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NCERT Exemplar Solution for Class 10 Mathematics: Circles (Part-IA)

Class 10 Mathematics chapter 9- Circles: NCERT Exemplar Problems and Solutions (Part-IA) is available here. This part comprises of solutions to multiple choice questions (1-5) form exercise 9.1 of NCERT Exemplar for Mathematics chapter 9. Every solution has been designed to give students an easy and apt explanation.

Aug 4, 2017 11:28 IST
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Class 10 Maths NCERT Exemplar, Circles NCERT Exemplar Problems, NCERT Exemplar Problems, Circles Class 10 NCERT ExemplarHere you get the CBSE Class 10 Mathematics chapter 9, Circles: NCERT Exemplar Problems and Solutions (Part-IA). This part of the chapter includes solutions to Question Number 1 to 5 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.

CBSE Class 10 Mathematics Syllabus 2017-2018

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:

Exercise 9.1

Multiple Choice Questions (Q. No. 1-5)

Question. 1 If radii of two concentric circles are 4 cm and 5 cm, then length of each chord of one circle which is tangent to the other circle, is

(a) 3 cm             

(b) 6 cm           

(c) 9 cm        

(d) 1 cm

Solution. (b)

Explanation:

Let C1 and C2 be the two concentric circles, with centre at O and respective radii being r1 = 4cm and r2 = 5cm.

Draw a chord AC of circle C2, to touch circle C1 at B.

Join OB.

 

Solution. (d)

Explanation:

Since, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

i.e.,    AOB + COD = 180°

⟹     ∠COD = 180° - ∠AOB

⟹     ∠COD = = 180° - 125° = 55°

Question. 3 In figure, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to

Solution. (c)

Explanation:

Here AC is the diameter of the circle.

∴               ∠ABC = 90°                      [ Angle in a semi-circle]

Now, in ΔACB, A + B + C = 180°      [Sum of all interior angles of a triangle is 180°]

⟹     ∠A + 90° + 50° = 180°

⟹             ∠A + 140 = 180

⟹                      ∠A = 180o - 140° = 40°

Or                   ∠OAB = 40°                                        …(i)

Here, OAAT

       OAT = 90°          

⟹       ∠OAB + ∠BAT = 90°

⟹                  BAT = 90° - 40° = 50°    [Using (i)]

Hence, the value of ∠BAT is 50°.

Question. 4 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 is drawn. Then, the area of the quadrilateral PQOR is

(a) 60 cm2         

(b) 65 cm2       

(c) 30 cm2         

(d) 32.5 cm2

Solution. (a)

Explanation:

Given conditions are described in the diagram given below:

Question. 5 At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A, is

(a) 4 cm    

(b) 5 cm    

(c) 6 cm    

(d) 8cm

Solution. (d)        

Explanation:

Given conditions are described in the diagram given below:

Now, as the perpendicular from centre to the chord of the same circle bisects that chord.

Hence, length of chord CD = 2EC = 2 × 4 = 8 cm

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