Question

The radius of the in circle of a triangle is 4cm and segment into which one side is divided by the point of contact are 6cm and 8 cm. Determine the other two sides of the triangle.​

Answer 1

$\huge \boxed{ \bf{answer}}$

$\rightarrow \bold{to \: find}$

let there is a circle having Centre or touches the side"AB" and "AC" of the triangle at the point E and F respectively .

Now,

let the length of the line segment AE be "x" .

$\rightarrow \bold{in \: triange \: {abc}}$

CF = CD = 6 ( tangnets on the circle from point C )

BE = BD = 6 ( tangnets on the circle from point B )

AE = AF = X ( tangents on the circle from the point A )

$\bold{now}$

AB =AE +EB

=> AB = X + 8

BC = BD + DC

=> BC =8+6=14

CA = CF + FA

=> CA = 6+X

$\bold{then}$

$s = \frac{ab + bc + ca}{2}$

$s = \frac{x + 8 + 14 + 6 + x}{2}$

$s = \frac{2x + 28}{2}$

$s = x + 14$

$\bold{area \: of \: triangle \: abc}$

$\sqrt{s} \times( (s - a)(s - b)(s - c))$

${ \sqrt{14} ((14 + x) - 14) \times ((14 + x) - (6 + x)) \times ((14 + x) - (8 + x))}$

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

$= (\sqrt{14} + x) \times x \times 2 \times 4 \times 2 \times 3$

$\bold{area \: of \: the \: triangle \: abc}$

$= 4 \sqrt{3x} (14 + x)....(1)$

Now area of ∆OBC

$\frac{1}{2} \times od \times bc$

$\frac{1}{2} \times 4 \times 14$

$\frac{56}{2}$

$= 28$

After This Other solution are in attachment .

Hope it helps u