Students can practice 10th Class Maths Study Material Telangana Chapter 9 Tangents and Secants to a Circle InText Questions to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions

Do This

Question 1.

Draw a circle with any radius. Draw four tangents at different points. How many tangents can you draw to this circle ? (AS_{3}, AS_{5}) (Page No. 226)

Solution:

Let ‘O’ be the centre of the circle with radius ‘OA’. l, m, n, p and q be the tangents to the circle at A, B, C, D and E. We can draw a tangent at each point on the circle, i.e., infintely many tangents can be drawn to a circle.

Question 2.

How many tangents you can draw to circle from a point away from it ? (AS_{3}) (Page No. 226)

Solution:

We can draw only two tangents from an exterior point.

Question 3.

In the below figure which are tangents to the circles ? (AS_{3}) (Page No. 226)

Solution:

P and M are the tangents to the circles.

Question 4.

Draw a circle and a secant PQ of the circle on a paper as shown below. Draw various lines parallel to the secant on both sides of it. What happens to the length of chord coming closer and closer to the center of the circle ?

What is the longest chord ? How many tangents can you draw to a circle, which are parallel to each other. (Page No. 227)

Solution:

The length of the chord increases as it comes closer to the centre of the circle.

i) What is the longest chord ? (Page No. 227)

Solution:

Diameter is the longest of all chords.

ii) How many tangents can you draw to a circle which are parallel to each other ? (Page No. 227)

Solution:

Only one tangent can be drawn parallel to a given tangent. To a circle, we can draw infinetely pairs of parallel tangents.

Try This

Question 1.

How can you prove the converse of the above theorem. “If a line in the plane of a circle is perpendicular to the radius at its end point on the circle, then the line is tangent to the circle”. (Page No. 228)

Solution:

Given : Circle with centre ‘O’, a point ‘A’ on the circle and the line AT’ perpendicular to OA’.

RTP : ‘AT’ is a tangent to the circle at A.

Construction :

Suppose ‘AT’ is not a tangent then ‘AT’ produced either way if necessary, will meet the circle again. Let it do so at R join OR

Proof : Since OA = OP (radii)

∴ ∠OAP = ∠OPA

But ∠OPA = 90°

∴ Two angles of a triangle are right angles which is impossible.

∴ Our supposition is false, hence AT is a tangent.

We can find some more results using the above theorem.

- Since there can be only one perpendicular OP at the point R it follows that one and only are tangent can be drawn to a circle at a given point on the circumference.
- Since there can be only one perpendicular to XY at the point R it follows that the perpendicular to a tangent at its point of contact passes through the centre.

Question 2.

How can you draw the tangent to a circle at a given point when the centre of the circle is not known ? (Page No. 229) Hint : Draw equal angles ∠QPX and ∠PRQ. Explain the construction.

Solution:

Steps of construction :

- Take a point ‘P’ and draw a chord PR through P
- Construct ∠PRQ and measure it.
- Construct ∠QPX at P equal to ∠PRQ.
- Extend PX on otherside. XY is the required tangent at R

Note : Angle between a tangent and chord is equal to angle in the alternate segment.

Try This

Question 1.

Use pythagorus theorem and write proof for the statement “the lengths of tangents drawn from an external point to a circle are equal”. (Page No. 231)

Solution:

Given : Two tangents PA and PB to a circle with centre ‘O’, from an exterior point P.

R.T.P : PA = PB

Proof : In ∆OAP; ∠OAP = 90°

∴ AP^{2} = OP^{2} – OA^{2} [∵ square of the hypotenuse is equal to the sum of squares on the other 2 sides → Pythagoras theorem]

⇒ (AP)^{2} = (OP)^{2} – (OB)^{2}

[∵ OA = OB, radii of the same circle]

⇒ AP^{2} = BP^{2}

[∴ In a ∆OBP; OB^{2} + BP^{2} = OP^{2}

⇒ BP^{2} = OP^{2} – OB^{2}]

⇒ AP^{2} = BP^{2}

∴ PA = PB (CPCT)

Hence proved.

Question 2.

Draw a pair of radii OA and OB such that ∠BOA = 120°. Draw the bisector of ∠BOA and draw lines perpendicular to OA and OB at A and B. These lines meet on the bisector of ∠BOA at a point which is the external point and the perpendicular lines are the required tangents. Construct and justify. (Page No. 235)

Solution:

Justification :

OA ⊥ PA

OB ⊥ PB

Also in ∆OAP, ∆OBP

OA = OB

∠OAP = ∠OBP

OP = OP

∴ ∆OAP ≅ ∆OBP

∴ PA = PB [Q.E.D]

Do This

Question 1.

Shankar made the following pictures also

What shapes can they be broken into that we can find area easily ? (Page No. 237)

Solution:

Into two rectangles.

A rectangle and a circle

A cone and a secant

A rectangle and 2 segments

Make some more pictures and think of the shapes they can be divided into different parts.

Solution:

A cone and segment

A rectangle and a segment

A square and four segments

A cone and segment

A rectangle and a segment A square and four segments.

Do This

Question 1.

Find the area of sector, whose radius is 7 cm with the given angle : (Page No. 239)

(i) 60°

(ii) 30°

(iii) 72°

(iv) 90°

(v) 120°

Solution:

Question 2.

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 10 minutes. (Page No. 239)

Solution:

Angle made by minute hand in 360°

1 min = \(\frac{360^{\circ}}{60^{\circ}}\) = 6°

Angle made by minute hand in

10 min = 10 × 6° = 60°

The area swept by minute hand is in the shape of a sector with radius.

r = 14 cm and angle x° = 60°

Area (A) = \(\frac{x^{\circ}}{360^{\circ}}\) × πr^{2}

= \(\frac{60^{\circ}}{360^{\circ}}\) × \(\frac{22}{7}\) × 14 × 14

= \(\frac{616}{6}\) ⇒ 102.66 cm^{2}

∴ Area swept by the minute hand in 10 minutes = 102.66 cm^{2}

Try This

Question 1.

How can you find the area of major seg¬ment using are a of minor segment ?

Solution:

Area of the major segment = (Area of the circle) – (Area of the minor segment)