From a point to a straight line, two oblique ones are drawn, the projections of which on a straight line are 9 cm and 16 cm

From a point to a straight line, two oblique ones are drawn, the projections of which on a straight line are 9 cm and 16 cm. Find the distance from a point to a straight line if one of the oblique is 5 cm more from the other.

In fact, two oblique lines drawn from one point to a straight line and their projections form a triangle ABC, the base of which is AC = 9 + 16 = 25 (cm), and the distance from a point to a straight line is the height of the resulting triangle BH. Thus, BH divides triangle ABC into two right-angled triangles ABH and BCH. Let us denote the hypotenuse of the triangle ABH x, and the hypotenuse of the triangle BCH x + 5 (since one of the inclined ones is 5 cm larger than the other).
In the triangle ABН, we find BH by the Pythagorean theorem:
BH = √ (x ^ 2 – 9 ^ 2).
In the triangle BCH, we find BH by the Pythagorean theorem:
BH = √ ((x + 5) ^ 2) – 16 ^ 2).
It turns out:
√ (x ^ 2 – 9 ^ 2) = √ ((x + 5) ^ 2) – 16 ^ 2);
x ^ 2 – 81 = x ^ 2 + 10x + 25 – 256;
10x = 256 – 25 – 81;
10x = 150;
x = 15 cm
So: the hypotenuse of the triangle ABН is equal to 15 cm.In the triangle ABН we find BH by the Pythagorean theorem:
BH = √ (x ^ 2 – 9 ^ 2);
BH = √ (225 – 81) = √144 = 12 (cm).
To check, let’s do the same in the BCH triangle:
BH = √ ((x + 5) ^ 2) – 16 ^ 2);
BH = √ (20 ^ 2 – 256) = √ (400 – 256) = √144 = 12 (cm).
Answer: BH = 12 cm