**RD Sharma Class 10 Solutions Chapter 10 Exercise 10.1: **In this exercise, you’ll learn about interesting concepts like secant and tangent to a circle. **RD Sharma Class 10 Solutions** is available to those who want to build a strong basis in these Maths chapters. Students can also use the **RD Sharma Solutions for Class 10 Maths Chapter 10 Circles** Exercise 10.1 PDF to get a better understanding of the concepts covered in this exercise.

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RD Sharma Class 10 Solutions Chapter 10 Exercise 10.1

### Access answers to RD Sharma Solutions Class 10 Maths Chapter 10 Exercise 10.1- Important Question with Answers

Question 1.

Fill in the blanks :

(i) The common point of a tangent and the circle is called ……….

(ii) A circle may have ………. parallel tangents.

(iii) A tangent to a circle intersects it in ……….. point(s).

(iv) A line intersecting a circle in two points is called a …………

(v) The angle between tangent at a point on a circle and the radius through the point is ………..

Solution:

(i) The common point of a tangent and the circle is called the point of contact.

(ii) A circle may have two parallel tangents.

(iii) A tangent to a circle intersects it in one point.

(iv) A line intersecting a circle in two points is called a secant.

(v) The angle between tangent at a point, on a circle and the radius through the point is 90°.

Question 2.

How many tangents can a circle have?

Solution:

A circle can have infinitely many tangents.

Question 3.

O is the center of a circle of radius 8 cm. The tangent at a point A on the circle cuts a line through O at B such that AB = 15 cm. Find OB.

Solution:

Radius OA = 8 cm, ST is the tangent to the circle at A and AB = 15 cm

OA ⊥ tangent TS

In right ∆OAB,

OB² = OA² + AB² (Pythagoras Theorem)

= (8)² + (15)² = 64 + 225 = 289 = (17)²

OB = 17 cm

Question 4.

If the tangent at a point P to a circle with center O cuts a line through O at Q such that PQ = 24 cm and OQ = 25 cm. Find the radius of the circle.

Solution:

OP is the radius and TS is the tangent to the circle at P

OQ is a line

OP ⊥ tangent TS

In right ∆OPQ,

OQ² = OP² + PQ² (Pythagoras Theorem)

=> (25)² = OP² + (24)²

=> 625 = OP² + 576

=> OP² = 625 – 576 = 49

=> OP² = (7)²

OP = 7 cm

Hence radius of the circle is 7 cm

We have provided complete details of RD Sharma Class 10 Solutions Chapter 10 Exercise 10.1. If you have any queries related to **CBSE** Class 10, feel free to ask us in the comment section below.

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The RD Sharma Class 10 Solutions Chapter 10 Exercise 10.1 answers were quite helpful. This makes it simple to clear any questions about arithmetic progression. Easily answer all of the questions that the Class 10 students have given for exercise.

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