Question

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 9
2 a altitude b apothem c area d volume 
3 a 2 b 3 c 4 d 6
4 a 3/2 b 3 c 6 d 9/2

Answer 1

I got it correct:
The height of the pyramid can be represented as 3x.
The area of an equilateral triangle with length x is x^2 sq.rt.3 / 4 units^2.
The area of the hexagon base is six times the area of the equilateral triangle.
The volume of the pyramid is 3/2 x^3 sq.rt.3 units^3.

Answer 2

Answer:

(a)

h=3x

(b)

$A=\frac{\sqrt{3} }{4} x^2$

(c)

$A=\frac{3\sqrt{3} }{2} x^2$

(d)

$V=\frac{3\sqrt{3} }{2} x^3$ units^3

Step-by-step explanation:

We are given a regular hexagon pyramid

Since, it is regular hexagon

so, value of edge of all sides must be same

The length of the base edge of a pyramid with a regular hexagon base is represented as x

so, edge of base =x

b=x

Let's assume each blank spaces as a , b , c, d

we will find value for each spaces

(a)

The height of the pyramid is 3 times longer than the base edge

so, height =3*edge of base

height=3x

h=3x

(b)

Since, it is in units^2

so, it is given to find area

we know that

area of equilateral triangle is

$=\frac{\sqrt{3} }{4} b^2$

h=3x

b=x

now, we can plug values

$A=\frac{\sqrt{3} }{4} x^2$

(c)

we know that

there are six such triangles in the base of hexagon

So,

Area of base of hexagon = 6* (area of triangle)

Area of base of hexagon is

$=6\times \frac{\sqrt{3} }{4} x^2$

$=\frac{3\sqrt{3} }{2} x^2$

(d)

Volume=(1/3)* (Area of hexagon)*(height of pyramid)

now, we can plug values

Volume is

$=\frac{1}{3}\times\frac{3\sqrt{3} }{2} x^2\times (3x)$

$V=\frac{3\sqrt{3} }{2} x^3$ units^3