joshfowler
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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

Respuesta :

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^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

[tex]=\frac{\sqrt{3} }{4} b^2[/tex]

h=3x

b=x

now, 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 hexagon

So,

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 values

Volume 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


MrRoyal

The true statements are:

  • The height of the pyramid can be represented as 3x.
  • The area of an equilateral triangle with length x is [tex]\frac{\sqrt 3}{4}x^2[/tex] units^2.
  • The area of the hexagon base is six times the area of the equilateral triangle.
  • The volume of the pyramid is [tex] \frac{3\sqrt 3}{2}x^3[/tex] units^3.

The length of base edge is given as:

[tex]Length=x[/tex]

The height is three times the length.

So, the expression that represents the height is:

[tex]Height =3x[/tex]

The area of an equilateral triangle is then calculated as:

[tex]A_t = \frac{\sqrt 3}{4}l^2[/tex]

Where l represents the side length of the triangle.

So, we have:

[tex]A_t = \frac{\sqrt 3}{4}x^2[/tex]

The area of the hexagonal base is 6 times the area of the triangle.

This is so because, an hexagon has 6 sides

So, we have:

[tex]A_h = \frac{\sqrt 3}{4}x^2 \times 6[/tex]

[tex]A_h = \frac{3\sqrt 3}{2}x^2 [/tex]

Lastly, the volume of the pyramid is then calculated as:

[tex]V = \frac 13 \times A_h \times h[/tex]

This gives

[tex]V = \frac 13 \times \frac{3\sqrt 3}{2}x^2 \times 3x[/tex]

So, we have:

[tex]V = \frac{3\sqrt 3}{2}x^2 \times x[/tex]

[tex]V = \frac{3\sqrt 3}{2}x^3[/tex]

Read more about volumes at:

https://brainly.com/question/16597042

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