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A right pyramid with a square base has a base edge length of 12 meters and slant height of 6 meters.

The apothem is meters.

The hypotenuse of ΔABC is the .

The height is meters.

The volume of the pyramid is cubic meters.

A right pyramid with a square base has a base edge length of 12 meters and slant height of 6 meters The apothem is meters The hypotenuse of ΔABC is the The heig class=

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Answer:

Step-by-step explanation:

Square base of the base edge = 12 meters and slant height = 6√2 meters

Apothem of the right pyramid will be = AC

1). Now AC = [tex]\sqrt{AB^{2}-BC^{2}}[/tex]

AC = [tex]\sqrt{(6\sqrt{2})^{2}-(6)^{2}}[/tex]

     = [tex]\sqrt{72-36}=\sqrt{36}[/tex]

     = 6

Now Apothem = 6 meters

2). Hypotenuse of Δ ABC = AB = 6√2 meters

3). Height AC = 6 meters

4). Volume of the pyramid = [tex]\frac{1}{3}(\text{Area of the base})(\text{Apothem})[/tex]

Volume = [tex]\frac{1}{3}(12)^{2}(6)[/tex]

             = 2×144

             = 288 meter²

Correct responses:

  • The apothem is 6 meters
  • The hypotenuse of ΔABC is 6·√2 meters
  • The height is 6 meter
  • The volume of the pyramid is 288 cubic meters

Methods used for the calculations

The given dimensions of the right pyramid having a square base are;

Base edge length, l = 12 meters

Slant height = 6·√3

Required:

Length of the apothem.

Solution:

The apothem, a, is the line drawn from the middle of the polygon to the midpoint of a side.

Therefore;

The apothem of the square base = [tex]\dfrac{l}{2} [/tex] = [tex]\overline{BC}[/tex]

Which gives;

[tex]a = \dfrac{12}{2} = 6[/tex]

  • The apothem is 6 meters

Required:

The hypotenuse of triangle ΔABC

Solution:

The hypotenuse of ΔABC = The slant height of the square pyramid = 6·√2 meters

Therefore;

  • The hypotenuse of ΔABC = 6·√2 meters

Required:

The height of the pyramid

Solution:

The height of the pyramid = The length of the side [tex]\overline{AC}[/tex] in right triangle

ΔABC, therefore, by Pythagorean theorem, we have;

[tex]\overline{AC}^2[/tex] = [tex]\overline{AB}^2[/tex] - [tex]\overline{BC}^2[/tex]

Which gives;

[tex]\overline{AC}^2[/tex] = (6·√2)² - 6² = 6²·((√2)² - 1) = 6² × 1  = 6²

[tex]\overline{AC}[/tex] = √(6²) = 6

  • The height of the pyramid, h = [tex]\overline{AC}[/tex] = 6 meters

Required:

The volume of the pyramid.

Solution:

[tex]The \ volume \ of \ a \ pyramid \ is, \ V = \mathbf{\dfrac{1}{3} \times Base \ area \times Height}[/tex]

[tex]V = \dfrac{1}{3} \times A \times h [/tex]

The base area of the square pyramid, A = 12 m × 12 m = 144 m²

Therefore;

[tex]V = \dfrac{1}{3} \times 144\, m^2 \times 6 \, m = 288 \, m^2[/tex]

  • The volume of the pyramid is V = 288 cubic meters

Learn more about the volume of solids here:

https://brainly.com/question/13338580

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