HELPP! (pic)

Since we have two different answers I might as well do it as well

$$\\x=\sqrt{(\sqrt{15}-\sqrt{3})^2+(\sqrt{15}-\sqrt{3})^2}\\\\
x=\sqrt{2*(15-2\sqrt{45}+3)}\\\\
x=\sqrt{2*(18-6\sqrt{5})}\\\\
x=\sqrt{36-12\sqrt{5}}\\\\$$

$${\sqrt{{\mathtt{36}}{\mathtt{\,-\,}}{\mathtt{12}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}}} = {\mathtt{3.027\: \!735\: \!832\: \!268\: \!483\: \!3}}$$

check

$${\sqrt{{\mathtt{2}}{\mathtt{\,\times\,}}{\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}}} = {\mathtt{3.027\: \!735\: \!832\: \!268\: \!483\: \!3}}$$

Seems anon is correct on this one.           Sorry Chris.   

The only way that x = 6 is if the other two sides are √15 - √3  and √15 + √3

Proof

√[(√15 - √3)^2 + (√15 + √3)^2 ] =

√(15 - 2√45 + 3) + (15 +2√45 + 3) =

√(30 + 6) = √36 = 6

Assuming the above is true....the area =

(1/2)[√15 - √3][√15 +√3] =

(1/2)[15 - 3] = (1/2)(12) = 6 sq units

And the perimeter is just [√15 - √3] + [√15 +√3] + 6  =  6 + 2√15

To find that x = 6, use the pythagorean theorm, $${{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = {{\mathtt{c}}}^{{\mathtt{2}}}$$

$${\mathtt{a}} = {\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}$$, $${\mathtt{b}} = {\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}$$

$${\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}} = {{\mathtt{c}}}^{{\mathtt{2}}}$$ replace $${{\mathtt{c}}}^{{\mathtt{2}}}$$ with $${{\mathtt{x}}}^{{\mathtt{2}}}$$

$${\sqrt{{\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}}} = {\mathtt{3.027\: \!735\: \!832\: \!268\: \!483\: \!3}}$$

The answer x = 6 is not correct.

To find the area of the triangle us the formula for area of a triangle, $${\mathtt{a}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{l}}{\mathtt{\,\times\,}}{\mathtt{w}}$$

$${\mathtt{l}} = {\mathtt{length}}$$, $${\mathtt{w}} = {\mathtt{width}}$$

$${\mathtt{a}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)$$

$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right) = {\mathtt{2.291\: \!796\: \!067\: \!500\: \!630\: \!9}}$$

The area of the triangle is approximately 2.30

To find the perimeter of the triangle, add up all the sides

$${\mathtt{p}} = \left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}{\sqrt{{\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}}}$$

$$\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}{\sqrt{{\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}}} = {\mathtt{7.309\: \!600\: \!909\: \!545\: \!562\: \!5}}$$

The perimeter of the triangle is approximately 7.31