Question

A triangle ABC is drawn to circumscribe a circle of radius 3 cm. such that the segments BD and DC into which BC is divided by the point of contact D are of length 9 cm. and 3 cm. respectively Find the sides AB and AC.

Answer 1

HELLO DEAR,

IN ∆ ABC,

CE = CD = 3cm  (Tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point C)

BF = BD = 9cm  (Tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point B)

AE = AF = x  (Again, tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point A)

Now, AB = AF + FB 

AB = (x + 9)cm

BC = BD + CD

BC = (9 + 3) = 12cm

AC = AE + EC

AC = (x + 3)cm

semi-perimeter of triangle ABC = sum of all sides)/2

s = (x + 9 + 12 + x + 3)/2

s = (2x + 24)/2

s = (x + 12)cm

using Heron's formula for calculating area of ∆ ABC,

area ∆ ABC = $\sf{\sqrt{s(s - a)(s - b)(s - c)}}$

$\sf{\Rightarrow \sqrt{(x + 12)(x + 12 - x - 9)(x + 12 - 12)(x + 12 - x - 3)}}$

$\sf{\Rightarrow \sqrt{(x + 12)(3)(x)(9)}}$

$\sf{\Rightarrow \sqrt{(x + 12)*27x}}$

$\sf{\Rightarrow \triangle ABC = \sqrt{27x^2 + 324x}}$-------------( 1 )

Also,

OF = OE = OD = 3CM (radius of circle)

NOW,

area ∆ABC = area∆AOC + area∆AOB + area∆BOC

∆ABC = (1/2 × OE × AC) + (1/2 × OF × AB) + (1/2 × OD × BC)

∆ABC = 1/2 × 3{x + 3 + x + 9 + 12}

∆ABC = 1/2 × 3{2x + 24}

∆ABC = (3x + 36)------( 2 )

from-----------( 1 ) & ---------------( 2 )

$\sf{\therefore , \Rightarrow \sqrt{27x^2 + 324x} = 3x + 36}$

on squaring both side;

$\sf{\Rightarrow 27x^2 + 324x = 9x^2 + 1296 + 216x}$

$\sf{\Rightarrow 27x^2 - 9x^2 + 324x - 216x - 1296 = 0}$

$\sf{\Rightarrow 18x^2 + 108 - 1296 = 0}$

$\sf{\Rightarrow x^2 + 6x - 72 = 0}$

$\sf{\Rightarrow x^2 + 12x - 6x - 72 = 0}$

$\sf{\Rightarrow x(x + 12) - 6(x + 12) = 0}$

$\sf{\Rightarrow (x - 6)(x + 12) = 0}$

x - 6 = 0 , x + 12 = 0

x = 6 , x = -12{neglect} because, side of triangle can't be negative

hence, AB = (x + 9)cm = (6 + 9)cm = 15cm

AC = (x + 3)cm = (6 + 3)cm = 9cm.

I HOPE ITS HELP YOU DEAR,

THANKS

Answer 2

Construction : Draw radius OB = 3cm

Proof : AC, BC and AB are tangents
BF = BD = 9cm—tangents from B
AF = AB—tangent from A
∴ CD = CB tangents from C
OD = OB = 3cm radius
OBCD forms a square of side 3cm
OD = OB = BC = CD = 3cm
∴ ∠BCD = 90°
BD = 9cm and DC = 3cm
OB = 3cm radius of circle
Let AF = AB = x
Applying Pythagoras
AB = BC + AC
(9 + x)² = (12)² + (3 + x)²
81 + 18x + x² = 144 + 9 + 6x + x²
12x = 72
x = 6cm
∴ AB = 9 + 6 = 15cm
AC = 6 + 3 = 9cm