## Revision Notes for ICSE Class 10 Maths Chapter 18 - Free PDF Download

## FAQs on ICSE Class 10 Mathematics Revision Notes Chapter 18 - Tangents and Intersecting Chords

**1. Are ICSE Class 10 Mathematics Revision Notes Chapter 18- Tangents And Intersecting Chords good to score better in class 10 examinations?**

Yes, by practicing ICSE Class 10 Mathematics Revision Notes Chapter 18- Tangents and Intersecting Chords, students will tend to learn more and more of the subject and thus score better grades in the Class 10 ICSE board examination. This is because, when students practice the worksheets, they learn about all the concepts of the subject deeply. So they are well prepared for the examination. They don't need to be in chaos before the early days of the examination. Also, the learning power of the students is enhanced.

**2. Where can I find useful study resources for ICSE Class 10 Mathematics Revision Notes Chapter 18- Tangents And Intersecting Chords?**

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**3. After studying chapter 18 of ICSE Class 10 Maths, what do you know about tangents?**

Tangent to a circle is a straight line that touches the circle at only one point. This point where the tangent touches the circle is called the point of tangency. The tangent to a circle is perpendicular to the radius at the point of tangency. The tangent of a circle is always perpendicular to the radius which joins the center of the circle to point P. You can find the tangent of the circle through the equation of the tangent will be of the form y = m x + c.

**4. Solve the exercise question with the concept discussed in Class 10 ICSE Maths Chapter 18,**

**If, **

**AP + BQ + CR = BP + CQ + AR.**

**Also, show that AP + BQ + CR = ½ x perimeter of triangle ABC.**

As from point B, BQ and BP are the tangents to the circle

We have, BQ = BP ………..(1)

Similarly, we also get

AP = AR ………….. (2)

And, CR = CQ …… (3)

Adding (1), (2) and (3) we get,

AP + BQ + CR = BP + CQ + AR ……… (4)

Now, adding AP + BQ + CR to both sides in (4), we get

2(AP + BQ + CR) = AP + PQ + CQ + QB + AR + CR

2(AP + BQ + CR) = AB + BC + CA

Therefore, we get

AP + BQ + CR = ½ x (AB + BC + CA)

I.e.

AP + BQ + CR = ½ x perimeter of triangle ABC

**5. Solve the exercise question with the concept discussed in Class 10 ICSE Maths Chapter 18:**

**Given in the question:**

**PQ is the tangent to the circle at A, DB is diameter and O is the centre of the circle. If ∠ADB = 30o and ∠CBD = 60o; you have to calculate:**

**i) ∠QAB **

**ii) ∠PAD**

**iii) ∠CDB**

**(i) Given, PAQ is a tangent and AB is the chord**

∠QAB = ∠ADB = 30^{o} (Angles in the alternate segment)

**(ii) OA = OD (radii of the same circle)**

So, ∠OAD = ∠ODA = 30^{o}

But, as OA ⊥ PQ

∠PAD = ∠OAP – ∠OAD = 90^{o} – 30^{o} = 60^{o}

**(iii) As BD is the diameter, we have**

∠BCD = 90^{o} (Angle in a semicircle)

Now in ∆BCD,

∠CDB + ∠CBD + ∠BCD = 180^{o}

∠CDB + 60^{o} + 90^{o} = 180^{o}

Thus, ∠CDB = 180^{o} – 150^{o} = 30^{o}