In a right-angled triangle, the height and median drawn to the hypotenuse are 24 cm and 25 cm

In a right-angled triangle, the height and median drawn to the hypotenuse are 24 cm and 25 cm. Find the perimeter of the triangle.

ABC – right-angled triangle, angle C = 90 degrees, CH = 24 cm – height, CM = 25 cm – median.
1. From the properties of a right-angled triangle, it is known that the median falling on the hypotenuse is equal to half of the hypotenuse. Then:
CM = AB / 2;
AB / 2 = 25;
AB = 25 * 2 (in proportion);
AB = 50 cm.
2. Height is the average proportional to the two segments of the hypotenuse formed by it, that is:
CH ^ 2 = AH * BH.
Let AH = x, and BH = y, then:
xy = 24 ^ 2;
xy = 576.
Let’s compose a system of linear equations:
xy = 576;
x + y = 50.
In the second equation, we express x through y:
x = 50 – y.
We substitute the resulting expression into the first equation of the system:
(50 – y) y = 576;
50y – y ^ 2 – 576 = 0;
y ^ 2 – 50y + 576 = 0.
Let’s solve the quadratic equation. Let’s find the discriminant:
D = b ^ 2 – 4ac;
D = 50 ^ 2 – 4 * 1 * 576 = 2500 – 2304 = 196.
y = (-b +/- √D) / 2a.
y1 = (- (- 50) + √196) / 2 * 1 = (50 + 14) / 2 = 64/2 = 32.
y2 = (- (- 50) – √196) / 2 * 1 = (50 – 14) / 2 = 36/2 = 18.
Then:
x1 = 50 – y1 = 50 – 32 = 18.
x2 = 50 – y2 = 50 – 18 = 32.
Thus, AH = 18 cm or 32 cm, BH = 32 cm or 18 cm.Let AH = 18 cm, and BH = 32 cm.
3. Each leg is the geometric mean of the hypotenuse and the projection of the leg onto the hypotenuse, that is:
AC ^ 2 = AB * AH;
BC ^ 2 = AB * BH.
Let’s find the lengths of the legs AC and BC:
AC = √AB * AH = √50 * 18 = √900 = 30 (cm);
BC = √AB * BH = √50 * 32 = √1600 = 40 (cm).
4. The ABC perimeter is:
P = AB + BC + AC;
P = 50 + 40 + 30 = 120 (cm).
Answer: P = 120 cm.