At the base of the SABCD pyramid there is a rhombus with diagonals AC = 8 and BD = 6.

| September 23, 2021 | education

At the base of the SABCD pyramid there is a rhombus with diagonals AC = 8 and BD = 6. The height of the SO pyramid passes through the point of intersection of the base diagonals and is equal to 1.8. Find the height SH of the SAB face.

The diagonals of the rhombus at the point of their intersection are divided in half, then ОВ = ОD = ВD / 2 = 8/2 = 4 cm, ОА = OC = АС / 2 = 6/2 = 3 cm.

Triangle AOB is rectangular, in which, according to the Pythagorean theorem, AB ^ 2 = OB ^ 2 + OA ^ 2 = 16 + 9 = 25.

AB = 5 cm.

Saov = ОА * ОВ / 2 = 3 * 4/2 = 6 cm2.

Saov = AB * OH / 2 = 5 * OH / 2.

5 * OH / 2 = 6.

OH = 12/5 = 2.4 cm.

From the right-angled triangle SHO, by the Pythagorean theorem, SH ^ 2 = SO ^ 2 + OH ^ 2 = 3.24 + 5.76 = 9.

SH = 3 cm.

Answer: The length of the SH height is 3 cm.



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