Answer:
Formula for Sine ratio:
[tex]\sin \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}[/tex]
As per the statement:
labelled the diagram as shown below
The base of a regular pyramid is a hexagon as shown in the diagram.
Let s be the side of the regular hexagon.
Using sine ratio on triangle ABC
[tex]\sin 60^{\circ} = \frac{a}{14}[/tex]
⇒[tex]\frac{\sqrt{3}}{2}= \frac{a}{14}[/tex]
Multiply both sides by 14 we have;
⇒[tex]7\sqrt{3} = a[/tex]
or
[tex]a = 7\sqrt{3}[/tex] cm
Using Pythagoras theorem.
[tex]\text{Hypotenuse side}^2 = \text{opposite side}^2+\text{Adjacent side}^2[/tex]
Apply the Pythagoras theorem to ABC:
[tex]14^2 = (7\sqrt{3})^2+(BC)^2[/tex]
⇒[tex]196 = 147+BC^2[/tex]
⇒[tex]49 = BC^2[/tex]
Simplify:
[tex]BC = \sqrt{49} = 7[/tex]
AD = 2 BC
⇒[tex]s = 2 \cdot 7 = 14 cm[/tex]
Area of the regular hexagon(A) is given by:
[tex]A =\frac{3\sqrt{3}}{2}s^2[/tex]
where, s is the side of the regular hexagon.
Substitute the given values we have;
[tex]A=\frac{3\sqrt{3}}{2} \cdot 14 \cdot 14 = 3\sqrt{3} \cdot 7 \cdot 14= 294\sqrt{3} cm^2[/tex]
Therefore, the area of the base of the pyramid is, [tex]294\sqrt{3} cm^2[/tex]
