find the surface area of a reg. triangular pyramid

Question

w/ base edge length 6ft & slant height 10 ft

Answer 1

We never find the surface area of the regular triangular pyramid, with both the measurements of base edge and slant height.

We can find out the surface area of the regular triangular pyramid(regular tetrahedron) with the base edges.

Surface area of the regular triangular pyramid(regular tetrahedron) = √3 a² , where, a is base edge length.

Base edge length of the regular triangular pyramid (a) = 6 ft.

So,

Surface area of the regular triangular pyramid(regular tetrahedron) = √3 * 6²

= 1.732 * 36

= 62.3538.

Therefore, the surface area of the regular triangular pyramid(regular tetrahedron) is 62.3538 ft ².

Comments

Surface Area of a Regular Pyramid: SA = B + 1/2 nbl, where B is the area of the base, and n is the number of triangles, b is the base edge length and l is the slant height.

The base edge length of regular triangular pyramid is 6 ft and its slant height is 10 ft.

The triangular pyramid has 4 faces, those are

The 3 side faces are triangles,

The base is also a triangles.

Here triangular pyramid is regular that means base is an equilateral triangle.

Let's draw it and find its area.

The height of the base (equilateral triangle) is h = √(6² - 3²) = √(36 - 9) = √(27) = 3√3.

The area of the base(equilateral triangle) is B = 1/2 · (base) (height).

B = 1/2 · (6) (3√3) = 9√3 ≅ 15.59 ft².

Surface area of the regular triangular pyramid : SA = B + 1/2 nbl.

B = 15.59 ft², n = 3, b = 6 ft and l = 10 ft.

SA = 15.59 + 1/2 (3)(6)(10)

SA = 15.59 + 90

SA = 105.59 ft².

Surface area of the regular triangular pyramid : SA = 105.59 ft².