The length of the base edge of a pyramid with a regular hexagon base is represented as x. The height of the pyramid is 3 times longer than the base edge. The height of the pyramid can be represented as_____ . The ____ of an equilateral triangle with length x is units2. The area of the hexagon base is_____ times the area of the equilateral triangle. The volume of the pyramid is_____ x3 units3.1. a 3 b 3+x c 3x d 92 a altitude b apothem c area d volume 3 a 2 b 3 c 4 d 64 a 3/2 b 3 c 6 d 9/2

Answer:(a)h=3x(b)[tex]A=\frac{\sqrt{3} }{4} x^2[/tex](c)[tex]A=\frac{3\sqrt{3} }{2} x^2[/tex](d)[tex]V=\frac{3\sqrt{3} }{2} x^3[/tex] units^3Step-by-step explanation:We are given a regular hexagon pyramidSince, it is regular hexagon so, value of edge of all sides must be sameThe length of the base edge of a pyramid with a regular hexagon base is represented as xso, edge of base =xb=xLet's assume each blank spaces as a , b , c, dwe will find value for each spaces(a)The height of the pyramid is 3 times longer than the base edgeso, height =3*edge of baseheight=3xh=3x(b)Since, it is in units^2so, it is given to find area we know that area of equilateral triangle is [tex]=\frac{\sqrt{3} }{4} b^2[/tex]h=3xb=xnow, we can plug values[tex]A=\frac{\sqrt{3} }{4} x^2[/tex](c)we know that there are six such triangles in the base of hexagonSo, Area of base of hexagon = 6* (area of triangle)Area of base of hexagon is [tex]=6\times \frac{\sqrt{3} }{4} x^2[/tex][tex]=\frac{3\sqrt{3} }{2} x^2[/tex](d)Volume=(1/3)* (Area of hexagon)*(height of pyramid)now, we can plug valuesVolume is [tex]=\frac{1}{3}\times\frac{3\sqrt{3} }{2} x^2\times (3x)[/tex][tex]V=\frac{3\sqrt{3} }{2} x^3[/tex] units^3