In a regular triangular pyramid, the side edge is 4 cm. And the side of the base is 6 cm.

| May 9, 2021 | education

In a regular triangular pyramid, the side edge is 4 cm. And the side of the base is 6 cm. Find the area of the entire surface.

The equilateral triangle ABC is located at the base of the regular triangular pyramid.

The area of ​​an equilateral triangle is equal to: Saws = a2 * √3 / 4, where a is the length of the side of the triangle, then Saws = 36 * √3 / 4 = 9 * √3 cm2.

The SCB triangle is isosceles, then SH is its height and median, which means CH = BC / 2 = 6/2 = 3 cm.

In a right-angled triangle SCH, by the Pythagorean theorem, SH ^ 2 = SC ^ 2 – CH ^ 2 = 16 – 9 = 7.

SH = √7 cm.

Determine the area of ​​the triangle SBC. Ssвс = ВС * SH / 2 = 6 * √7 / 2 = 3 * √7 cm2.

Then Sside = 3 * Ssvs = 9 * √7 cm2.

Spov = S main + S side = 9 * √3 + 9 * √7 = 9 * (√3 + √7) cm2.

Answer: The total surface area is 9 * (√3 + √7) cm2.



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