Circle With Triangle

The question tells us we want to find the area of the shaded region, which is equal to the area of the triangle $ABC$ minus the area of the circle with center $O$ which inscribed the circle. To do so, we need the following information:
( i ) The area of the circle with centre $O$ ,
( ii ) The area of the triangle $ABC$ .

To find ( i ), we need to know the diameter or radius of this circle. However since this circle is inscribed inside the triangle $ABC$ and there is no other information besides that, let us first find ( ii ).

Recall the identity,

$\text{Area of triangle} = r \times \text{ semiperimeter of triangle} \qquad \qquad (1)$

Where $r$ denotes the inradius of the triangle $ABC$ . In this case, it's the circle with centre $O$ .

Since we are given all the three sides of the triangle already ( $AC=14,AB=13, BC=15$ ), then Heron's formula comes to mind.

$(\text{Area of triangle} )^2 = s(s-a)(s-b)(s-c)$

Where $a,b$ and $c$ denotes the sides of the triangle $ABC$ and $s$ denotes its semiperimeter (half of the perimeter), $s = \dfrac{a+b+c}2 = \dfrac{13+14+15}2 = 21$ . And so,

$(\text{Area of triangle} )^2 = 21(21-13)(21-14)(21-15) \Rightarrow \text{Area of triangle} = 84$

(Note that we only take positive value only because it represents an area)

Substituting into the identity, $(1)$ gives us $84 = r \times 21 \Rightarrow r =4$ .

To summarize, we managed to calculate:
( i ) The area of the circle with centre $O$ to be $\pi r^2 = 16 \pi$ ,
( ii ) The area of the triangle $ABC$ to be $84$ .

Hence the area, $A$ is equal to $84 - 16\pi \approx 33.73$ , and so $\lfloor A \rfloor = \boxed{33}$ .

The area of the green region is equal to the area of the triangle minus the area of the circle. The radius of the inscribed circle can be computed as $r=\dfrac{2A}{P}$ where $A$ and $P$ are the area and perimeter of the triangle, respectively. The area of the triangle can be computed using the Heron's Formula. We have

$s=\dfrac{a+b+c}{2}=\dfrac{13+14+15}{2}=21$

$A=\sqrt{s(s-a)(s-b)(s-c)(s-d)}=\sqrt{21(21-13)(21-14)(21-15)}=84$

$P=a+b+c=13+14+15=42$

So the radius of the inscribed circle is

$r=\dfrac{2(84)}{42}=4$

Finally, the area of the green region is

$A_{GREEN}=84-\pi (4^2) \approx$ $\boxed{33}$