There are two right-angled triangles shown in the image. In the first triangle, the hypotenuse is divided into two segments of length 9 and 16. In the second triangle, one leg is of length 10 and the hypotenuse is of length 25. For both triangles, some side lengths are marked with question marks, indicating that they need to be calculated. Calculate all the missing side lengths for both triangles.

Solution:

This problem involves right-angled triangles and the application of the Pythagorean theorem and geometric mean theorems.

Triangle № 1:

• Let the given triangle be ABC, with the right angle at C. Let CD be the altitude to the hypotenuse AB.
• We are given that the hypotenuse AB is divided into segments AD = 9 and DB = 16.
• By the geometric mean theorem (altitude theorem), the altitude CD is the geometric mean of the segments of the hypotenuse:
• \[ CD^2 = AD \times DB \]
• \[ CD^2 = 9 \times 16 \]
• \[ CD^2 = 144 \]
• \[ CD = \sqrt{144} \]
• \[ CD = 12 \]
• By the geometric mean theorem (leg theorem), the leg AC is the geometric mean of the hypotenuse AB and the segment of the hypotenuse adjacent to AC (which is AD):
• \[ AC^2 = AB \times AD \]
• \[ AC^2 = (9 + 16) \times 9 \]
• \[ AC^2 = 25 \times 9 \]
• \[ AC^2 = 225 \]
• \[ AC = \sqrt{225} \]
• \[ AC = 15 \]
• Similarly, the leg BC is the geometric mean of the hypotenuse AB and the segment of the hypotenuse adjacent to BC (which is DB):
• \[ BC^2 = AB \times DB \]
• \[ BC^2 = (9 + 16) \times 16 \]
• \[ BC^2 = 25 \times 16 \]
• \[ BC^2 = 400 \]
• \[ BC = \sqrt{400} \]
• \[ BC = 20 \]
• Alternatively, we can use the Pythagorean theorem on the smaller right triangles:
• In triangle ADC:
• \[ AC^2 = AD^2 + CD^2 \]
• \[ AC^2 = 9^2 + 12^2 \]
• \[ AC^2 = 81 + 144 \]
• \[ AC^2 = 225 \]
• \[ AC = 15 \]
• In triangle BDC:
• \[ BC^2 = DB^2 + CD^2 \]
• \[ BC^2 = 16^2 + 12^2 \]
• \[ BC^2 = 256 + 144 \]
• \[ BC^2 = 400 \]
• \[ BC = 20 \]

Triangle № 2:

• Let the given triangle be PQR, with the right angle at Q. Let QS be the altitude to the hypotenuse PR.
• We are given one leg PQ = 10 and the hypotenuse PR = 25.
• We can find the other leg QR using the Pythagorean theorem:
• \[ PQ^2 + QR^2 = PR^2 \]
• \[ 10^2 + QR^2 = 25^2 \]
• \[ 100 + QR^2 = 625 \]
• \[ QR^2 = 625 - 100 \]
• \[ QR^2 = 525 \]
• \[ QR = \sqrt{525} \]
• \[ QR = \sqrt{25 \times 21} \]
• \[ QR = 5\sqrt{21} \]
• Now we can find the segments of the hypotenuse. Using the leg theorem for PQ:
• \[ PQ^2 = PR \times PS \]
• \[ 10^2 = 25 \times PS \]
• \[ 100 = 25 \times PS \]
• \[ PS = \frac{100}{25} \]
• \[ PS = 4 \]
• The other segment of the hypotenuse is SR:
• \[ SR = PR - PS \]
• \[ SR = 25 - 4 \]
• \[ SR = 21 \]
• We can verify this using the leg theorem for QR:
• \[ QR^2 = PR \times SR \]
• \[ (5\sqrt{21})^2 = 25 \times 21 \]
• \[ 525 = 525 \]
• Finally, we can find the altitude QS using the altitude theorem:
• \[ QS^2 = PS \times SR \]
• \[ QS^2 = 4 \times 21 \]
• \[ QS^2 = 84 \]
• \[ QS = \sqrt{84} \]
• \[ QS = \sqrt{4 \times 21} \]
• \[ QS = 2\sqrt{21} \]

Summary of missing values:

• Triangle № 1:
• Altitude to hypotenuse: 12
• One leg: 15
• Other leg: 20
• Triangle № 2:
• Other leg (QR): $$5\sqrt{21}$$
• Segment of hypotenuse (PS): 4
• Segment of hypotenuse (SR): 21
• Altitude to hypotenuse (QS): $$2\sqrt{21}$$

Ответ:

• Triangle № 1: The missing sides are 12, 15, and 20.
• Triangle № 2: The missing sides are $$5\sqrt{21}$$, 4, 21, and $$2\sqrt{21}$$.