Question

Q32. In the given figure, if radius of circle is 3cm. Find the perimeter of ∆ABC.​

Answer 1

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

★ Given that

• Radius of in-circle, r = 3 cm

★ Let assume that incircle of radius r touches BC, AB, AC at D, E, F respectively.

We know,

★ Radius and tangent are perpendicular to each other.

So,

OD, OE, OF are perpendiculars to BC, BA, AC respectively.

Now,

It is given that

$\rm :\longmapsto\:OC = OB = 3 \sqrt{5} \: cm$

Now,

$\bf :\longmapsto\:In \: \triangle \: ODC$

★ Using Pythagoras Theorem,

$\rm :\longmapsto\: {OC}^{2} = {OD}^{2} + {DC}^{2}$

$\rm :\longmapsto\: {(3 \sqrt{5} )}^{2} = {3}^{2} + {DC}^{2}$

$\rm :\longmapsto\: 45 = 9 + {DC}^{2}$

$\rm :\longmapsto\: 45 - 9 = {DC}^{2}$

$\rm :\longmapsto\: 36 = {DC}^{2}$

$\bf\implies \:DC = 6 \: cm$

We know,

★ Length of tangents drawn from external point are equal.

$\bf\implies \:DC = CF = 6 \: cm$

Similarly,

$\bf\implies \:DB = BE = 6 \: cm$

Now,

★ Let assume that AF = x cm

• So,

• AF = AE = x cm.

So,

★ Dimensions of triangle ABC are as

• AB = 6 + x cm

• BC = 12 cm

• CA = 6 + x cm

Now, we evaluate area of triangle ABC,

$\bf :\longmapsto\:Ar_{(\triangle ABC)}$

$\rm \: \: = \: Ar_{(\triangle AOC)} + Ar_{(\triangle AOB)} + Ar_{(\triangle BOC)}$

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

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

$\rm \: \: = \: \dfrac{1}{2}(3)(6 + x + 6 + x + 12)$

$\rm \: \: = \: \dfrac{1}{2}(3)(24 + 2x)$

$\rm \: \: = \: \dfrac{1}{2}(3)(2)(12 + x)$

$\rm \: \: = \: 3(12 + x)$

$\bf\implies \:Ar_{(\triangle ABC)} = 3(12 + x) - - - (1)$

Again,

Evaluate the area of triangle ABC using Heron's Formula,

★ Dimensions of triangle ABC are as

• c = AB = 6 + x cm

• a = BC = 12 cm

• b = CA = 6 + x cm

★ Semi-perimeter (s) of triangle ABC is

$\bf :\longmapsto\:s = \dfrac{1}{2}(a + b + c)$

$\rm \: \: = \: \dfrac{1}{2}(6 + x + 12 + 6 + x)$

$\rm \: \: = \: \dfrac{1}{2}(24 +2x)$

$\rm \: \: = \: 12 + x$

$\bf\implies \:s = 12 + x$

Now,

★ We know area of triangle ABC using Heron's formula is

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

$\rm \: \: = \: \sqrt{(12 + x)(12 + x - 6 - x)(12 + x - 6 - x)(12 + x - 12)}$

$\rm \: \: = \: \sqrt{(12 + x)(6)(6)x}$

$\rm \: \: = \: \sqrt{36x(12 + x)}$

$\bf\implies \:Ar_{(\triangle ABC)} = \sqrt{36x(12 + x)} - - - (2)$

From equation (1) and (2), we have

$\rm :\longmapsto\: \sqrt{36x(12 + x)} = 3(12 + x)$

On squaring both sides, we get

$\rm :\longmapsto\:36x(12 + x) = 9 {(12 + x)}^{2}$

$\rm :\longmapsto\:4x = 12 + x$

$\rm :\longmapsto\:4x - x = 12$

$\rm :\longmapsto\:3x = 12$

$\bf\implies \:x = 4 \: cm$

So,

★ Dimensions of triangle ABC are as

• AB = 6 + x = 6 + 4 = 10 cm

• BC = 12 cm

• CA = 6 + x = 6 + 4 =10 cm

Hence,

$\bf :\longmapsto\:Perimeter_{(\triangle ABC)} = AB + BC + CA$

$\rm :\longmapsto\:Perimeter_{(\triangle ABC)} = 10 + 12 + 10$

$\bf :\longmapsto\:Perimeter_{(\triangle ABC)} = 32 \: cm$