Question

if xy = 24, xz = 22, and jq = 9, find the radius of the circumscribed circle of δxyz. the radius of the circumscribed circle of δxyz is \boxed{}. (simplify your answer.)

Explanation:

Step1: Identify the circumcenter

The point \( Q \) is the circumcenter (intersection of perpendicular bisectors) of \( \triangle XYZ \). \( JQ \) is the distance from circumcenter to \( XY \), but we can use the perpendicular bisector on \( XZ \). The length of \( XK=\frac{XZ}{2}=\frac{22}{2} = 11\) (since \( K \) is the midpoint of \( XZ \) as \( QK \) is perpendicular bisector).

Step2: Use the right triangle

Let the radius be \( r \) (distance from \( Q \) to any vertex, say \( X \)). Consider right triangle \( XJQ \)? Wait, no, better: \( J \) is midpoint of \( XY \) (since \( JQ \) is perpendicular bisector of \( XY \), as marks show \( XJ = JY \)). So \( XJ=\frac{XY}{2}=\frac{24}{2}=12 \), \( JQ = 9 \), and \( XQ \) is the radius \( r \).

Step3: Apply Pythagorean theorem

In right triangle \( XJQ \), \( XQ^2=XJ^2 + JQ^2 \). So \( r^2=12^2 + 9^2 \). Calculate \( 12^2 = 144 \), \( 9^2 = 81 \), sum is \( 144 + 81 = 225 \). Then \( r=\sqrt{225}=15 \). Wait, but wait: Wait, is \( XQ \) the radius? Wait, the circumradius is the distance from circumcenter to a vertex. Since \( Q \) is circumcenter, \( XQ = YQ = ZQ = r \). And \( J \) is midpoint of \( XY \), so \( XJ = 12 \), \( JQ = 9 \), right angle at \( J \), so \( XQ=\sqrt{12^2 + 9^2}=15 \). But wait, check \( XZ \): midpoint \( K \), \( XK = 11 \), \( QK \) is perpendicular. Let's see if \( QK \) can be found, but maybe my initial assumption is wrong? Wait, no, the circumcenter is the intersection of perpendicular bisectors. So \( JQ \) is perpendicular bisector of \( XY \), \( QK \) perpendicular bisector of \( XZ \), \( LQ \) perpendicular bisector of \( YZ \). So \( Q \) is circumcenter. Then \( XQ \) is radius. So using \( XJ = 12 \), \( JQ = 9 \), right triangle, so \( XQ = 15 \). But wait, let's check with \( XZ \): \( XK = 11 \), if \( XQ = 15 \), then \( QK=\sqrt{15^2 - 11^2}=\sqrt{225 - 121}=\sqrt{104}\), which is okay. So the radius is 15? Wait, but wait, maybe I mixed up inradius and circumradius? Wait, no: inradius is distance from incenter to side, circumradius is distance from circumcenter to vertex. The diagram has perpendicular bisectors, so \( Q \) is circumcenter. So \( XQ \) is circumradius. So calculation: \( XJ = 12 \), \( JQ = 9 \), right triangle, so \( r = \sqrt{12^2 + 9^2} = 15 \).

Answer:

15