From a point to a plane, two inclined lengths of 4 cm and 6 cm and a perpendicular are drawn.

From a point to a plane, two inclined lengths of 4 cm and 6 cm and a perpendicular are drawn. Oblique projections are referred to as 2 to 3 (2: 3). Calculate the length of the perpendicular and the projection of the smaller of the oblique.

From point A to the plane, inclined AB = 4 cm and AC = 6 cm and perpendicular AH are drawn. The projection of the inclined AB HВ refers to the projection of the inclined AC CH as 2/3, that is:
ВН / CH = 2/3.
1. Triangles AНВ and AНС are rectangular. In the AНВ triangle:
AH = √ (AB ^ 2 – BH ^ 2) (by the Pythagorean theorem).
In the ANS triangle:
AH = √ (AC ^ 2 – CH ^ 2) (by the Pythagorean theorem).
Then the equality is true:
√ (AB ^ 2 – BH ^ 2) = √ (AC ^ 2 – CH ^ 2);
AB ^ 2 – BH ^ 2 = AC ^ 2 – CH ^ 2;
4 ^ 2 – BH ^ 2 = 6 ^ 2 – CH ^ 2;
16 – BH ^ 2 = 36 – CH ^ 2;
– BH ^ 2 = 36 – 16 – CH ^ 2;
BH ^ 2 = CH ^ 2 – 20;
BH = √ (CH ^ 2 – 20).
Substitute this expression into the equality ВН / CH = 2/3:
√ (CH ^ 2 – 20) / CH = 2/3;
(CH ^ 2 – 20) / CH ^ 2 = 4/9;
9CH ^ 2 – 180 = 4CH ^ 2;
5CH ^ 2 = 180;
CH ^ 2 = 180/5;
CH ^ 2 = 36;
CH = √36;
CH = 6 cm.
Let’s find the length of the ВН:
BH = √ (6 ^ 2 – 20) = √ (36 – 20) = √16 = 4 (cm).
BH = 4 cm – a projection of a smaller oblique AB.
2. Let’s find the length of the perpendicular АН:
AH = √ (4 ^ 2 – 4 ^ 2) = √ (16 – 16) = √0 = 0 (cm).
Since in triangles ANV and ANS, the lengths of the legs BH and CH are equal to the lengths of the hypotenuses AB and AC, respectively, and the leg AH = 0, then these triangles are degenerate.
Answer: BH = 4 cm, AH = 0 cm.