Question

question 3 (1 point)
find the value for x.
image of a right triangle with an altitude to the hypotenuse, labeled with segments x, 4, and 8
blank 1:

Explanation:

Step1: Recall the geometric mean theorem (altitude-on-hypotenuse theorem) for right triangles. The theorem states that in a right triangle, the length of the altitude drawn to the hypotenuse is the geometric mean of the lengths of the two segments that the hypotenuse is divided into. Also, each leg of the right triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg. In formula terms, if we have a right triangle with hypotenuse divided into segments of length \( x \) and \( 4 \), and the altitude to the hypotenuse is \( 8 \), but wait, actually, the correct relation for the leg (the segment adjacent to \( x \)) is \( x^2 = 8 \times (8 + 4) \)? No, wait, let's re-examine. Wait, the triangle: the large triangle is right-angled, and there's a smaller right triangle inside with altitude \( 8 \), and the segments on the hypotenuse are \( x \) and \( 4 \). Wait, actually, the geometric mean theorem (leg rule) states that \( \text{leg}^2 = \text{hypotenuse segment} \times \text{entire hypotenuse} \)? No, no. Wait, the altitude to the hypotenuse: in a right triangle, when you draw an altitude from the right angle to the hypotenuse, it creates two smaller similar right triangles, each similar to the original triangle and to each other. So, the ratio of corresponding sides should be equal. So, the triangle with side \( x \) (a leg of the smaller triangle) and the altitude \( 8 \), and the other smaller triangle with side \( 8 \) and \( 4 \). Wait, actually, the correct proportion is \( \frac{x}{8} = \frac{8}{4} \). Let's derive that. Since the two smaller triangles are similar (all right triangles with a common angle), the ratio of the leg of one triangle to the leg of the other should be equal. So, the triangle with hypotenuse segment \( x \) and leg \( 8 \) (altitude) is similar to the triangle with leg \( 8 \) and hypotenuse segment \( 4 \). So, corresponding sides: \( x \) (hypotenuse of smaller triangle) corresponds to \( 8 \) (leg of larger smaller triangle), and \( 8 \) (leg of smaller triangle) corresponds to \( 4 \) (hypotenuse of larger smaller triangle). Wait, maybe better to use the geometric mean: the length of the leg (the segment adjacent to \( x \)) is the geometric mean of \( x \) and \( (x + 4) \)? No, no, the altitude is the geometric mean of the two segments. Wait, altitude \( h = \sqrt{xy} \), where \( x \) and \( y \) are the segments of the hypotenuse. But here, the altitude is \( 8 \), and one segment is \( 4 \), the other is \( x \). Wait, no, that can't be, because \( 8 = \sqrt{x \times 4} \) would give \( 64 = 4x \), \( x = 16 \). Wait, that makes sense. Wait, the altitude to the hypotenuse is the geometric mean of the two segments. So \( h^2 = x \times 4 \). Wait, \( h = 8 \), so \( 8^2 = x \times 4 \). So \( 64 = 4x \), so \( x = 16 \). Let's check. So the two segments of the hypotenuse are \( x = 16 \) and \( 4 \), so the entire hypotenuse is \( 20 \). Then the leg of the large triangle would be \( \sqrt{16 \times 20} \)? Wait, no, the leg is the geometric mean of the hypotenuse segment and the entire hypotenuse. Wait, maybe I mixed up. Let's use similar triangles. The large right triangle (let's call it \( \triangle ABC \), right-angled at \( C \)), with altitude \( CD \) to hypotenuse \( AB \), where \( D \) divides \( AB \) into \( AD = x \) and \( DB = 4 \), and \( CD = 8 \). Then \( \triangle ACD \sim \triangle CBD \sim \triangle ABC \). So, in \( \triangle ACD \) and \( \triangle CBD \), the ratio of corresponding sides: \( \frac{AD}{CD} = \frac{CD}{DB} \). So \( \frac{x}{8} = \frac{8}{4} \). Cross-multiplying: \( 4x = 64 \), so \( x = 16 \). Yes, that's the correct proportion. So the formula is \( AD \times DB = CD^2 \)? No, wait, \( \frac{AD}{CD} = \frac{CD}{DB} \) implies \( AD \times DB = CD^2 \)? Wait, no, \( \frac{x}{8} = \frac{8}{4} \) cross-multiplied is \( 4x = 64 \), so \( x = 16 \). So that's the solution.

Step2: Solve for \( x \) using the proportion from similar triangles. We have \( \frac{x}{8} = \frac{8}{4} \). Cross-multiplying gives \( 4x = 8 \times 8 \), so \( 4x = 64 \). Dividing both sides by 4, we get \( x = \frac{64}{4} = 16 \).

Answer:

\( \boxed{16} \)