Tangent to Circle

NCERT Exercise

10.2 Part 6

Question: 12 - A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

Answer: We have; CD = 6 cm and DB = 8 cm and we need to find AC and AB

For this, we need to use the area of triangle ABC (using Heron’s Formula) and equate the area with sum of areas of triangles AOB, COD and COA. This would help in finding the value of AE or AF.

CD = CE = 6 cm

DB = BF = 8 cm

AE = AF = x

(These are tangents respectively from points C, B and A.)

Perimeter 2S = AB + AC + BC = x + 8 + x + 6 + 14 = 2x + 28

Or, S = x + 14

s- a = x + 14 – x – 8 = 6

s – b = x + 14 – x – 6 = 8

s – c = x + 14 – 14 = x

As per Heron’s formula, Area of triangle

=sqrt(s(s-a)(s-b)(s-c))

=sqrt((x+14)xx6xx8xxtext(x))

=sqrt(48x(x+14))

Or, Area2 =48x(x+14)
=48x^2+672x -------(1)

Now; area of following triangles can be given as follows:

Area of triangle AOB = ½ xx AB xx OF

= ½ xx (x + 8) 4 = 2x + 16

Area of triangle AOC = ½ xx AC xx OF

= ½ xx (x + 6) 4 = 2xx + 12

Area of triangle COB = ½ xx BC xx OF

= ½ xx 14 xx 4 = 28

Area of triangle ABC = Area (∆AOB) + Area (∆AOC) + Area (∆COB)

= 2x + 16 + 2x + 12 + 28

= 4x + 56

Or, Area2 = (4x + 56)^2

= 16x^2 + 448x + 3136 ……. (2)

From equations (1) and (2), we get:

16x^2 + 448x + 3136 = 48x^2 + 672x

Or, 16x^2 + 448x + 3136 - 48x^2 - 672x = 0

Or, - 32 x^2 – 224x + 3136 = 0

Or, x^2 + 7x – 98 = 0

Or, x^2 + 14x – 7x – 98 = 0

Or, x(x + 14) – 7(x + 14) = 0

Or, (x + 14)(x – 7) = 0

Hence, x = - 14 and x = 7

Discarding the negative value; we have AE = 7 cm.

Hence; AB = 7 + 8 = 15 cm

And AC = 7 + 6 = 13 cm

Question: 13 - Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer: Construction: Draw a circle with centre O. Draw a quadrilateral ABCD which touches the circle at four points.

To Prove: ∠AOB + ∠DOC = 180°

In Δ AOS and ΔAOP;

AS = AP (tangents from A)

AO = AO (common side)

OS = OP (radii)

Hence; Δ AOS ≅ ΔAOP

Hence; ∠1 = ∠8

Similarly, following can be proved:

∠2 = ∠3

∠4 = ∠5

∠6 = ∠7

Now;

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

Or, (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360°

Or, 2(∠1) + 2(∠2) + 2(∠5) + 2(∠6) = 360°

Or, ∠1 + ∠2 + ∠5 + ∠6 = 180°

Or, (∠1 + ∠2) + (∠5 + ∠6) = 180°

Or, ∠AOB + ∠DOC = 180° Proved

Similarly, ∠AOD + ∠BOC` can be proved