Question

1. c is the circumcenter of δxyz. find the lengths indicated.

cz =
xy =
cl =
zn =
cy =
zl =
khoa pham © 2025

Explanation:

Step1: Recall Circumcenter Properties

The circumcenter \( C \) of a triangle is equidistant from all vertices (\( CZ = CX = CY \)) and is the intersection of the perpendicular bisectors, so it bisects the sides (e.g., \( L \) bisects \( ZX \), \( M \) bisects \( XY \), \( N \) bisects \( ZY \)).

Step2: Find \( CZ \)

First, use the right triangle with legs \( 13 \) ( \( CN \)) and \( 84 \) ( \( NY \), since \( N \) bisects \( ZY \), \( ZN = NY = 84 \)? Wait, no, \( N \) is on \( ZY \), and \( C \) is circumcenter, so \( CN \) is part of the perpendicular bisector? Wait, actually, let's look at \( \triangle CNY \): \( CN = 13 \), \( NY = 84 \), so \( CY = \sqrt{13^2 + 84^2} \)? Wait, no, maybe \( ZY \) is a side, and \( N \) is the midpoint? Wait, the diagram shows \( ZN = 84 \)? Wait, no, the label is \( 84 \) from \( N \) to \( Y \), so \( ZN = NY = 84 \)? Wait, no, the length from \( N \) to \( Y \) is \( 84 \), so \( ZN = NY = 84 \)? Wait, maybe \( L \) is midpoint of \( ZX \), so \( ZL = LX = 75 \), so \( ZX = 150 \). \( M \) is midpoint of \( XY \), so \( XM = MY = 80 \), so \( XY = 160 \). \( N \) is midpoint of \( ZY \), so \( ZN = NY = 84 \), so \( ZY = 168 \). Now, \( C \) is circumcenter, so \( CZ = CX = CY \). Let's compute \( CY \): in \( \triangle CMY \), \( CM \) is? Wait, no, \( C \) to \( Y \): since \( M \) is midpoint of \( XY \), \( XM = 80 \), and \( CX = CY \). Wait, maybe use \( \triangle CNY \): \( CN = 13 \), \( NY = 84 \), so \( CY = \sqrt{13^2 + 84^2} = \sqrt{169 + 7056} = \sqrt{7225} = 85 \). So \( CZ = CY = 85 \).

Step3: Find \( ZN \)

Since \( N \) is the midpoint of \( ZY \) (circumcenter is on perpendicular bisector), \( ZN = NY = 84 \).

Step4: Find \( XY \)

\( M \) is the midpoint of \( XY \), so \( XY = 2 \times XM = 2 \times 80 = 160 \).

Step5: Find \( CY \)

As calculated earlier, \( CY = 85 \) (since \( CZ = CX = CY \), and using \( \triangle CNY \) with legs \( 13 \) and \( 84 \), hypotenuse \( 85 \)).

Step6: Find \( CL \)

\( L \) is midpoint of \( ZX \), so \( ZL = LX = 75 \). \( C \) is circumcenter, so \( CL \) is the distance from \( C \) to \( ZX \). In \( \triangle CLX \), \( CX = 85 \) (from \( CZ = 85 \)), \( LX = 75 \), so \( CL = \sqrt{CX^2 - LX^2} = \sqrt{85^2 - 75^2} = \sqrt{(85 - 75)(85 + 75)} = \sqrt{10 \times 160} = \sqrt{1600} = 40 \). Wait, no, \( CL \) is the perpendicular distance? Wait, maybe \( CL \) is calculated as \( \sqrt{CZ^2 - ZL^2} = \sqrt{85^2 - 75^2} = 40 \).

Step7: Find \( ZL \)

\( L \) is midpoint of \( ZX \), so \( ZL = \frac{ZX}{2} \). \( ZL = 75 \) (given \( LZ = 75 \), so \( ZL = 75 \)).

Answer:

\( CZ = 85 \)
\( ZN = 84 \)
\( XY = 160 \)
\( CY = 85 \)
\( CL = 40 \)
\( ZL = 75 \)