Question

practice!

1. if z is the circumcenter of $\triangle tuv$, find each missing measure.

a) $tu = $
b) $vy = $
c) $uz = $
d) $wv = $
e) $tv = $

1. if h is the circumcenter of $\triangle bcd$, find each missing measure.

a) $gd = $
b) $bc = $
c) $eh = $
d) $fd = $
e) $cd =

Answer:

To solve these problems, we use the property of the circumcenter: the circumcenter of a triangle is the intersection of the perpendicular bisectors of the triangle's sides, so it is equidistant from all three vertices, and it bisects each side (i.e., the distance from the circumcenter to a vertex is the same for all vertices, and the segments from the circumcenter to the midpoint of a side are equal, and the circumcenter bisects each side into two equal parts).

Problem 1: \( Z \) is the circumcenter of \( \triangle TUV \)

We analyze each part:

a) \( TU \)

From the diagram, \( TU \) is given as \( 19 \)? Wait, no—wait, the segment from \( T \) to \( U \) is labeled \( 19 \)? Wait, actually, the diagram shows \( x \) and \( 19 \), but maybe \( TU = 19 \)? Wait, no—wait, the circumcenter \( Z \) implies that \( Z \) is on the perpendicular bisector of \( TU \), but actually, looking at the diagram, \( TU \) is a side, and if \( Z \) is the circumcenter, but maybe \( TU \) is given as \( 19 \)? Wait, maybe the diagram has \( TU = 19 \), so \( TU = 19 \).

b) \( VY \)

\( Y \) is the midpoint of \( UV \) (since \( Z \) is the circumcenter, it lies on the perpendicular bisector of \( UV \), so \( Y \) is the midpoint). Thus, \( VY = YU \), and \( UV = 34 \), so \( VY = \frac{34}{2} = 17 \).

c) \( UZ \)

The circumcenter is equidistant from all vertices, so \( UZ = VZ = TZ \). From the diagram, \( VW = 15 \), \( WZ = 21 \)? Wait, no—wait, \( VZ \) is a segment from \( V \) to \( Z \), but actually, \( Z \) is equidistant from \( T \), \( U \), \( V \). Wait, maybe \( VZ = TU \)? No, wait, let's check the lengths. Wait, \( UV = 34 \), so \( VY = 17 \) (midpoint). Then \( UZ \) should be equal to \( VZ \). Wait, maybe \( VZ \) is calculated via Pythagoras? Wait, \( VW = 15 \), \( WZ = 21 \)? No, maybe \( VZ \) is \( \sqrt{15^2 + 21^2} \)? Wait, no—wait, the diagram shows \( VW = 15 \), \( WZ = 21 \), but maybe \( Z \) is the circumcenter, so \( UZ = VZ = TZ \). Wait, maybe \( UZ = 19 \)? No, wait, \( TU = 19 \), and \( Z \) is equidistant from \( T \) and \( U \), so \( TZ = UZ \), but \( TU = 19 \) is a side. Wait, maybe I misread. Let's re-express:

Wait, the circumcenter \( Z \) implies \( ZT = ZU = ZV \). From the diagram, \( TU = 19 \) (so \( ZT = ZU \), but \( TU \) is a side, not a segment from \( Z \). Wait, maybe the length \( x \) is \( 19 \), so \( TU = 19 \).

d) \( WV \)

Wait, \( W \) is the midpoint of \( TV \) (since \( Z \) is on the perpendicular bisector of \( TV \)), so \( WV = WT \). But the diagram shows \( WV = 15 \)? Wait, the diagram labels \( WV = 15 \), so \( WV = 15 \).

e) \( TV \)

Since \( W \) is the midpoint (because \( Z \) is the circumcenter, so \( ZW \) is the perpendicular bisector of \( TV \)), \( TV = 2 \times WV = 2 \times 15 = 30 \).

Problem 2: \( H \) is the circumcenter of \( \triangle BCD \)

We analyze each part:

a) \( GD \)

\( G \) is the midpoint of \( BD \) (since \( H \) is the circumcenter, so \( HG \) is the perpendicular bisector of \( BD \)). Thus, \( GD = BG \), and \( BD = 24 \) (from the diagram: \( BG = 24 \)? Wait, no—\( BD \) is labeled \( 24 \), so \( GD = \frac{24}{2} = 12 \). Wait, the diagram shows \( G \) is the midpoint, so \( BG = GD = 12 \). Wait, the diagram has \( BG = 24 \)? No, the diagram says \( G \) is at \( 24 \), so \( BD = 24 \), so \( GD = \frac{24}{2} = 12 \).

b) \( BC \)

\( E \) is the midpoint of \( BC \) (since \( H \) is the circumcenter, so \( HE \) is the perpendicular bisector of \( BC \)). Thus, \( BC = 2 \times EC \), and \( EC = 7 \) (from the diagram), so \( BC = 2 \times 7 = 14 \).

c) \( EH \)

We know \( BC = 14 \), so \( EC = 7 \). \( H \) is the circumcenter, and \( BH = CH = DH \). From the diagram, \( BH = 11 \) (segment \( BH = 11 \)), so \( CH = 11 \). Then \( EH \) is the distance from \( E \) to \( H \) in \( \triangle ECH \), where \( EC = 7 \), \( CH = 11 \). By Pythagoras:
\( EH = \sqrt{CH^2 - EC^2} = \sqrt{11^2 - 7^2} = \sqrt{121 - 49} = \sqrt{72} = 6\sqrt{2} \)? Wait, no—wait, maybe \( EH \) is calculated as \( \sqrt{11^2 - 7^2} \)? Wait, no, maybe \( BH = 11 \), \( EH \) is a segment. Wait, maybe the diagram has \( EH \) as a missing measure, and \( CH = 11 \), \( EC = 7 \), so \( EH = \sqrt{11^2 - 7^2} = \sqrt{72} = 6\sqrt{2} \approx 8.485 \), but maybe I misread. Wait, the diagram shows \( H \) to \( F \) is \( 3 \), but maybe \( EH \) is \( \sqrt{11^2 - 7^2} \). Wait, no—maybe \( BH = 11 \), \( EH \) is \( \sqrt{11^2 - 7^2} = \sqrt{72} = 6\sqrt{2} \), but maybe the diagram has \( EH = \sqrt{11^2 - 7^2} \).

d) \( FD \)

\( F \) is the midpoint of \( CD \) (since \( H \) is the circumcenter, so \( HF \) is the perpendicular bisector of \( CD \)). Thus, \( FD = FC \). From the diagram, \( HF = 3 \), and \( DH = BH = 11 \) (since \( H \) is equidistant from all vertices). Then \( FD \) is the distance from \( F \) to \( D \) in \( \triangle FHD \), where \( DH = 11 \), \( HF = 3 \). By Pythagoras:
\( FD = \sqrt{DH^2 - HF^2} = \sqrt{11^2 - 3^2} = \sqrt{121 - 9} = \sqrt{112} = 4\sqrt{7} \approx 10.583 \), but maybe \( FD = FC \), and \( CD \) is calculated later. Wait, maybe \( FD = \sqrt{11^2 - 3^2} = \sqrt{112} = 4\sqrt{7} \).

e) \( CD \)

Since \( F \) is the midpoint of \( CD \), \( CD = 2 \times FD \). From part (d), \( FD = \sqrt{11^2 - 3^2} = \sqrt{112} \), so \( CD = 2 \times \sqrt{112} = 2 \times 4\sqrt{7} = 8\sqrt{7} \approx 21.166 \).

Summary of Answers (Problem 1)

a) \( TU = 19 \)
b) \( VY = 17 \)
c) \( UZ = 19 \) (assuming \( UZ = TU \), or via circumcenter equidistance)
d) \( WV = 15 \)
e) \( TV = 30 \) (since \( WV = 15 \), \( TV = 2 \times 15 = 30 \))

Summary of Answers (Problem 2)

a) \( GD = 12 \)
b) \( BC = 14 \)
c) \( EH = \sqrt{11^2 - 7^2} = \sqrt{72} = 6\sqrt{2} \approx 8.49 \) (or exact form \( 6\sqrt{2} \))
d) \( FD = \sqrt{11^2 - 3^2} = \sqrt{112} = 4\sqrt{7} \approx 10.58 \)
e) \( CD = 2 \times FD = 8\sqrt{7} \approx 21.17 \)

(Note: Some answers depend on the exact diagram labels, so adjust if the diagram has different measurements. The key is using the circumcenter property: perpendicular bisector, so midpoints and equal distances to vertices.)