Question

what is the value of x? 12 units 15 units 20 units 24 units

Explanation:

Step1: Recall geometric mean theorem

In a right triangle, the altitude to the hypotenuse relates the segments of the hypotenuse. Also, the leg of a right triangle is the geometric mean of the hypotenuse and the adjacent segment. Here, triangle \( SRQ \) is right - angled at \( R \), and \( RT \) is the altitude to hypotenuse \( SQ \). So, \( x^{2}=ST\times SQ \)

Step2: Calculate \( SQ \)

We know that \( ST = 9 \) and \( TQ=16 \), so \( SQ=ST + TQ=9 + 16=25 \)

Step3: Solve for \( x \)

Using the formula \( x^{2}=ST\times SQ \), substitute \( ST = 9 \) and \( SQ = 25 \)? Wait, no, wait. Wait, actually, the correct formula is that in a right triangle, if we have a right triangle with hypotenuse \( c \), and a segment of the hypotenuse \( a \), and the leg \( x \), then \( x^{2}=a\times c \). Wait, no, let's re - examine the diagram. The right triangle is \( \triangle SRQ \) with right angle at \( R \), and \( \triangle STR \) is also right - angled. The correct geometric mean formula is that the length of a leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the adjacent segment of the hypotenuse. So, \( SR^{2}=ST\times SQ \). Wait, \( SQ=ST + TQ=9 + 16 = 25 \), \( ST = 9 \), then \( SR^{2}=9\times25 \)? No, that can't be. Wait, I made a mistake. Let's look again. The two segments of the hypotenuse are \( ST = 9 \) and \( TQ = 16 \), and the leg is \( SR=x \). The correct formula is \( x^{2}=ST\times SQ \)? No, the correct formula is \( x^{2}=ST\times SQ \) where \( SQ=ST + TQ \)? Wait, no, the correct formula is that in a right triangle, when you draw an altitude from the right angle to the hypotenuse, then each leg is the geometric mean of the hypotenuse and the adjacent segment. So, \( SR^{2}=ST\times SQ \), where \( SQ=ST + TQ=9 + 16 = 25 \), \( ST = 9 \), then \( SR^{2}=9\times25=225 \)? No, that would give \( SR = 15 \), but wait, \( 9\times25 = 225 \), \( \sqrt{225}=15 \). Wait, let's check again.

Wait, the hypotenuse \( SQ=ST + TQ = 9+16 = 25 \). The leg \( SR \) (which is \( x \)) satisfies \( x^{2}=ST\times SQ \)? No, the correct formula is \( x^{2}=ST\times SQ \)? Wait, no, the formula is: In right triangle \( \triangle SRQ \), with altitude \( RT \) to hypotenuse \( SQ \), then \( SR^{2}=ST\times SQ \), \( RQ^{2}=TQ\times SQ \), and \( RT^{2}=ST\times TQ \).

So, \( SQ=9 + 16=25 \), \( ST = 9 \), so \( x^{2}=9\times25 \)? No, \( 9\times25 = 225 \), \( x=\sqrt{225}=15 \)? Wait, but let's check the answer options. 15 is one of the options. Wait, but let's re - derive.

Wait, the right triangle is \( \triangle SRQ \), right - angled at \( R \). The altitude from \( R \) to \( SQ \) is \( RT \). Then, by the geometric mean theorem (also known as the altitude - on - hypotenuse theorem), we have:

\( SR^{2}=ST\times SQ \)

We know that \( ST = 9 \), \( TQ = 16 \), so \( SQ=ST + TQ=9 + 16 = 25 \)

Substitute into the formula: \( x^{2}=9\times25 \)? No, \( 9\times25=225 \), so \( x=\sqrt{225}=15 \)

Answer:

15 units