From this point, a perpendicular is dropped onto the plane and two oblique ones are drawn.

From this point, a perpendicular is dropped onto the plane and two oblique ones are drawn. One oblique is 6 longer than the other. Their projections on the plane are 27 and 15, respectively. Find the length of the perpendicular.

Suppose that there is a certain point A, from which the perpendicular AO is dropped onto the plane (that is, on the plane itself directly under the point A point O lies). According to the condition of the problem, 2 inclined AB and AC are lowered from point A to the plane. It is known that AB is more than AC by some 6 units. Suppose that it is 6 cm larger. In this case, the equality is true: AB = AC + 6. It is also known that BO = 27 units (suppose that 27 cm), and CO = 15 units (or centimeters). In this case, we get 2 right-angled triangles: ABO and AСO.

Therefore, in order to find the value of AO, we need to apply the Pythagorean theorem, which says that the sum of the squares of the legs of a right-angled triangle is equal to the square of its hypotenuse.

Let us express the AO value using the lengths of the sides of both triangles. We get the following system of equations:

AO ^ 2 = AC ^ 2 – CO ^ 2 = AC ^ 2 – 225.

AO ^ 2 = AB ^ 2 – BO ^ 2 = (AC + 6) ^ 2 – 27 ^ 2 = (AC ^ 2 + 12 * AC + 36) – 729 = AC ^ 2 + 12 * AC – 693.

In this case, the equality is true:

AC ^ 2 – 225 = AC ^ 2 + 12 * AC – 693

12 * AC = 693 – 225 = 468.

AC = 468/12 = 39 cm.

In this case, AO ^ 2 = AC ^ 2 – 225 = 39 ^ 2 – 225 = 1521 – 225 = 1296 cm ^ 2

Hence AO = 36 cm.

Let’s check if we solved the problem correctly. We must have the following equality: AB ^ 2 – BO ^ 2 = AO ^ 2.

AB = AC + 6 = 39 + 6 = 45 cm.

AB ^ 2 – BO ^ 2 = 45 ^ 2 – 27 ^ 2 = 2025 – 729 = 1296 cm ^ 2

This means that AO = 36 cm.

The problem was solved correctly.

Answer: The length of the perpendicular AO = 36 cm.