The base of the DABC pyramid is a regular triangle ABC, the side of which is equal to p. The edge DA is perpendicular to the plane ABC, and the plane DBC makes an angle of 30 degrees with the plane ABC. Find the side area and the total surface area of the pyramid.
At the base of the pyramid is an equilateral triangle with a side length equal to b cm.
Determine the area of the base of the cone.
Sb = b ^ 2 * √3 / 4 cm2.
Determine the height AH of the isosceles triangle ABC
AH = b * √3 / 2 cm.
Triangle ADH is rectangular, since, by condition, AD is perpendicular to ABC, and angle DHA = 30.
Then AD = AH * tg30 = b * √3 / 2 * (√3 / 3) = b / 2 cm.
Let us determine the areas of the side faces ADC and ADB.
Sads = Sadv = AD * AC / 2 = (b / 2) * b / 2 = b2 / 4 cm2.
Determine the height DH of the triangle BDC. DH = AH / Cos30 = (b * √3 / 2) / √3 / 2 = b see.
Then Sdvs = BC * DH / 2 = b * b / 2 = b ^ 2/2 cm2.
Sside = 2 * Sads + Sdvs = b ^ 2/2 + b ^ 2/2 = b ^ 2 cm2.
Spov = b ^ 2 + b ^ 2 * √3 / 4 = b ^ 2 * (1 + √3 / 4) cm2.
Answer: Side = b ^ 2 cm2, Sпов = b ^ 2 * (1 + √3 / 4) cm2.