Such Triangle

Imgur

Rotate $\triangle ABC$ about point $C$ to get $\triangle BB'C$ .

Point $P$ is also mapped to point $P'$ .

Connect $PP'$ ; note that $CP=CP'$ . Also, $\angle PCP'=90^{\circ}$

Thus, $PP'=\sqrt{6^2+6^2}=6\sqrt{2}$ .

Now, notice that $\triangle BPP'$ is right, in particular, $\angle BPP'=90^{\circ}$ because $(6\sqrt{2})^2+(7)^2=(11)^2$ .

Thus, $\angle BPC=45^{\circ}+90^{\circ}=135^{\circ}$ .

Now using LoC on $\triangle BPC$ wrt $\angle BPC$ , we see that $s^2=6^2+7^2-2\cdot 6\cdot 7\cdot \cos 135^{\circ}=85+42\sqrt{2}$ .

Thus, $s=\sqrt{85+42\sqrt{2}}$ and our answer is $85+42=\boxed{127}$ .

Let B, A & P be (a,0), (0,a) & (x,y) resply. Then (a-x)^2+y^2=49, x^2+(y-a)^2=121, x^2+y^2=36 which can be solved simultaneously to obtain a=(85+42(2)^(1/2))^(1/2) or a+b =85 +42 =127

Just putting One Top 's solution in better way.

---

Let point $C,A,B,P$ be $(0,0),(0,a),(a,0),(x,y)$ respectively.

Now we have following equations:

• $x^2+y^2=6^2$
• $(x-a)^2+y^2=7^2$
• $x^2+(y-a)^2=11^2$

Starting with 2nd equation, $x^2-2ax+a^2+y^2=49\Rightarrow a^2-13=2ax\\\Rightarrow x=\dfrac{a^2-13}{2a}$

Moving to 3rd equation, $x^2+y^2-2ay+a^2=121\Rightarrow a^2-85=2ay\\\Rightarrow y=\dfrac{a^2-85}{2a}$

Substituting $x$ and $y$ in first equation,

$\dfrac{(a^2-13)^2+(a^2-85)^2}{4a^2}=36$

Bash bash bash

$a^4-170a^2+3697=0\Rightarrow a^2=85\pm\sqrt{3528}=85\pm 42\sqrt{2}$

$85+42=\boxed{127}$