## AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Optional Exercise

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise Textbook Questions and Answers.

### 10th Class Maths 9th Lesson Tangents and Secants to a Circle Optional Exercise Textbook Questions and Answers

Question 1.

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line – segment joining the points of contact at the centre.

Answer:

Given: A circle with centre ‘O’.

Two tangents \(\overleftrightarrow{\mathrm{PQ}}\) and \(\overleftrightarrow{\mathrm{PT}}\) from an external point P. Let Q, T be the points of contact.

R.T.P: ∠P and ∠QOT are supplementary.

Proof: OQ ⊥ PQ

[∵ radius is perpendicular to the tangent at the point of contact] also OT ⊥ PT

∴ ∠OQP + ∠OTP = 90° + 90° = 180° Nowin oPQOT,

∠OTP + ∠TPQ + ∠PQO + ∠QOT

= 360° (angle sum property)

180° + ∠P + ∠QOT = 360°

∠P + ∠QOT = 360°- 180° = 180° Hence proved. (Q.E.D.)

Question 2.

PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (See figure). Find the length of TP.

Answer:

Given: PQ = 8

⇒ PR = 4

⇒ PO^{2} = PR^{2} + OR^{2}

⇒ 25 = 16 + OR^{2}

⇒ OR = 3

Now let RT = x and PT in △OPT, ∠P = 90°

∴ OT is hypotenuse.

∴ OT^{2} = OP^{2} + PT^{2}

(Pythagoras theorem)

(3 + x)^{2} = 5^{2} + y^{2} …….. (1)

and in △PRT, ∠R = 90°

∴ \(\overline{\mathrm{PT}}\) is hypotenuse.

∴ PT^{2} = PR^{2} + RT^{2}

y^{2} = 4^{2} + x^{2} …….. (2)

Now putting the value of y^{2} = 4^{2} + x^{2} in equation (1) we got

(3 + x)^{2} = 5^{2} + x^{2} + 4^{2}

9 + x^{2} + 6x = 25 + 16 + x^{2}

6x = 25 + 16 – 9 = 25 + 7 = 32

⇒ x = \(\frac{32}{6}\) = \(\frac{16}{3}\)

Now from equation (2), we get

Question 3.

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer:

Given: Let a circle with centre ‘O’ touches the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R and S respectively.

R.T.P: ∠AOB + ∠COD = 180°

∠AOD + ∠BOC = 180°

Construction: Join OP, OQ, OR and OS.

Proof: Since the two tangents drawn from an external point of a circle subtend equal angles.

At the centre,

∴ ∠1 = ∠2

∠3 = ∠4 (from figure)

∠5 = ∠6

∠7 = ∠8

Now, ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°

[∵ Sum of all the angles around a point is 360°]

So, 2 (∠2 + ∠3 + ∠6 + ∠7) = 360°

and 2 (∠1 + ∠8 + ∠4 + ∠5) = 360°

(∠2 + ∠3) + (∠6 + ∠7) = \(\frac{360}{2}\) = 180°

Also, (∠1 + ∠8) + (∠4 + ∠5) = \(\frac{360}{2}\) = 180°

So, ∠AOB + ∠COD = 180°

[∵ ∠2 + ∠3 = ∠AOB;

∠6 + ∠7 = ∠COD

∠1 + ∠8 = ∠AOD

and ∠4 + ∠5 = ∠BOC [from fig.]]

and ∠AOD + ∠BOC = 180°

Question 4.

Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

Answer:

Steps of construction:

- Draw a line segment AB of length 8 cm.
- With A and B as centres and 4 cm, 3 cm as radius draw two circles.
- Draw the perpendicular bisectors \(\stackrel{\leftrightarrow}{\mathrm{XY}}\) of AB. Let \(\stackrel{\leftrightarrow}{\mathrm{XY}}\) and AB meet at M.
- Taking M as centre and MA or MB as radius draw a circle which cuts the circle with centre A at P and Q and circle with centre B at R, S.
- Join BP, BQ and AR, AS.

Question 5.

Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.

Answer:

Steps of construction:

- Draw AABC such that AB = 6 cm; ∠B = 90° and BC – 8 cm.
- Drop a perpendicular BD from B on AC.
- Draw the circumcircle to ABCD. Let ‘E’ be its centre.
- Join AE and draw its perpendicular bisector \(\stackrel{\leftrightarrow}{\mathrm{XY}}\). Let it meet AE at M.
- Taking M as centre and MA or ME as radius draw a circle, which’ cuts the circumcircle of △BCD at P and B.
- Join AP and extend AB, which are the required tangents.

Question 6.

Find the area of the shaded region in the figure, given in which two circles with centres A and B touch each other at the point C. If AC = 8 cm. and AB = 3 cm.

Answer:

Given: Two circles with centres A and B, whose radii are 8 cm and 5 cm.

[∵ AC = 8 cm, AB = 3 cm ⇒ BC = 8 – 3 = 5 cm]

Area of the shaded region = (Area of the larger circle) – (Area of the smaller circle)

Question 7.

ABCD is a rectangle with AB = 14 cm. and BC = 7 cm. Taking DC, BC and AD as diameters, three semicircles are drawn as shown in the figure. Find the area of the shaded region.

Answer:

Given AB = 14 cm, AD = BC = 7 cm Area of the shaded and unshaded region

= (2 × Area of the semi-circles with AD as diameter) + Area of the rectangle