00d7 \\(\\frac{1}{2}\\) base \u00d7 height
\n= RP \u00d7 OR
\n= 12 \u00d7 5
\n= 60 cm2<\/sup><\/p>\nAt one end A of diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the cirde. The length of the chord CD parallel to XY and at a distance 8 cm from A is:<\/h2>\n
\n(B) 5cm
\n(C) 6cm
\n(D) 8cm
\nAnswer:
\n(D) 8cm<\/p>\n
\nXAY is tangent and AO is radius at contact point A of circle.
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-5.png?resize=266%2C192&ssl=1\")
\nAO = 5 cm
\n\u2220OAY = 90\u00b0
\n\u2234 CD is another chord at distance (perpendicular) of 8 cm from A and CMD || XAY meets AB at M.
\njoin OD.
\nOD = 5 cm
\nOM = 8 – 5 = 3 cm
\n\u2220OMD = \u2220OAY = 90\u00b0
\nNow, in right angled \u2206OMD
\nMD2 = OD2 – MO2
\n= 52 – 32
\n= 25 – 9
\n= 16
\n\u21d2MD = 4 cmWe know thaat perpendiculars from centre O of circle bisect the chorde.
\nCD = 2MD
\n= 2 \u00d7 4
\n= 8 cm.
\nHence, length of chord, CD = 8 cm.<\/p>\n<\/p>\nIn the given figure, AT is a tangent to the circle with centre ‘O’ such that OT = 4 cm and \u2220OTA = 30\u00b0. Then AT is equal to:<\/h2>\n
\n(B) 2 cm
\n(C) 2\u221a3 cm
\n(D) 4\u221a3 cm
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-6.png?resize=280%2C180&ssl=1\")
\nAnswer:
\n(C) 2\u221a3 cm<\/p>\n
\nJoin OA. OA is radius and AT is tangen at contact point A.
\n\u2234\u2220OAT = 90\u00b0,
\nGiven that, OT = 4 cm
\nNow, \\(\\frac{A T}{4}\\) = \\(\\vec{a}\\) = cos 30\u00b0
\n\u21d2 AT = 4 \u00d7 \\(\\frac{\\sqrt{3}}{2}\\) = 2\u221a3 cm.<\/p>\nIn the given figure, ‘O’ is the centre of circle, PQ is a chord and the tangent PR at P makes an angle of 50\u00b0 with PQ, then POQ is equal to:<\/h2>\n
\n(B) 80\u00b0
\n(C) 90\u00b0
\n(D) 75\u00b0
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-7.png?resize=269%2C190&ssl=1\")
\nAnswer:
\n(A) 100\u00b0<\/p>\n
\nOP is radius and PR is tangent at P.
\nso, \u2220OPR = 90\u00b0
\n\u2220OPQ + 50 = 90\u00b0
\n\u2220OPQ = 90 – 50\u00b0
\n\u2220OPQ = 40\u00b0
\nIn OPQ, OP = OQ
\n[Radii of same circle]
\n\u2234\u2220Q = \u2220OPQ = 40\u00b0
\n[ Angle opposite to equal sides are equal]
\nBut, \u2220POQ = 180\u00b0 – \u2220P – \u2220QO= 180\u00b0 – 40\u00b0 – 40\u00b0
\n= 180 – 80\u00b0 = 100\u00b0
\n\u21d2 \u2220POQ = 100\u00b0 .<\/p>\nIn the given figure, if PA and PB are tangents to the circle with centre O such that \u2220APB = 50\u00b0, then \u2220OAB is equal to:<\/h2>\n
\n(B) 30\u00b0
\n(C) 40\u00b0
\n(D) 50\u00b0
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-8.png?resize=282%2C164&ssl=1\")
\nAnswer:
\n(A) 25\u00b0<\/p>\n
\nIn \u2206OAB, we have OA = OB [Radii of same circle ]
\n\u2234\u2220OAB = \u2220OBA [Angle opposite to equal sides are equal]
\nAs OA and PA are radius and tangent respectively at contact point A.
\nSo, \u2220OAP = 90\u00b0
\nSimilariy, \u2220OBP = 90\u00b0
\nNow, in quadrilateral PAOB,
\n\u2220P + \u2220A + \u2220O +\u2220B = 360\u00b0
\n\u21d2 50\u00b0 + 90 +\u2220 O + 90 = 360\u00b0
\n\u21d2 \u2220O = 360\u00b0 – 90\u00b0 – 90\u00b0 – 50\u00b0
\n\u21d2 \u2220O = 130\u00b0
\nAgain, in \u2206OAB,
\n\u2220O + \u2220OAB +\u2220 OBA = 180\u00b0
\n\u21d2 130\u00b0 +\u2220OAB + \u2220OAB = 180\u00b0 [\u2234 \u2220OBA = \u2220OAB]
\n\u21d2 2 \u2220OAB = 180\u00b0 – 130\u00b0 = 50v
\n\u21d2 \u2220OAB = 25\u00b0
\nHence,\u2220OAB = 25\u00b0<\/p>\n
\n(A) Both A and R are true and R is the correct explanation of A
\n(B) Both A and R are true but R is NOT the correct explanation of A
\n(C) A is true but R is false
\n(D) A is false and R is True<\/p>\nAssertion (A): The length of the tangent drawn from a point 8 cm away from the centre of circle of radius 6 cm is 2 \u221a7 cm.
\nReason (R): If the angle between two radii of a circle is 130\u00b0, then the angle between the tangents at the end points of radii at their point of intersection is 50\u00b0.<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
\nIn case of assertion:
\nLength of the tangent =\\(\\sqrt{d^{2}-r^{2}}\\)
\n= \\(\\sqrt{(8)^{2}-(6)^{2}}\\)
\n= \\(\\sqrt{64-36}\\)
\n= \\(\\sqrt{28}\\) = 2\u221a7 cm
\n\u2234 Assertion is correct.
\nIn case of reason:
\nSince, sum of the angles between radii and between intersection point of tangent is 180\u00b0. Angle at the point of intersection of tangents = 180\u00b0 -130\u00b0 = 50\u00b0.
\n\u2234 Reason is correct.
\nHence, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\n<\/p>\nAssertion (A): If a chord AB subtends an angle of 60\u00b0 at the centre of a circle, then the angle between the tangents at A and B is also 60\u00b0.
\nReason (R): The length of the tangent from an external point P on a circle with centre O is always less than OP.<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nIn case of assertion:
\nChord AB subtends \u222060\u00b0 at O
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-9.png?resize=240%2C222&ssl=1\")
\n\u2234 \u2220OAP = 90\u00b0
\nSimilarly, \u2220OBP = 90\u00b0
\nIn quadrilateral OAPB,
\n\u2220O + \u2220P + \u2220OAP + \u2220OBP = 360\u00b0
\n\u21d2 60 + \u2220P + 90\u00b0 +90\u00b0 = 360\u00b0
\n\u21d2 \u2220P = 360\u00b0 – 240\u00b0
\n\u21d2 \u2220P = 120\u00b0
\n\u2234 Assertion is correct.
\nIn case of reason:
\nPT and OT are the tangent and radius, respectivly at contact point T.
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-10.png?resize=323%2C166&ssl=1\")
\nSo, OTP = 90\u00b0
\n\u21d2 \u2206OPT is right angled triangle.
\nAgain, in \u2206OPT
\n\u2234 \u2220T > \u2220O
\n\u2234 \u2220OP > \u2220PT
\n[ side opposite to greater angle is larger]
\n\u2234 Reason is correct.
\nHence, assertion is incorrect but reason is correct.<\/p>\nAssertion (A): The tangent to the circumcircle of an isosceles AABC at A, in which AB = AC, is parallel to BC.
\nReason (R): PQ is a tangent drawn from an external point P to a circle with centre O. QOR is the diameter of the circle. If \u2220POR = 120\u00b0, then the measure of \u2220OPQ is 60\u00b0.<\/h2>\n
\n(C) A is true but R is false<\/p>\n
\nIn case of assertion:
\ngiven that, \u2206ABC, inscribed in a circle in which AB = AC
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-11.png?resize=181%2C206&ssl=1\")
\nPAQ is tangent at A.
\nAB is chord.
\n\u2234\u2220 PAB = \u2220C
\nAs, we know that , \u2220PAB formed by chord AB with tangente segment.
\nIn \u2206ABC,
\nAB = AC [ Given]
\n\u2234 \u2220B = \u2220C [\u2234Angles opposite to equai sides are equal]……(ii)
\nFrom (i) and (ii), \u2220B = \u2220PAB
\nThese are alternate interior angles.
\nTherefore, PAQ || BC.
\n\u2234 Assertion is correct.
\nIn case of reason:
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-12.png?resize=289%2C150&ssl=1\")
\nCiven that,
\nPQ is a tangent, \u2220POR = 120\u00b0
\n\u2220POR = \u2220OQP + \u2220OPQ [ Exterior angle sum property]
\n\u2220OPQ = 120 – 90\u00b0 = 30\u00b0
\n\u2220OPQ = 30\u00b0
\n\u2234 Reason is correct.
\nHence, assertion is correct but reason is incorrect.<\/p>\nAssertion (A): PQ is a tangent to a circle with centre O at point P. If AOPQ is an isosceles triangle, then \u2220OQP = 45\u00b0.
\nReason (R): If two tangents inclined at 60\u00b0 are drawn to a circle of radius 3 cm, then the length of each tangent is 3\u221a3 cm.<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
\nIn case of assertion:
\nAs we know that \u2220OPQ = 90\u00b0 ( Angle between tangent and radius)
\nLet \u2220PQO be x\u00b0, then\u2220QOP = x\u00b0
\nSince\u2220 OPQ is an isosceles triangle.
\n(given)(OP = OQ)
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-13.png?resize=286%2C178&ssl=1\")
\nIn \u2206OPQ,
\n\u2220OPQ + \u2220PQO + \u2220QOP = 180
\n(property of the sum of angles of a triangle)
\n\u2234 90\u00b0 + x\u00b0 + x\u00b0 = 180\u00b0
\n\u21d2 2x\u00b0 = 180\u00b0 – 90\u00b0 = 90\u00b0
\n\u21d2 x = \\(\\frac{90}{2}\\) = 45.
\nHence, OQP is 45<\/p>\n
\nIn case of reason:
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-14.png?resize=259%2C172&ssl=1\")
\nIn \u2206PAO,
\ntan 30 = \\(\\frac{A O}{P A}\\) (Using trigonometry)
\n\\(\\frac{1}{\\sqrt{3}}\\) = \\(\\frac{3}{P A}\\)
\nPA = 3 \u221a3 cm.
\n\u2234 Reason is correct.
\nHence, assertion is correct but reason is incorrect.<\/p>\n
\nI. Read the following text and answer the following question on the basis of the same:
\nThere is a circular filed of radius 5 m. Kanabh, Chikoo and Shubhi are playing with ball, in which Kanabh and Chikoo are standing on the boundary of the circle. The distance between Kanabh and Chikoo is 8 m. From Shubhi point S, two tangents are drawn as shown in the figure. Give the answer of the following questions.<\/p>\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-15-1.png?resize=282%2C220&ssl=1\")
<\/p>\nWhat is the relation between the lengths of SK and SC?<\/h2>\n
\n(B) SK = SC
\n(C) SK > SC
\n(D) SK < SC.
\nAnswer:
\n(B) SK = SC<\/p>\n
\nWe know that the lengths of tangents drawn from an external point to a circle are equal. So SK and SC are tangents to a circle with centre O.
\n\u2234SK = SC<\/p>\nThe length (distance) of OR is:<\/h2>\n
\n(B) 4 m
\n(C) 5 m
\n(D) 6 m.
\nAnswer:
\n(A) 3 m<\/p>\n
\nIn question 1, we have proved SK = SC
\nThen ASKC is an isosceles triangle and SO is the angle bisector of ZKSC. So, OS X KC.
\n\u2234 OS bisects KC, gives KR = RC = 4 cm
\nNow, OR = \\(\\sqrt{O K^{2}-K R^{2}}\\)
\n(By using Pythagoras theorem)
\n= \\(\\sqrt{5^{2}-4^{2}}=\\) = \\(\\sqrt{25-16}\\)
\n= \\(\\sqrt{25-16}\\)
\n= \\(\\sqrt{9}\\)
\n= 3 m<\/p>\nThe sum of angles SKR and OKR is:<\/h2>\n
\n(B) 30\u00b0
\n(C) 90\u00b0
\n(D) none of these
\nAnswer:
\n(C) 90\u00b0
\n\u2220SKR + \u2220OKR = \u2220OKR
\n= 90 (Radius is \u22a5 to tangent)<\/p>\nThe distance between Kanabh and Shubhi is:<\/h2>\n
\n(B) \\(\\frac{13}{3} \\mathrm{~m}\\)
\n(C) \\(\\frac{16}{3} \\mathrm{~m}\\)
\n(D) \\(\\frac{20}{3} \\mathrm{~m}\\)
\nAnswer:
\n(D) \\(\\frac{20}{3} \\mathrm{~m}\\)<\/p>\n
\n\u2206SER and \u2206RKO,
\n\u2220RKO = \u2220KSR
\n\u2220SRK = \u2220ORK
\n\u2206KSR ~ \u2206OKR . (By AA Similarity)
\n\\(\\frac{S K}{K O}\\) = \\(\\frac{R K}{R O}\\)
\nThen \\(\\frac{S K}{5}\\) = \\(\\frac{4}{3}\\)
\n(RO = 3 m, proved in Question 2)
\n\u21d2 3SK = 20
\n\u21d2 Sk = \\(\\frac{20}{3}\\)
\nHence, the distance between Kanabh and Shubhi is \\(\\frac{20}{3}\\)<\/p>\nWhat is the mathematical concept related to this question ?<\/h2>\n
\n(B) Area
\n(C) Circle
\n(D) none of these
\nAnswer:
\n(C) Circle<\/p>\n
\nThe mathematical concept (Circle) is related to this question.<\/p>\n
\nABCD is a playground. Inside the playground a circular track is present such that it touches AB at point P, BC at Q, CD at R and DA at S.
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-16.png?resize=285%2C201&ssl=1\")
<\/p>\nIf DR = 5 m, then DS is equal to:<\/h2>\n
\n(B) 11 m
\n(C) 5 m
\n(D) 18 m
\nAnswer:
\n(C) 5 m<\/p>\n
\n\u2234 DR =5 m (given)
\nDR = DS (Length of tangents are equal)
\nDS = 5 m.<\/p>\n<\/p>\nThe length of AS is:<\/h2>\n
\n(B) 13 m
\n(C) 14 m
\n(D) 12 m
\nAnswer:
\n(A) 18 m<\/p>\n
\nWe have AD = 23 m.
\nands DS = 5 m (Proved in Question 1)
\n\u2234 AS = AD – DS
\n= (23 – 5)m = 18 m.<\/p>\nThe length of PB is:<\/h2>\n
\n(B) 11 m
\n(C) 13 m
\n(D) 20 m
\nAnswer:
\n(B) 11 m<\/p>\n
\nWe have,
\nAB = 29 m
\nBut AS = AP (lengths of tangents are equal)
\nand AS = 18 m(proved in Question 2)
\n\u2234 AP = 18 m
\nNOw, PB = AB – AP
\n= (29 – 18) m
\n= 11 m<\/p>\nWhat is the angle of OQB?<\/h2>\n
\n(B) 30\u00b0
\n(C) 45\u00b0’
\n(D) 90\u00b0
\nAnswer:
\n(D) 90\u00b0<\/p>\n
\n\u2220OQB = 90\u00b0
\n(Radius is \u2019 to tangent)<\/p>\nWhat is the diameter of given circle?<\/h2>\n
\n(B) 33 m
\n(C) 20 m
\n(D) 30 m
\nAnswer:
\n(A) 22 m<\/p>\n
\nPB = 11 m (proved in Question 3) PB = BQ (lengths oftangents are equal)
\nBut PB = BQ
\n\u2234BQ = 11 m
\nor = OQ = QB = 11m
\nHence, diameter = 2r = 2 \u00d7 11 = 22 m.<\/p>\n
\nA Ferris wheel (or a big wheel in the United Kingdom) is an amusement ride consisting of a rotating upright wheel with multiple passenger-carrying components (commonly referred to as passenger cars, cabins, tubs, capsules, gondolas, or pods) attached to the rim in such a way that as the wheel turns, they are kept upright, usually by gravity.After taking a ride in Ferris wheel, Aarti came out from the crowd and was observing her friends who were enjoying the ride. She was curious about the different angles and measures that the wheel will form. She forms the figure as given below.
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-17.png?resize=248%2C162&ssl=1\")
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-18.png?resize=263%2C157&ssl=1\")
<\/p>\nIn the given figure find \u2220ROQuestion<\/h2>\n
\n(B) 100\u00b0
\n(C) 150\u00b0
\n(D) 90\u00b0
\nAnswer:
\n(C) 150\u00b0<\/p>\n
\n\u2220ORP = 90\u00b0 & \u2220OQP [\u2235 radius of circle is perpendicular to tangent]
\n\u2234 \u2220ROQ + \u2220ORP + \u2220OQP + \u2220QPR = 360\u00b0 \u2220ROQ + 90\u00b0 + 90\u00b0 + 30\u00b0 = 360\u00b0 \u2220ROQ + 210\u00b0 = 360\u00b0 \u2220ROQ = 360\u00b0-210\u00b0
\n\u2220ROQ = 150\u00b0<\/p>\nFind \u2220RQP.<\/h2>\n
\n(B) 60\u00b0
\n(C) 30\u00b0
\n(D) 90\u00b0
\nAnswer:
\n(A) 75\u00b0<\/p>\n
\n\u2220OQR = \u2220ORQ
\n\u2220RPQ = 150
\nand \u2220ROQ + \u2220OQR + \u2220ORQ = 180
\n150 + 2\u2220ORQ =180
\n2\u2220ORQ = 30
\n\u2220ORQ = 15
\n\u2220OQR = ORQ = 1
\n\u2220RQP = \u2220OQP – \u2220OQR= 90 – 15<\/p>\n<\/p>\nFind \u2220RSQuestion<\/h2>\n
\n(B) 75
\n(C) 100\u00b0
\n(D) 30
\nAnswer:
\n(B) 75<\/p>\nFind \u2220ORP<\/h2>\n
\n(B) 70
\n(C) 100\u00b0
\n(D) 60
\nAnswer:
\n(A) 90\u00b0<\/p>\n
\n\u2220ORP = 90\u00b0
\nBecause, radius of circle is perpendicular to tangent.<\/p>\n
\nVarun has been selected by his School to design logo for Sports Day T-shirts for students and staff. The logo design is as given in the figure and he is working on the fonts and different colours according to the theme. In given figure, a circle with centre O is inscribed in a AABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. The lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively.
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-19.png?resize=255%2C233&ssl=1\")
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-20.png?resize=157%2C160&ssl=1\")
<\/p>\nFind the length of AD.<\/h2>\n
\n(B) 8
\n(C) 5
\n(D) 9
\nAnswer:
\n(A) 7<\/p>\n
\nGiven, AB = 12 cm
\nBC = 8 cm
\nCA = 10 cm
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-10-Circles-21.png?resize=318%2C261&ssl=1\")
\nIn \u2206ABC
\nx + y =12 cm -(i)
\nx + z = 10 cm …(ii)
\ny + z = 8 cm -(Mi)
\nAdding (i), (ii) and (iii)
\n2(x + y + z) =30 cm
\nx + y + z =15 cm …(iv)
\nfrom eQuestion (iii) and (iv)
\nx = 7 cm
\nAD = x = 7 cm<\/p>\nFind the Length of BE.<\/h2>\n
\n(B) 5
\n(C) 2
\n(D) 9
\nAnswer:
\n(B) 5<\/p>\n
\nWe have
\nx + y + z = 15 cm
\nand x + z = 10 cm
\nBy solving above equations
\nwe get y = 5 cm
\ni.e., BE = y = 5 cm<\/p>\nFind the length of CF.<\/h2>\n
\n(B) 5
\n(C) 2
\n(D) 3
\nAnswer:
\n(D) 3<\/p>\n
\nWe have
\nx y + z = 15 cm
\nand x + y =12 cm
\nBy solving above equations,
\nwe get z = 3 cm
\ni.e., CF = z = 3 cm<\/p>\n<\/p>\nIf radius of the circle is 4 cm, find the area of \u2206OAB.<\/h2>\n
\n(B) 36
\n(C) 24
\n(D) 48
\nAnswer:
\n(C) 24<\/p>\n
\nArea of \u2206OAB = \\(\\frac{1}{2}\\) \u00d7 AB \u00d7 OD
\n= \\(\\frac{1}{2}\\) \u00d7 12 \u00d7 4
\n= 24 cm2<\/sup><\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"