From point M to the plane, two inclined 12 cm and 24 cm are drawn, the projections of which are 1: 7.

From point M to the plane, two inclined 12 cm and 24 cm are drawn, the projections of which are 1: 7. Find the distance from point M to the plane.

Let’s build a perpendicular OM to the plane, then the triangles AOM and BOM are rectangular.

Let ОВ = X cm, then ОА = 7 * X cm.

In a right-angled triangle BOM, according to the Pythagorean theorem:

OM ^ 2 = BM ^ 2 – OB ^ 2 = 144 – X ^ 2.

In a right-angled triangle AOM, according to the Pythagorean theorem:

ОМ ^ 2 = АМ ^ 2 – ОА ^ 2 = 576 – 49 * X ^ 2.

Let’s equate equalities 1 and 2.

144 – X ^ 2 = 576 – 49 * X ^ 2.

48 * X ^ 2 = 576 – 144 = 43 ^ 2.

X ^ 2 = OB ^ 2 = 43 ^ 2/48 = 9 cm.

OB = 3 cm.

ОМ ^ 2 = 144 – 9 = 135.

ОМ = 3 * √15 cm.

Answer: From point M to plane 3 * √15 cm.