Question

what is the value of x? 20 units 24 units 15 units 12 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 the right triangle is the geometric mean of the hypotenuse and the adjacent segment. So, \( x^2 = (9 + 16)\times16 \)? Wait, no, wait. Wait, the right triangle \( \triangle SRQ \) with right angle at \( R \), and \( RT \) is the altitude to hypotenuse \( SQ \). Wait, \( SQ = 9 + 16 = 25 \)? Wait, no, \( ST = 9 \), \( TQ = 16 \), so \( SQ = 9 + 16 = 25 \). Then, by geometric mean theorem, \( RQ^2 = TQ \times SQ \)? Wait, no, \( RQ \) is a leg, \( TQ \) is the segment adjacent to \( RQ \) on the hypotenuse. Wait, the formula is: in right triangle \( \triangle SRQ \), right - angled at \( R \), and \( RT \perp SQ \), then \( RQ^2=TQ\times SQ \). Wait, \( SQ = ST + TQ=9 + 16 = 25 \), \( TQ = 16 \). So \( RQ^2=16\times25 \)? No, that can't be. Wait, no, I made a mistake. Wait, the correct formula: if we have a right triangle, and we draw an altitude from the right angle to the hypotenuse, then each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg. So, \( RQ^2=TQ\times SQ \)? Wait, \( SQ=ST + TQ = 9 + 16=25 \), \( TQ = 16 \), so \( RQ^2=16\times25 \)? Then \( RQ=\sqrt{16\times25}=\sqrt{16}\times\sqrt{25}=4\times5 = 20 \)? Wait, but let's check again. Wait, the right triangle: \( \angle R = 90^{\circ} \), \( \angle RTQ = 90^{\circ} \), so \( \triangle RTQ\sim\triangle SRQ \) (by AA similarity, since \( \angle Q \) is common and both are right - angled). So, the ratio of corresponding sides: \( \frac{RQ}{SQ}=\frac{TQ}{RQ} \), so \( RQ^2=TQ\times SQ \). \( SQ = 9 + 16 = 25 \), \( TQ = 16 \), so \( RQ^2=16\times25 = 400 \), then \( RQ=\sqrt{400}=20 \).

Step2: Calculate \( x \)

We have \( RQ = x \), \( TQ = 16 \), \( SQ=9 + 16 = 25 \). From the similarity of \( \triangle RTQ \) and \( \triangle SRQ \), we have the proportion \( \frac{x}{25}=\frac{16}{x} \) (cross - multiplying gives \( x^{2}=16\times25 \)). Then \( x=\sqrt{16\times25}=\sqrt{16}\times\sqrt{25}=4\times5 = 20 \).

Answer:

20 units