The height of a regular quadrangular pyramid is 10 cm. The diagonal section is equal to the base.
The height of a regular quadrangular pyramid is 10 cm. The diagonal section is equal to the base. Find the total surface area of the pyramid.
The diagonal section of a regular quadrangular pyramid is an isosceles triangle, in which the base is equal to the diagonal of the base of the pyramid, the height is equal to the height of the pyramid. The area of such a section can be found as half of the product of the height and the diagonal of the base: Ssection = 0.5 * h * d.
At the base of this pyramid lies a square, its area is equal to half of the square of the diagonal: Sbn = 0.5 * d ^ 2.
By condition, the area of the diagonal section is equal to the area of the base of the pyramid, which means:
Ssec = Sbn;
0.5 * h * d = 0.5 * d ^ 2;
h = d – therefore, the height of this pyramid is equal to the diagonal of the base.
Sb = 0.5 * d ^ 2 = 0.5 * 10 ^ 2 = 0.5 * 100 = 50 cm2.
On the other hand, the area of the base is equal to the square of the side of the base, hence the side of the base of the pyramid is a = √50 = 5√2 cm.
The side faces of this pyramid are equal isosceles triangles, the bases of which are equal sides of the square that lies at the base of the pyramid.
The segment drawn from the center of the base of the pyramid to the middle of any of its sides is equal to half of the side.
Consider a right-angled triangle formed by this segment, the height of the pyramid and the height of the side face. The height of the side face can be found as the square root of the sum of the squares of the height of the pyramid and half of the side of the base:
hface = √ (10 ^ 2 + (5√2 / 2) ^ 2) = √ (100 + 25/2) = √ (225/2) = 15 / √2 cm.
The area of the side face is defined as half of the product of the side of the base of the pyramid by the height of the side face: Sgr = 0.5 * a * hface = 0.5 * 5√2 * 15 / √2 = 75/2 = 37.5 cm2.
The total surface area of the pyramid is: Sful = Sbn + 4 * Sgr = 50 + 4 * 37.5 = 50 + 150 = 200 cm2.
