Question

what is the value of x?
12 units
15 units
20 units
25 units

Explanation:

Step1: Recall the geometric mean theorem (altitude-on-hypotenuse theorem)

In a right triangle, the length of the altitude drawn to the hypotenuse is the geometric mean of the lengths of the two segments of the hypotenuse. Also, the length of each leg is the geometric mean of the length of the hypotenuse and the length of the adjacent segment. Here, in right triangle \( \triangle SRQ \) with right angle at \( R \), and \( RT \) is the altitude to hypotenuse \( SQ \). The segments of the hypotenuse are \( ST = 9 \) and \( TQ = 16 \), and \( RT=x \). Wait, actually, the geometric mean theorem for the altitude: \( RT^{2}=ST\times TQ \)? No, wait, the altitude to the hypotenuse: \( RT^{2}=ST\times TQ \)? Wait, no, the correct formula is that in right triangle \( \triangle SRQ \), right - angled at \( R \), and \( RT\perp SQ \) ( \( T \) is on \( SQ \) ), then \( RT^{2}=ST\times TQ \)? Wait, no, actually, the length of the leg: for example, \( SR^{2}=ST\times SQ \) and \( RQ^{2}=TQ\times SQ \), and the altitude \( RT^{2}=ST\times TQ \). Wait, let's check the lengths. \( ST = 9 \), \( TQ = 16 \), so \( SQ=ST + TQ=9 + 16 = 25 \). Then, by the geometric mean theorem (altitude to hypotenuse), \( RT^{2}=ST\times TQ \)? Wait, no, that's not right. Wait, the correct formula is that the altitude to the hypotenuse of a right triangle is the geometric mean of the two segments into which it divides the hypotenuse. So \( RT=\sqrt{ST\times TQ}\)? Wait, no, wait, let's re - derive.

In right triangle \( \triangle SRQ \), \( \angle R = 90^{\circ} \), \( RT\perp SQ \). Then \( \triangle STR\sim\triangle RTQ\sim\triangle SRQ \) (by AA similarity, since all right triangles and share an acute angle). So from \( \triangle STR\sim\triangle RTQ \), we have \( \frac{ST}{RT}=\frac{RT}{TQ} \), which implies \( RT^{2}=ST\times TQ \). Wait, but \( ST = 9 \), \( TQ = 16 \), so \( RT^{2}=9\times16 = 144 \), then \( RT=\sqrt{144}=12 \)? But wait, that contradicts? Wait, no, maybe I mixed up the segments. Wait, no, the hypotenuse is \( SQ=9 + 16 = 25 \). Wait, another part of the geometric mean theorem: the length of each leg is the geometric mean of the hypotenuse and the adjacent segment. Wait, maybe I made a mistake. Wait, let's look at the triangle again. The right angle is at \( R \), so \( SR \) and \( RQ \) are the legs, \( SQ \) is the hypotenuse. \( RT \) is the altitude from \( R \) to \( SQ \), hitting \( SQ \) at \( T \). Then, the two segments of the hypotenuse are \( ST = 9 \) and \( TQ = 16 \). Then, the length of the altitude \( RT \) is given by \( RT=\sqrt{ST\times TQ} \)? Wait, \( 9\times16 = 144 \), \( \sqrt{144}=12 \). But wait, the options have 12 as an option. Wait, but let's check again. Wait, maybe the formula is \( RT^{2}=ST\times TQ \), so \( x^{2}=9\times16=144 \), so \( x = 12 \)? But wait, no, wait, the hypotenuse is \( 25 \), and the legs: \( SR=\sqrt{ST\times SQ}=\sqrt{9\times25}=\sqrt{225} = 15 \), \( RQ=\sqrt{TQ\times SQ}=\sqrt{16\times25}=\sqrt{400}=20 \). But the question is about \( x \), the altitude. So according to the geometric mean theorem (altitude to hypotenuse), \( x^{2}=ST\times TQ \), so \( x=\sqrt{9\times16}=\sqrt{144} = 12 \). Wait, but let's confirm the theorem. The geometric mean (altitude) theorem states that in a right triangle, the altitude drawn to the hypotenuse is the geometric mean between the segments into which it divides the hypotenuse. So if the hypotenuse is divided into segments of length \( a \) and \( b \), then the altitude \( h \) satisfies \( h^{2}=a\times b \). Here, \( a = 9 \), \( b = 16 \), so \( h^{2}=9\times16 = 144 \), so \( h = 12 \).

Step2: Calculate \( x \)

We know that \( ST = 9 \) and \( TQ = 16 \). Using the geometric mean theorem for the altitude to the hypotenuse in a right triangle: \( x^{2}=ST\times TQ \)
Substitute \( ST = 9 \) and \( TQ = 16 \) into the formula: \( x^{2}=9\times16=144 \)
Take the square root of both sides: \( x=\sqrt{144}=12 \)

Answer:

12 units