At the base of a straight pyramid, a rectangle with sides 6m 8m, the lateral edge is 13 m, find the height.

Since there is a rectangle at the base of the pyramid, the ACD triangle is rectangular.

Let us find by the Pythagorean theorem the diagonal AC at the base of the pyramid.

AC ^ 2 = AD ^ 2 + CD ^ 2 = 82 + 62 = 64 + 36 = 100.

AC = 10 m.

Since the diagonals in the rectangle are divided in half at the point of intersection, then AO = CO = AC / 2 = 10/2 = 5 m.

KO is the height of the pyramid, so the triangle KOC is rectangular at the vertex O, then by the Pythagorean theorem KO ^ 2 = KC ^ 2 – OS ^ 2 = 13 ^ 2 – 5 ^ 2 = 169 – 25 = 144.

KO = 12 m.

Answer: The height of the pyramid is 12 m.