Question

⭐⭐ 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.

❤️USE HERON'S FORMULA❤️​

Answer 1

$\huge{\orange{\underline{\red{\mathbf{SOLUTION}}}}}$

Let ᐃABC be the given triangle circumscribing the given circle with centre 'O' and radius 3 cm.

i.e., the circle touches the sides BC, CA and AB at D, E, F respectively.

it is given that BD = 9cm

CD = 3cm.

∵ Lengths of two tangents drawn from an external point to a circle are equal.

∴ BF = BD = 9 cm

CD = CE = 3cm

AF = AE = x cm say

∴ The sides of the triangle are

12 cm, (9 + x) cm, (3 + x) cm

Perimeter = 2S = 12 + 9 + x + 3 + x

⇒ 2S = 24 + 2x

or S = 12 + x

S - a = 12 + x - 12 = x

S - b = 12 + x - 3 - x = 9

S - c = 12 + x - 9 - x = 3

∴ Area of the triangle

USING HERON'S FORMULA :-

ᐃABC = $\sqrt{s(s - a)(s - b)(s - b)}$

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

=$\sqrt{27( {x }^{2} + 12x) }$ ━━━━▶ ( 1 )

But, ᐃABC = ᐃOBC + ᐃOCA + ᐃOAB

= 1/2 BC×OD + 1/2 × CA×OE + 1/2 AB×OF

= 1/2(12×3)+1/2(3+x)×3+1/2(9+x)×3

= 1/2 [36 + 9 + 3x + 27 + 3x]

= 1/2 [72 + 6x] = 36 + 3x ━━━━▶( 2 )

From ( 1 ) and ( 2 ),

$\sqrt{27( {x}^{2} + 12x)} = \: 36 + 3x$

Squaring on both sides we get,

$27(x^2+12x)=(36+3x)^2$

$27x^2+324x=1296+9x^2+216x$

⇒$18x^2+108x-1296=0$

⇒$x^2+6x-72=0$

⇒$x^2+12x-6x-72=0$

⇒$x(x+12)-6(x+12)=0$

⇒$(x-6)(x+12)=0$

⇒$x=6\:or\:-12$

But 'x' can't be negative hence, x = 6

∴ AB = 9 + 6 = 15 cm

AC = 3 + 6 = 9 cm.