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What is the surface area of a regular tetrahedron for which each edge is 7 cm long and the slant height is 6.1 cm? Round your answer to the nearest whole number.

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Final answer:

The surface area of a regular tetrahedron with each edge measuring 7 cm and the slant height of 6.1 cm is approximately 85 cm².

Step-by-step explanation:

A regular tetrahedron is a three-dimensional shape with four equilateral triangle faces. To find the surface area of a regular tetrahedron, we can calculate the area of each face and then sum them up. Since each edge of the tetrahedron is 7 cm long and the slant height is 6.1 cm, we can use the formula for the area of an equilateral triangle to find the area of each face:

A = (√3 / 4) * side^2

Substituting the value of the side (7 cm) into the formula, we get:

A = (√3 / 4) * 7^2

Calculating this expression, the area of each face is approximately 21.217 cm².

Since the tetrahedron has four faces, the total surface area is 4 * 21.217 = 84.868 cm². Rounding this to the nearest whole number, the surface area is approximately 85 cm².

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