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A regular square pyramid has a height of 6 and a base with sides of length 12.
A) What is the volume of the pyramid?
B) What is the lateral area of the pyramid?
C) What is the total surface are of the pyramid?


Sagot :

Answer:

288, 203.65, and 347.65

Step-by-step explanation:

There's a mess of equations that I won't bother with - find it on G0ogle to verify my answer.

However, for the first one...

The bases' side lengths are 12 each, so 144... then multiply it by 6, so 864... then divide the result by 3, so 288.

Answer:

  A) 288 units³

  B) 144√2 ≈ 203.65 units²

  C) 144(1+√2) ≈ 347.65 units²

Step-by-step explanation:

A)

The volume of a pyramid is given by the formula ...

  V = 1/3Bh

where B is the area of the base, and h is the height. The base is a square with side length 12, so its area is

  A = s² = 12² = 144

Then the volume is ...

  V = 1/3(144)(6) = 288 . . . . cubic units

__

B)

To find the area of one of the triangular faces of the pyramid, we must first find the slant height. The slant height is the hypotenuse of a right triangle with legs equal to the pyramid height and half the base side length. Those are both 6 units, so the slant height is the hypotenuse of an isosceles right triangle with legs that are 6:

  h = √(6² +6²) = 6√2

The area of the four faces is 4 times the area of one triangular face. So, the lateral area is ...

  LA = 4(1/2)(12)(6√2) = 144√2 ≈ 203.65 . . . . square units

__

C)

The total surface area of the pyramid is the sum of the lateral area and the base area:

  SA = 144√2 +144 = 144(1+√2) ≈ 347.65 . . . . square units

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