<PRE><B>Question 574820</B>
Several ways to find the area. You can construct an altitude from one vertex to the opposite side, use 30-60-90 triangles, et cetera. For simplicity, I would use the area formula


*[tex \LARGE Area = \frac{1}{2}ab \sin C] where C is the included angle between sides a and b. This becomes


*[tex \LARGE Area = \frac{1}{2}x^2 \sin 60^{\circ}]


sin 60 = sqrt(3)/2, so


*[tex \LARGE Area = \frac{\sqrt{3}}{4}x^2]


Perimeter of the same triangle is just x+x+x, or 3x.</PRE>