Question

Q19⟩⟩ In the given figure, a ∆ABC is drawn to circumscribe a circle of radius 4cm such that the segments BD and DC are of lengths 8cm and 6cm respectively. Find the lengths of sides AB and AC.​

Answer 1

$\large\underline{\sf{Solution-}}$

★ Let assume that the incircle of radius 4 cm touches the sides of triangle BC, AB, AC at D, E, F respectively.

★ Construction :- Join OE, OF, OA, OB, OC

★ Now, OD = OE = OF = r = 4 cm

Now,

We know that,

★ Radius is perpendicular to tangent.

• ★ OD is perpendicular to BC

• ★ OE is perpendicular to AB

• ★ OF is perpendicular to AC.

According to statement,

• CD = 6 cm

• DB = 8 cm

We know,

★ Length of tangents drawn from external point are equal,

Therefore,

• CD = CF = 6 cm

• BD = BE = 8 cm.

Let assume that

• AF = x cm

So,

• AE = AF = x cm

Hence,

★ Sides of triangle ABC are,

• AC = 6 + x cm

• BC = 14 cm

• AB = 8 + x cm

★ We evaluate area of triangle ABC by Heron's Formula,

Let assume that sides of triangle ABC be represented as

• c = AB = 8 + x cm

• a = BC = 14 cm

• b = AC = 6 + x cm

We know,

Semi - Perimeter of a triangle (s) is

$\bf :\longmapsto\:s = \dfrac{a + b + c}{2}$

$\rm :\longmapsto\:s = \dfrac{14 + 6 + x + 8 + x }{2}$

$\rm :\longmapsto\:s = \dfrac{28+ 2x}{2}$

$\bf\implies \:s = 14 + x$

Now,

★ Area of triangle ABC is

$\rm :\longmapsto\:Area_{(\triangle ABC)} = \sqrt{s(s - a)(s - b)(s - c)}$

$\rm \: = \: \: \: \sqrt{(14 + x)(14 + x - 6 - x)(14 + x - 8 - x)(14 + x - 14)}$

$\rm \: = \: \: \: \sqrt{(14 + x)(6)(8)(x)}$

$\bf\implies \:Area_{(\triangle ABC)} = \sqrt{x(x + 14)(48)} - - (1)$

Again,

$\rm :\longmapsto\:Area_{(\triangle ABC)} = Area_{(\triangle AOB)} + Area_{(\triangle BOC)} + Area_{(\triangle AOC)}$

$\rm \: = \: \: \:\dfrac{1}{2}(r)AB + \dfrac{1}{2}(r)BC + \dfrac{1}{2}(r)CA$

$\rm \: = \: \: \:\dfrac{1}{2}(r)\bigg(AB + BC + CA\bigg)$

★ Now, on substituting the values of r, AB, BC, CA, we get

$\rm \: = \: \: \:\dfrac{1}{2}(4)\bigg(8 + x + 14 + 6 + x\bigg)$

$\rm \: = \: \: \:2(28 + 2x)$

$\rm \: = \: \: \:4(14 + x)$

$\bf\implies \:Area_{(\triangle ABC)} = 4(14 + x) - - (2)$

★ On equating equation (1) and equation (2), we get

$\rm :\longmapsto\: \sqrt{x(x + 14)(48)} = 4(14 + x)$

★ On squaring both sides, we get

$\rm :\longmapsto\: x(x + 14)(48)= 16(14 + x)^{2}$

$\rm :\longmapsto\:3x = x + 14$

$\rm :\longmapsto\:3x - x = 14$

$\rm :\longmapsto\:2x = 14$

$\bf\implies \:x = 7$

Hence,

$\bf :\longmapsto\:AB = 8 + x = 8 + 7 = 15 \: cm$

$\bf :\longmapsto\:AC = 6 + x = 6 + 7 = 13 \: cm$