Solution:

$$\eqalign{ {R_2} = \frac{{abc}}{{4\vartriangle }} \cr \vartriangle = \sqrt {S\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr} $$
$$ = \sqrt {12\left( {\sqrt 5 + 1} \right)\left( {12} \right) \times 12 \times 12\left( {\sqrt 5 - 1} \right)} $$
where a = 12$$\sqrt 5 $$ , b = 12$$\sqrt 5 $$  c = 24
$$\eqalign{ S = \frac{{a + b + c}}{2} \cr S = \frac{{24\sqrt 5 + 24}}{2} \cr S = 12\left( {\sqrt 5 + 1} \right) \cr {R_2} = \frac{{12\sqrt 5 \times 12\sqrt 5 \times 24}}{{4 \times 12 \times 12 \times 2}} \cr {R_2} = \frac{{30}}{2} \cr {R_2} = 15\,{\text{cm}} \cr} $$

Solution:
Two triangles are similar if their corresponding sides are proportional

Solution:
According to question,

Given: BAC is a right angle triangle
AD ⊥ BC
AD = 6 cm
BD = 4 cm
BC = ?
In ΔBAD
$$\eqalign{ AB = \sqrt {B{D^2} + A{D^2}} \cr AB = \sqrt {{4^2} + {6^2}} \cr AB = \sqrt {52} \,cm \cr} $$
ΔBAC ∼ ΔBDA


$$\eqalign{ \therefore \frac{{BC}}{{AB}} = \frac{{AB}}{{BD}} \cr \therefore \frac{{BC}}{{\sqrt {52} }} = \frac{{\sqrt {52} }}{4} \cr BC = \frac{{52}}{4} \cr BC = 13\,cm \cr} $$

Solution:

∠C = 180° - 120° = 60°
∠A + ∠B + ∠C = 180°
∠A + $$\frac{1}{2}$$ ∠A + 60° = 180°
$$\frac{3}{2}$$ ∠A = 120°
∠A = 80°

Solution:
According to question,

Given :
AB = AC
∠C = ∠A = 2x - 20°
∠B = x°
As we know that
∠A + ∠B + ∠C = 180°
(2x - 20)° + x + (2x - 20)° = 180°
5x = 220°
x = 44°
∴ ∠B = 44°

Solution:
According to question,

ABC is a right angle triangle
By using Pythagoras theorem
AC² = BC² + AB²
5² = 3² + 4²
25 = 9 + 16
25 = 25 (satisfied)
∴ Smallest length of right angle triangle is 3 units

Solution:
∠1 + ∠2 = ∠3 [PS is bisector.] - - - - - - (1) ∠Q = 90° - ∠1 ∠R = 90° -∠2 - ∠3 So, ∠Q - ∠R = (90° - ∠1) - (90° - ∠2 - ∠3) ∠Q - ∠R = ∠2 + ∠3 - ∠1 ∠Q - ∠R = ∠2 + (∠1 + ∠2) -∠1[using equation 1] ∠Q - ∠R = 2∠2 $$\frac{1}{2}$$ × (∠Q - ∠R) = ∠TPS