Isosceles trapezoid inscribed in a semicircle

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An isosceles trapezoid $ABCD$ is inscribed in a semicircle of diameter $6$, with the major base $AB$ coinciding with the diameter. The angles adjacent to the major base are $60^\circ$. Determine the perimeter and area of the trapezoid.

Solution

Place $A(-3;0)$, $B(3;0)$, centre $O(0;0)$, radius $3$. Side $AD$ leaves $A$ at $60^\circ$: $D=(-3+t\cos60^\circ;\,t\sin60^\circ)$. Imposing $D$ on the circle $x^2+y^2=9$: $\left(-3+\tfrac t2\right)^2+\tfrac{3t^2}{4}=9\ \Longrightarrow\ t^2-3t=0\ \Longrightarrow\ t=3.$ So the slant side $AD=3$, and $D=\left(-\tfrac32;\,\tfrac{3\sqrt3}{2}\right)$, $C=\left(\tfrac32;\,\tfrac{3\sqrt3}{2}\right)$. The minor base is $DC=3$, the height $h=\dfrac{3\sqrt3}{2}$. Perimeter: $6+3+3+3=15$. Area: $\mathcal A=\dfrac12(6+3)\cdot\dfrac{3\sqrt3}{2}=\dfrac{27\sqrt3}{4}\approx11{,}69$. $\boxed{2p=15,\qquad \mathcal A=\frac{27\sqrt3}{4}\approx11{,}69.}$