One of the legs of a right-angled triangle is 15 cm, and the projection of the other leg onto

One of the legs of a right-angled triangle is 15 cm, and the projection of the other leg onto the hypotenuse is 9 cm. Find the hypotenuse.

Dan △ ABC: ∠C = 90 °, AC = 15 cm and BC – legs, AB – hypotenuse.
From the vertex C we draw the height CK to the hypotenuse AB, then the segments AK and BK are the projections of the legs AC and BC, so BK = 9 cm.
Let’s denote the segment AK as x, then the length of the hypotenuse is:
AB = AK + BK = x + 9.
From the properties of a right-angled triangle, it is known that each leg is the geometric mean of the hypotenuse and the projection of the leg onto the hypotenuse, that is:
AC² = AB * AK.
Then:
(x + 9) * x = 15²;
x² + 9 * x = 225;
x² + 9 * x – 225 = 0.
Let’s solve the quadratic equation. Let’s find the discriminant:
D = b² – 4 * a * c = 9² – 4 * 1 * (- 225) = 81 + 900 = 981.
x = (- b +/- √ D) / 2 * a.
x₁ = (- 9 + √ 981) / 2 * 1 = (- 9 + 3 √ 109) / 2.
x₂ = (- 9 – √ 981) / 2 * 1 = (- 9 – 3 √ 109) / 2 – this value of x does not make sense, since the length of the side of a triangle cannot be a negative number.
Thus, AK = x = (- 9 + 3 √ 109) / 2 cm.
Let’s find the length of the hypotenuse:
AB = (- 9 + 3 √ 109) / 2 + 9 = (- 9 + 3 √ 109 + 18) / 2 = (9 + 3 √ 109) / 2 (cm).
Answer: AB = (9 + 3 √ 109) / 2 cm.