Question

What will be the ratio of the radii of the circumcircle and the in circle of a triangle whose sides are 5, 6 and 7 cm?​

Answer 1

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

Given that, sides of a triangle are 5 cm, 6 cm and 7 cm

Let assume that sides of a triangle as

a = 5 cm

b = 6 cm

c = 7 cm

Let first evaluate the semi Perimeter (s) of the triangle.

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

$\rm \: s = \dfrac{5 + 6 + 7}{2}$

$\rm \: s = \dfrac{18}{2}$

$\rm\implies \:s = 9 \: cm$

Now, we have to find area of triangle.

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

So, on substituting the values, we get

$\rm \: \triangle \: = \: \sqrt{9(9 - 5)(9 - 6)(9 - 7)}$

$\rm \: \triangle \: = \: \sqrt{9(4)(3)(2)}$

$\rm\implies \:\rm \: \triangle \: = \: 6 \sqrt{6} \: {cm}^{2}$

We know,

Radius of incircle, r is given by

$\boxed{\tt{ \: \: \: r \: = \: \frac{\triangle}{s} \: \: \: }}$

So, on substituting the values, we get

$\rm\implies \:r = \dfrac{6 \sqrt{6} }{9} = \dfrac{2 \sqrt{6} }{3} \: cm$

Now, we know,

Circumradius, R is given by

$\boxed{\tt{ \: \: R \: = \: \dfrac{abc}{4\triangle} \: \: \: }}$

So, on substituting the values, we get

$\rm \: R = \dfrac{5 \times 6 \times 7}{4 \times 6 \sqrt{6} }$

$\rm\implies \:R = \dfrac{35}{4 \sqrt{6} } \: cm$

Now, Consider

$\rm \: R \: : \: r$

$\rm \: = \: \dfrac{35}{4 \sqrt{6} } \: : \: \dfrac{2 \sqrt{6} }{3}$

$\rm \: = \: 35 \: : \: 16$

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ADDITIONAL INFORMATION

$\rm \: \dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC} = 2R$

Now,

$\rm\implies \:a = 2R \: sinA$

$\rm\implies \:a = 2R \: \times \: 2 \: sin \dfrac{A}{2} \: cos \dfrac{A}{2}$

$\rm \: a = 4R \times \sqrt{ \dfrac{(s - b)(s - c)}{bc} } \times \sqrt{ \dfrac{s(s - a)}{bc} }$

$\rm \: a = 4R \times \sqrt{ \dfrac{s(s - a)(s - b)(s - c)}{bc \times bc} }$

$\rm\implies \:R = \dfrac{abc}{4\triangle}$