# 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.

Here 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 *C*_{1} and *C*_{2 }be the two concentric circles, with centre at O and respective radii being *r*_{1} = 4cm and *r _{2} *= 5cm.

Draw a chord *AC *of circle *C*_{2}, to touch circle *C*_{1} 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 Δ*ACB, *∠*A + *∠*B + *∠*C* = 180° [Sum of all interior angles of a triangle is 180°]

⟹ ∠*A* + 90° + 50° = 180°

⟹ ∠*A* + 140 = 180

⟹ ∠*A* = 180^{o} - 140° = 40°

Or ∠*OAB *= 40° …(i)

Here, *OA* ⏊ *AT*

⟹* *∠*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 cm^{2}

(b) 65 cm^{2}

(c) 30 cm^{2}

(d) 32.5 cm^{2}

**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* = 2*EC *= 2 × 4 = 8 cm

**CBSE Class 10 NCERT Textbooks & NCERT Solutions**

**NCERT Solutions for CBSE Class 10 Maths**

**NCERT Exemplar Problems and Solutions Class 10 Science: All Chapters**