Given: ΔABC that is drawn to circumscribe a circle with radius r = 4 cm and BD = 6 cm DC = 8 cm

To Find: AB and AC

Now,

As we know tangents drawn from an external point to a circle are equal.

Then,

FB = BD = 6 cm [Tangents from same external point B]

DC = EC = 8 cm [Tangents from same external point C]

AF = EA = x (let) [Tangents from same external point A]

Using the above data we get

AB = AF + FB = x + 6 cm

AC = AE + EC = x + 8 cm

BC = BD + DC = 6 + 8 = 14 cm

Now we have heron's formula for area of triangles if its three sides a, b and c are given

ar = √(s(s – a)(s – b)(s – c))

Where,

So for ΔABC

a = AB = x + 6

b = AC = x + 8

c = BC = 14 cm

And

ar(ΔABC) = √((x + 14)(x + 14 – (x + 6))(x + 14 – (x + 8))(x + 14 – 14))

= √((x + 14)(8)(6)(x)) [1]

ar(ABC) = ar(AOB) + ar(BOC) + ar(AOC)

at, tangent at a point on the circle is perpendicular to the radius through point of contact,

So, we have

OF ⏊ AB, OE ⏊ AC and OD ⏊ BC

Therefore, AOB, BOC and AOC are right – angled triangles.

And area of right angled triangle =1/2 × Base × Height

Using the formula,

Using [1] we have,

Squaring both side

⇒ 48x(x + 14) = (2x + 6 + 28 + 2x + 16)²

⇒ 48x² + 672x = (56 + 4x)²

⇒ 48x² + 672x = (4(14 + x))²

⇒ 48x² + 672x = 16(196 + x² + 28x)

⇒ 3x² + 42x = 196 + x² + 28x

⇒ 2x² + 14x – 196 = 0

⇒ x² + 7x – 98= 0

⇒ x² + 14x – 7x – 98 = 0

⇒ x(x + 14) – 7(x + 14) = 0

⇒ (x – 7)(x + 14) = 0

⇒ x = 7 or x = – 14 cm

Negative value of x is not possible, as length can't be negative

Therefore,

x = 7 cm

⇒ AB = x + 6 = 7 + 6 = 13 cm

⇒ AC = x + 8 = 7 + 8 = 15 cm