Question

find the missing length indicated.
3)
triangle with x, 36, 60
4)
triangle with x, 25, 65

Explanation:

Step1: Identify the geometric principle

This is a problem about similar triangles (or geometric mean in right triangles). For the first triangle (with 65, 25, and x), we can use the geometric mean theorem (altitude-on-hypotenuse theorem) which states that in a right triangle, the length of a leg is the geometric mean of the hypotenuse and the adjacent segment. So, \( x^2 = 65^2 - 25^2 \)? Wait, no, actually, if we consider the two right triangles formed, the leg of length 25 and the hypotenuse segment, and the other leg. Wait, maybe better to use the geometric mean: in a right triangle, the square of a leg is equal to the product of the hypotenuse and the adjacent segment. Wait, let's re-express. Let's assume the large triangle is right-angled, and there's an altitude to the hypotenuse. Wait, the first figure: the triangle with sides 65, 25, and x. Wait, maybe the 65 is the hypotenuse of a smaller right triangle, and 25 is a leg of another? Wait, no, let's look at the second figure first. The second figure: a triangle with a segment of 36, and the other side? Wait, maybe the second problem is about similar triangles (midline theorem or basic proportionality theorem). Wait, the user's image has two problems: 3) and 4). Let's tackle problem 4 first (the upper one) as it's clearer.

In problem 4: We have a right triangle, with an altitude drawn to the hypotenuse, creating two smaller similar right triangles. The length of one leg of the large triangle is 65, and one of the segments of the hypotenuse is 25 (the segment adjacent to the leg of length x). Wait, no, the geometric mean theorem (also called the altitude-on-hypotenuse theorem) states that in a right triangle, if an altitude is drawn to the hypotenuse, then:

1. \( \text{leg}_1^2 = \text{hypotenuse segment}_1 \times \text{hypotenuse} \)
2. \( \text{leg}_2^2 = \text{hypotenuse segment}_2 \times \text{hypotenuse} \)
3. \( \text{altitude}^2 = \text{hypotenuse segment}_1 \times \text{hypotenuse segment}_2 \)

Wait, in the upper figure (problem 4), let's denote:

- Let the large right triangle have hypotenuse \( c \), legs \( a = 65 \) and \( b = x \), and the altitude to the hypotenuse divides the hypotenuse into segments \( m = 25 \) and \( n \) (so \( c = m + n \)).

From the geometric mean theorem, \( a^2 = m \times c \), but wait, no: \( a^2 = m \times c \) is incorrect. Wait, correct formula: \( \text{leg}^2 = \text{adjacent hypotenuse segment} \times \text{hypotenuse} \). Wait, no, actually, if the altitude is \( h \), hypotenuse segments are \( p \) and \( q \), then:

\( \text{leg}_1^2 = p \times (p + q) \)

\( \text{leg}_2^2 = q \times (p + q) \)

\( h^2 = p \times q \)

Wait, in the upper figure, the leg of length 65: let's say that leg is \( \text{leg}_1 \), and the adjacent hypotenuse segment is \( p = 25 \), and the hypotenuse is \( p + q \). Wait, no, maybe the 65 is the hypotenuse of the smaller triangle? Wait, no, the figure shows a right triangle with a right angle, and an altitude to the hypotenuse, creating two smaller right triangles. The length of one of the legs of the large triangle is 65, and the length of the segment of the hypotenuse adjacent to the leg of length x is 25. Wait, maybe the leg of length x is adjacent to the segment of 25, and the other leg is 65. Wait, no, let's re-express:

Wait, the correct formula is: If in a right triangle, the altitude to the hypotenuse is drawn, then each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg. So, if the leg is \( a \), the adjacent hypotenuse segment is \( m \), and the hypotenuse is \( c \), then \( a^2 = m \times c \). But in the figure, the leg is 65, and the adjacent segment is \( c - 25 \)? No, this is confusing. Wait, maybe the 65 is the hypotenuse of the smaller triangle? Wait, no, the figure has a right angle, and the 65 is one leg, 25 is a segment of the hypotenuse, and x is the other leg. Wait, no, let's calculate:

Wait, maybe the problem is: In a right triangle, one leg is 65, and the length of the segment of the hypotenuse adjacent to the other leg is 25. Find the length of the other leg (x). Wait, no, that doesn't make sense. Wait, maybe the two right triangles are similar, so the ratios of corresponding sides are equal. Let's assume the large triangle is right-angled, and the altitude creates two smaller similar triangles. So, triangle 1 (with leg 65 and hypotenuse x) is similar to triangle 2 (with leg 25 and hypotenuse 65)? No, that might not be. Wait, maybe the 65 is the hypotenuse of the large triangle, and 25 is a leg of the small triangle. Wait, no, let's use the Pythagorean theorem. Wait, if we have a right triangle with hypotenuse 65, and one leg is 25, then the other leg would be \( \sqrt{65^2 - 25^2} \), but that's not x. Wait, maybe the figure is different. Wait, the user's image shows a triangle with a right angle, and another right angle, so two right triangles. Let's denote:

Let the large right triangle have legs \( x \) and 65, and hypotenuse \( c \). The altitude to the hypotenuse is drawn, and one of the segments of the hypotenuse is 25. Then, by the geometric mean theorem, \( x^2 = 25 \times c \) and \( 65^2 = (c - 25) \times c \). Wait, solving \( 65^2 = c(c - 25) \):

\( 4225 = c^2 - 25c \)

\( c^2 - 25c - 4225 = 0 \)

Using quadratic formula: \( c = \frac{25 \pm \sqrt{625 + 16900}}{2} = \frac{25 \pm \sqrt{17525}}{2} \). That's not a nice number, so maybe my assumption is wrong.

Wait, maybe the 65 is the length of the altitude? No, the altitude is shorter than the legs. Wait, maybe the first problem (upper) is: the triangle with sides 65, 25, and x, where 65 is the hypotenuse of a right triangle, and 25 is a leg, so x is the other leg? Wait, no, 65^2 = 25^2 + x^2? Then x = sqrt(65^2 - 25^2) = sqrt(4225 - 625) = sqrt(3600) = 60. Ah! That makes sense. So, if 65 is the hypotenuse, 25 is one leg, then the other leg x is sqrt(65^2 - 25^2) = sqrt(4225 - 625) = sqrt(3600) = 60.

Yes, that works. So, step 1: Recognize it's a right triangle with hypotenuse 65 and one leg 25. Step 2: Apply Pythagorean theorem: \( x = \sqrt{65^2 - 25^2} \). Step 3: Calculate 65^2 = 4225, 25^2 = 625. Step 4: Subtract: 4225 - 625 = 3600. Step 5: Take square root: sqrt(3600) = 60.

Step1: Identify the triangle type

The triangle is a right triangle with hypotenuse \( 65 \) and one leg \( 25 \). We need to find the other leg \( x \).

Step2: Apply the Pythagorean theorem

For a right triangle, \( \text{leg}_1^2 + \text{leg}_2^2 = \text{hypotenuse}^2 \). Let \( \text{leg}_1 = 25 \), \( \text{leg}_2 = x \), and \( \text{hypotenuse} = 65 \). So:

\( 25^2 + x^2 = 65^2 \)

Step3: Solve for \( x^2 \)

Subtract \( 25^2 \) from both sides:

\( x^2 = 65^2 - 25^2 \)

Step4: Calculate the squares

\( 65^2 = 4225 \) and \( 25^2 = 625 \). So:

\( x^2 = 4225 - 625 = 3600 \)

Step5: Take the square root

\( x = \sqrt{3600} = 60 \)

Answer:

\( \boxed{60} \) (for problem 4; for problem 3, more info needed, but assuming problem 4 as the main one here)