Question

if xy = 30, xz = 24, jq = 8, and the radius of the circumscribed circle of δxyz is 17, find qk. qk = (simplify your answer. round to one decimal place as needed.)

Explanation:

Step1: Identify the inradius and circumradius

The circumradius \( R = 17 \), \( JQ = 8 \) (which is the inradius \( r \)? Wait, no, \( Q \) is the incenter? Wait, the diagram has perpendiculars from \( Q \) to the sides, so \( Q \) is the incenter, and \( JQ, QK, QL \) are inradii? Wait, no, the problem says "circumscribed circle" (circumcircle) radius is 17. Wait, maybe \( Q \) is the circumcenter? Wait, no, the perpendiculars suggest incenter, but the problem mentions circumscribed circle. Wait, maybe there's a mix-up. Wait, let's recall the formula for the distance between incenter and circumcenter, but maybe not. Wait, alternatively, maybe \( Q \) is the circumcenter? No, circumcenter is the intersection of perpendicular bisectors. Wait, the diagram has \( K \) on \( XZ \), \( J \) on \( XY \), \( L \) on \( YZ \), with \( QK \perp XZ \), \( JQ \perp XY \), \( QL \perp YZ \), so \( Q \) is the incenter, and \( JQ = QK = QL = r \) (inradius). But the problem says "radius of the circumscribed circle is 17", so \( R = 17 \). Wait, maybe we need to use the formula for the distance between incenter (\( I \)) and circumcenter (\( O \)): \( d^2 = R(R - 2r) \), but here maybe \( Q \) is the incenter, and we need to find \( QK \), but \( JQ = 8 \), so if \( Q \) is the incenter, then \( QK = JQ = 8 \)? But that can't be, because the problem gives circumradius. Wait, maybe I misinterpret the diagram. Wait, maybe \( Q \) is the circumcenter? No, circumcenter is on perpendicular bisectors. Wait, \( XZ = 24 \), so the midpoint of \( XZ \) is \( K \) if \( QK \) is a perpendicular bisector? Wait, the diagram has \( XK = KZ \) (since \( XZ \) has two marks), so \( K \) is the midpoint of \( XZ \), so \( QK \) is the perpendicular bisector of \( XZ \), so \( Q \) is the circumcenter! Ah, that makes sense. So \( Q \) is the circumcenter, so \( QX = QY = QZ = R = 17 \). \( XZ = 24 \), so \( XK = \frac{XZ}{2} = 12 \) (since \( K \) is the midpoint, as \( XZ \) is bisected). Then in right triangle \( QKX \), \( QX = 17 \), \( XK = 12 \), so by Pythagoras, \( QK = \sqrt{QX^2 - XK^2} = \sqrt{17^2 - 12^2} = \sqrt{289 - 144} = \sqrt{145} \approx 12.0 \)? Wait, no, \( JQ = 8 \), what's \( JQ \)? \( J \) is on \( XY \), \( JQ \perp XY \), \( XY = 30 \), so midpoint of \( XY \) is \( J \)? Wait, \( XY \) has three marks, maybe \( XJ = JY = 15 \) (since \( XY = 30 \)). Then in right triangle \( QJX \), \( QX = 17 \), \( XJ = 15 \), so \( JQ = \sqrt{QX^2 - XJ^2} = \sqrt{17^2 - 15^2} = \sqrt{289 - 225} = \sqrt{64} = 8 \), which matches the given \( JQ = 8 \). So that confirms \( Q \) is the circumcenter, \( K \) is the midpoint of \( XZ \), so \( XK = 12 \), \( QX = 17 \), so in right triangle \( QKX \), \( QK = \sqrt{QX^2 - XK^2} = \sqrt{17^2 - 12^2} = \sqrt{289 - 144} = \sqrt{145} \approx 12.0 \)? Wait, no, \( \sqrt{145} \approx 12.0416 \), rounded to one decimal is 12.0? But wait, \( JQ = 8 \) was found using \( XY = 30 \), midpoint \( J \), \( XJ = 15 \), \( QX = 17 \), so \( JQ = 8 \), which matches. Now, for \( XZ = 24 \), midpoint \( K \), \( XK = 12 \), \( QX = 17 \), so \( QK = \sqrt{17^2 - 12^2} = \sqrt{289 - 144} = \sqrt{145} \approx 12.0 \) (wait, \( \sqrt{144} = 12 \), \( \sqrt{145} \approx 12.04 \), so 12.0 when rounded to one decimal? Wait, no, \( 17^2 = 289 \), \( 12^2 = 144 \), \( 289 - 144 = 145 \), \( \sqrt{145} \approx 12.0416 \), so 12.0? But that seems off. Wait, maybe I made a mistake. Wait, \( Q \) is the circumcenter, so \( QX = 17 \), \( XK = 12 \), right triangle, so \( QK = \sqrt{17^2 - 12^2} = \sqrt{145} \approx 12.0 \). But the problem says "JQ = 8", which we used for \( XY \), and that worked. So \( QK \) is the length from circumcenter to \( XZ \), which is the distance from \( Q \) to \( XZ \), calculated as above.

Step1: Find \( XK \)

Since \( K \) is the midpoint of \( XZ \) (as \( XZ \) is bisected, two marks), \( XK = \frac{XZ}{2} = \frac{24}{2} = 12 \).

Step2: Apply Pythagoras in \( \triangle QKX \)

\( QX \) is the circumradius, so \( QX = 17 \). In right triangle \( QKX \), \( QK = \sqrt{QX^2 - XK^2} \).

Substitute \( QX = 17 \) and \( XK = 12 \):

\( QK = \sqrt{17^2 - 12^2} = \sqrt{289 - 144} = \sqrt{145} \approx 12.0 \) (rounded to one decimal).

Wait, but \( JQ = 8 \) was for \( XY \), let's check that: \( XY = 30 \), so \( XJ = 15 \), \( QX = 17 \), so \( JQ = \sqrt{17^2 - 15^2} = \sqrt{289 - 225} = \sqrt{64} = 8 \), which matches. So that confirms \( Q \) is the circumcenter, \( K \) is midpoint of \( XZ \), so \( QK \) is calculated as above.

Answer:

\( \boxed{12.0} \)