SOLUTION:
\((A)\) Since we know the bottom left and bottom right angles of this trapezoid, we can construct two different triangles with angles \(60 ^{\circ}\) and \(90 ^{\circ}\). The trianges have angles \(90 ^{\circ}\)and \(60 ^{\circ}\) and since the angles of a triangle sum up to \(180 ^{\circ}\), it is a " \(90 ^{\circ}, 60 ^{\circ}, 30^{\circ}\) triange". " \(90 ^{\circ}, 60 ^{\circ}, 30^{\circ}\) trianges" have the following properties:
Hypotenuse = \(2x\)
Larger Triange Side = \(x\sqrt{3}\)
Smaller Triange Side = \(x\)
In the trapezoid, the larger triangle side equals the altitude thus the altidute equals \( \sqrt{3}\) since the hypotenuse equals \(2\) and therefore \((2x = 2), x = 1\). Since the bottom base of the trapezoid merely equals \(x + x + 2 = 4\) and we already have the top base, we can find the perimeter. The perimeter equals the sum of both bases and the two other sides of the trapezoid thus the perimeter equals \(4 + 2 + 2 + 2 = \boxed{10}\)
\((B)\) As you have most likely realized, the altitude isn't utilized whatsoever in the first problem, however, it is needed in order to compute the area. The formula of the area of a trapezoid is \((\frac{a+b}{2})h\)(\(a\) and \(b\) are the bases of a trapezoid and \(h\) is the altitude). Since we already know the bases and altitude of this specific trapezoid, we can quite feasibly compute the area. \((\frac{a+b}{2})h = (\frac{4 + 2}{2})\sqrt{3} = (\frac{6}{2})\sqrt{3} = \boxed{3\sqrt{3}}\)
RB - ∃
