Question

what is the area of trapezoid abcd with bases $overline{ab}$ and $overline{cd}$ if $mangle c = mangle d=60^{circ}, ab = bc = 8$ cm. answer: $a=square$ $cm^{2}$

Explanation:

Step1: Draw heights

Draw two heights from points A and B perpendicular to CD, let the feet of the perpendiculars be E and F respectively. Since ABCD is a trapezoid with \(AB\parallel CD\), \(AE = BF\) and \(EF=AB = 8\mathrm{cm}\).

Step2: Analyze right - triangles

In right - triangle BCF, \(\angle C = 60^{\circ}\), \(BC = 8\mathrm{cm}\). We know that \(\cos C=\frac{CF}{BC}\) and \(\sin C=\frac{BF}{BC}\). Since \(\cos60^{\circ}=\frac{1}{2}\), then \(CF=\frac{1}{2}BC = 4\mathrm{cm}\), and since \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\), then \(BF = 4\sqrt{3}\mathrm{cm}\). Similarly, in right - triangle ADE, \(DE = 4\mathrm{cm}\).

Step3: Calculate the length of the lower base

\(CD=DE + EF+FC\). Substituting the values we found, \(CD=4 + 8+4=16\mathrm{cm}\).

Step4: Calculate the area of the trapezoid

The area formula of a trapezoid is \(A=\frac{(a + b)h}{2}\), where \(a\) and \(b\) are the lengths of the bases and \(h\) is the height. Here, \(a = 8\mathrm{cm}\), \(b = 16\mathrm{cm}\), \(h = 4\sqrt{3}\mathrm{cm}\). So \(A=\frac{(8 + 16)\times4\sqrt{3}}{2}=48\sqrt{3}\mathrm{cm}^{2}\).

Answer:

\(48\sqrt{3}\)