Question

2.b > 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 from points A and B

Let's draw $AE\perp CD$ and $BF\perp CD$ at points E and F respectively. Since $AB\parallel CD$, $EF = AB = 8$ cm.

Step2: Analyze right - triangle BCF

In right - triangle $BCF$, $\angle C=60^{\circ}$, $BC = 8$ cm. We know that $\cos\angle C=\frac{CF}{BC}$ and $\sin\angle C=\frac{BF}{BC}$. So, $CF=\cos60^{\circ}\times BC=\frac{1}{2}\times8 = 4$ cm, and $BF=\sin60^{\circ}\times BC=\frac{\sqrt{3}}{2}\times8 = 4\sqrt{3}$ cm.

Step3: Find the length of CD

Since the trapezoid is isosceles (because $\angle C=\angle D = 60^{\circ}$), $DE=CF = 4$ cm. Then $CD=DE + EF+CF=4 + 8+4=16$ 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, $a = AB = 8$ cm, $b = CD = 16$ cm, and $h = BF = 4\sqrt{3}$ cm. So, $A=\frac{(8 + 16)\times4\sqrt{3}}{2}=48\sqrt{3}$ $cm^{2}$.

Answer:

$48\sqrt{3}$