NCERT Exemplar Solution for Class 10 Mathematics: Circles (Part-IB)
Here you get the NCERT Exemplar Problems and Solutions for CBSE Class 10 Mathematics chapter 9- Circles: (Part-IB). This part is a continuation of Part-IA and comprises of solutions question number 6 to10 form exercise 9.1 of NCERT Exemplar for Mathematics chapter 9. All these questions are very important for CBSE Class 10 Maths Exam 2017-2018.
Here you get the CBSE Class 10 Mathematics chapter 9, Circles: NCERT Exemplar Problems and Solutions (Part-IB). This part of the chapter includes solutions to Question Number 6 to 10 from Exercise 9.1 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Circles. This exercise comprises only the Multiple Choice Questions (MCQs) framed from various important topics in the chapter. Each question is provided with a detailed solution. Check out Part-IA for solutions to Q. No. 1-5 of Exercise 9.1 of NCERT Exemplar Problems Class 10 Maths.
NCERT exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.
Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Circles:
Multiple Choice Questions (Q. No. 6-10)
Question. 6 In figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ÐOTA = 30°. Then, AT is equal to
Question. 7 In figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is equal to
As, the tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ ∠OPR = 90°
⟹ ∠OPQ + ∠QPR = 90°
⟹ ∠OPQ = 90° - 50° = 40° [Given, ∠QPR = 50°]
Now, in ΔOPQ,
OP = OQ = Radius of circle
⟹ ∠OQP = ∠OPQ = 40° [Angles opposite to equal sides of a triangle are also equal]
∠O + ∠P + ∠Q = 180° [Sum of angles of a triangle = 180°]
⟹ ∠O = 180° - (40° + 40°) [∵ ∠P = ∠Q = 40°]
⟹ ∠O = 180° - 80° = 100°
Or ∠POQ = 100°
Question. 8 In figure, if PA and PB are tangents to the circle with centre O such that ÐAPB = 50°, then ÐOAB is equal to
As, we know that the tangents drawn from an external point to a circle are equal
∴ PA = PB
Thus, in ΔPAB,
∠PBA = ∠PAB [Angles opposite to equal sides of a triangle are also equal]
Let ∠PBA = ∠PAB = θ
Now, in ΔPAB,
∠P + ∠A + ∠B = 180° [Sum of angles of a triangle = 180°]
⟹ 50° + θ + θ = 180°
⟹ 2θ = 180° – 50° = 130°
⟹ θ = 65°
Now, as we know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ OA ⏊ PA
⟹ ∠PAO = 90°
⟹ ∠PAB + ∠OAB = 90°
⟹ 65° + ∠OAB = 90°
⟹ ∠OAB = 90° - 65° = 25°
Question. 9 If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm, then the length of each tangent is
Let PQ and PR be the two tangents drawn from external point P with angle between them being 60°.
Join OQ and OR.
We know that any two tangents drawn to a circle from an external point are equally inclined to the line segment, joining the centre to that point.
Question. 10 In figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to