Question

The base of a solid right pyramid is a regular hexagon with

Answer 1

Answer: Last option.

Step-by-step explanation:

Observe the base of the pyramid, which is a regular hexagon. You can see that there is a right triangle.

The area of the triangle can be calculated with this formula:

$A_t=\frac{bh}{2}$

Where "b" is the base and "h" is the height.

You can say that the apothem is the height of the right triangle, then, you need to find the base applying the Pythagorean Theorem:

$(2x)^2=(x\sqrt{3})^2+b^2\\\\b=\sqrt{(2x)^2-(x\sqrt{3})^2} \\\\b=\sqrt{4x^2-3x^2}\\\\b=\sqrt{x^2}\\\\b=x$

Then, the area of the triangle is:

 $A_t=\frac{x(x\sqrt{3})}{2}=\frac{x^2\sqrt{3})}{2}$

Since the base of the pyramid is a regular hexagon, then you can multiply by 12 the area of the right triangle calculated above, in order to find the area of the hexagon. Then you get:

$A_h=12(\frac{x^2\sqrt{3})}{2})=6x^2\sqrt{3}\ units^2$

Answer 2

Answer:

It is 6x^2 ‘3, idk what to use as a replacement for square root but anyways its D ^w^

Step-by-step explanation: