Answer:
Let's call the length of the hypotenuse SD as x.
Since the altitude from T to SD divides SD into two parts, let the length of the shorter part be y. Then the length of the longer part is x-y.
Using similar triangles, we have:
TU/TS = ST/TD
Substituting the values we have:
TU/(3√5) = √5/UD
TU = (3/5)UD
Using the Pythagorean theorem in triangle TUS, we have:
TU² + (3√5)² = TS²
(3/5 UD)² + 45 = ST²
9/25 UD² + 45 = ST²
Using the Pythagorean theorem in triangle TUD, we have:
TU² + UD² = TD²
(3/5 UD)² + UD² = x²
9/25 UD² + UD² = x²
34/25 UD² = x²
UD² = (25/34)x²
Substituting the value of UD² in the equation ST² = 9/25 UD² + 45, we get:
ST² = 9/25 (25/34)x² + 45
ST² = 45/34 x² + 45
Since y = x-y-1, we have y = (x-1)/2.
Using the Pythagorean theorem in triangle TUD, we have:
(1/4) (x-1)² + UD² = x²
(1/4) (x² - 2x + 1) + (25/34)x² = x²
(1/4)(x²) + (25/34)x² - (1/2)x + (1/4) = 0
(59/68)x² - (1/2)x + (1/4) = 0
Using the quadratic formula, we get:
x = [1/2 ± √(1/4 - 4(59/68)(1/4))]/(2(59/68))
x = [1/2 ± (3√34)/17]/(59/34)
x = 17/59 ± 6√34/59
Since x is the hypotenuse SD, we have:
UD² = (25/34) x²
UD² = (25/34) [(17/59 ± 6√34/59)²]
UD² = 136/59 ± 204√34/295
Therefore, the exact lengths of the two parts of the hypotenuse are:
SD = x = 17/59 ± 6√34/59
SU = x-y = (x-1)/2 = 8/59 ± 3√34/59
UD = y = (x-1)/2 = 8/59 ± 3√34/59
TU = (3/5) UD = (3/5) [8/59 ± 3√34/59] = 24/295 ± 9√34/295
ST² = 45/34 x² + 45 = 45/34 [(17/59 ± 6√34/59)²] + 45
ST = √[45/34 [(17/59 ± 6√34/59)²] + 45]