Solving For Sides I

Expanding, we get $x^3 - 30x^2 + 281x - 780 = 30$ which becomes $x^3 - 30x^2 + 281x - 810 = 0$ . We know that $r_1, r_2, r_3$ are the three unique solutions to this equation and thus this equation must be equivalent to $(x - r_1)(x - r_2)(x - r_3) = 0$ . Thus we have $x^3 - 30x^2 + 281x - 810 = (x - r_1)(x - r_2)(x - r_3) = 0$ which expands to $x^3 - 30x^2 + 281x - 810 = x^3 - (r_1 + r_2 + r_3)x^2 + (r_1r_2 + r_1r_3 + r_2r_3)x - r_1r_2r_3$ . Setting the coefficients of $x^3, x^2, x, 1$ on both sides equal to each other, we have $r_1 + r_2 + r_3 = 30, r_1r_2 + r_1r_3 + r_2r_3 = 281, r_1r_2r_3 = 810$ .

We know that $A = \sqrt{(s)(s - r_1)(s - r_2)(s - r_3)}$ by Heron's Formula and thus $A^2 = (s)(s - r_1)(s - r_2)(s - r_3)$ where $s = \frac{r_1 + r_2 + r_3}{2}$ . By our above work, we have $s = \frac{r_1 + r_2 + r_3}{2} = \frac{30}{2} = 15$ . So $A^2 = (s)(s - r_1)(s - r_2)(s - r_3) = (15)(15 - r_1)(15 - r_2)(15 - r_3) = (15)(-r_1r_2r_3 +(15)(r_1r_2 + r_1r_3 + r_2r_3) - (225)(r_1 + r_2 + r_3) + 3375 = (15)(-810 + (15)(281) - (225)(30) + 3375) = (15)(-810 + 4215 - 6750 + 3375) = (15)(30) = \boxed{450}$ .

Topics to research: Vieta's Formulas, Heron's Formula

Expanding and simplying gives $x^3 - 30x^2 + 281x - 810 = 0$ . Trial and error shows that $x = 10$ is a solution. Factor it out gives $(x-10)(x^2- 20x+81) = 0$ which yields the other two roots $10 \pm \sqrt{19}$

Let $r_1 = 10 + \sqrt{19}, r_2 = 10 - \sqrt{19}, r_3 = 10$

One of the variations of Heron's formula is $A = \frac {1}{4} \sqrt{ 4a^2 b^2 - (a^2 + b^2 - c^2)^2 }$ , then

$\begin{aligned} A^2 & = & \frac {1}{16} \left ( 4(10^2 - 19)^2 - \left ( (10 + \sqrt{19})^2 + (10 - \sqrt{19})^2 - 10^2 \right )^2 \right ) \\ & = & \frac {1}{16} \left ( 26244 - \left ( 2(10^2 + 19) - 100 \right )^2 \right ) \\ & = & \boxed{450} \\ \end{aligned}$

We need to solve the LHS of the equation to get the values of $x$ .

$(x-5)(x-12)(x-13)=x^{3}-30x^{2}+281x-810$

By using remainder theorem, we will find the value of $x$ .

Let $p(x)=x^{3}-30x^{2}+281x-810=0$

$p(10)=10^{3}-30\times10^{2}+281\times10-810$

$p(10)=1000-3000+2810-810$

$p(10)=3810-3810$

$p(10)=0$

So, one value of $x = 10$

So, $(x-10)$ is a factor of $(x^{3}-30x^{2}+281x-810)$

$(x-10)(x^{2}-20x+81)=0$

So, the two other roots are $10+\sqrt{19}$ and $10-\sqrt{19}$

Now sides of triangle are: $r_{1}=10$ , $r_{2}=10+\sqrt{19}$ and $r_{3}=10-\sqrt{19}$

Now, Area $(A) = \sqrt{s(s-r_{1})(s-r_{2})(s-r_{3})}$ where $s=\frac{r_{1}+r_{2}+r_{3}}{2}$

$s=\frac{10+(10+\sqrt{19})+(10-\sqrt{19})}{2}$

$s=\frac{30}{2}=15$

Now, $A = \sqrt{s(s-r_{1})(s-r_{2})(s-r_{3})}$

$A^{2} = s(s-r_{1})(s-r_{2})(s-r_{3})$

$A^{2} = 15(15-10)(15-(10+\sqrt{19}))(15-(10-\sqrt{19}))$

$A^{2} = 15(15-10)(15-10-\sqrt{19})(15-10+\sqrt{19})$

$A^{2} = 15\times5(5-\sqrt{19})(5+\sqrt{19})$

$A^{2} = 75\times(5^{2}-(\sqrt{19})^{2})$

$A^{2} = 75\times(25-19)$

$A^{2} = 75\times6$

$A^{2} = 450$

So, $A^{2}=450$

Thus, the answer is: $A^{2}=\boxed{450}$

By expanding , we get $x^3-30x^2+281x-810 = 0.$

By trial and error method , $10$ is a root.Therefore $x-10$ is a factor.

By dividing,we get , $(x-10)(x^2-20x+81) = 0$ ,which yields the other two roots $10+\sqrt{19}$ and $10-\sqrt{19}.$

Now , $r_1 = 10 , r_2 = 10+\sqrt{19} , r_3 = 10-\sqrt{19}.$

Now, $A=\sqrt{s(s-r_1)(s-r_2)(s-r_3)}$ , where $s=\dfrac{r_1+r_2+r_3}{2}$

$s=\dfrac{10+(10+\sqrt{19})+(10-\sqrt{19}) }{2} = 15.$

Therefore , $A=\sqrt{15(15-10)(15-(10+\sqrt{19}))(15-(10-\sqrt{19}))}$

$A^2=15(5)(5-\sqrt{19})(5+\sqrt{19})$

$A^2=75(5^2-(\sqrt{19})^2)$

$A^2=75(25-19)=75*6= \boxed{450}$

Therefore the required answer is $\boxed{450} .$