Irrational Triangle

Relevant wiki: Pythagorean Theorem

Notice that the numbers in the roots can be written as the sum of two numbers. We can then put the triangle $ABC$ into a $4\times 4$ square as shown below:

By Pythagorean Theorem , we can evaluate the triangle's sides:

$AB^2 = 13 = 3^2 + 2^2$

$BC^2 =17 = 1^2 + 4^2$

$CA^2 = 20 = 4^2 + 2^2$

From the image, we can see that $AB$ is the hypotenuse of the right triangle with other sides of $2$ & $3$ ; $BC$ with other sides of $1$ & $4$ ; and $CA$ with other sides of $2$ & $4$ .

Thus, the area of triangle $ABC$ = $(4\cdot 4) -\frac{1}{2}(2\cdot 3 + 1\cdot 4 + 2\cdot 4) = \boxed{7}$ .

By cosine rule , we have:

$\begin{aligned} 20 & = 13 + 17 - 2\sqrt{13 \cdot 17} \cos B \\ \implies \cos B & = \frac {10}{2\sqrt{13 \cdot 17}} = \frac 5{\sqrt {221}} \\ \sin B & = \sqrt{1-\cos^2 B} = \sqrt {1-\frac {25}{221}} = \sqrt{\frac {196}{221}} = \frac {14}{\sqrt{221}} \end{aligned}$

The area of $\triangle ABC$ is given by $A = \dfrac {ac \sin B}2 = \dfrac {\sqrt{13}\cdot \sqrt{17}}2 \cdot \dfrac {14}{\sqrt{221}} = \boxed{7}$

Trace the triangle height from vertex B to H. Denote BH and AH by $h$ and $x$ , respectively. Then, by Pythagorean Theorem, the following system holds: $13=h^2+x^2$ $17=h^2+(\sqrt{20} -x)^2$ which gives (squared terms disappear) $h=7\times \sqrt5 : 5$ and hence area is equal to $\sqrt{20} \times 7\sqrt5:5:2 =7$ .

Using Heron's formula , Area

$= \sqrt{\left(\dfrac{a + b + c}{2}\right)\left(\dfrac{a + b - c}{2}\right)\left(\dfrac{a - b + c}{2}\right)\left(\dfrac{-a + b + c}{2}\right)}$

$= \dfrac{1}{4} \sqrt{\{(a + b)^2 - c^2\}\{c^2 - (a - b})^2\}$

$=\dfrac{1}{4} \sqrt{(a^2 + b^2 + 2ab - c^2)(c^2 - a^2 - b^2 + 2ab)}$

$= \dfrac{\sqrt{4a^2b^2 - (a^2 + b^2 - c^2)}}{4}$

Putting $a = 13, b = 17, c = 20$

Area $=\boxed{7}$

Sum of sides / 2 = 6.1 = S

A = sqrt(S (S-A) (S-B)*(S-C)) = sqrt(6.1X2.49X1.98X1.63)) = 7.001