Question

1. if g is the circumcenter of (\triangle abc), find each measure.

a) (ad=)
b) (fc=)
c) (eb=)
d) (ag=)
e) (eg=)

1. if z is the circumcenter of (\triangle qrs), find each measure.

Explanation:

Step1: Recall the property of circum - center

The circum - center of a triangle is equidistant from the vertices of the triangle. Also, the perpendicular bisectors of the sides of the triangle pass through the circum - center.

Step2: Find \(AD\)

Since \(G\) is the circum - center and \(CD\) is a perpendicular bisector of \(AB\) (by the property of circum - center), \(AD = BD\). Given \(AB=18\), and \(E\) is the mid - point of \(AB\) (because the perpendicular bisector from the circum - center to a side bisects the side). But if we consider the length related to \(AC = 30\) and using the property of circum - center (the perpendicular bisector), we know that \(AD=\frac{1}{2}AB\). Wait, no, actually, since \(G\) is the circum - center and \(CD\) is the perpendicular bisector of \(AB\), \(AD=\frac{1}{2}AB\). But wait, no, looking at the figure (assuming the standard circum - center perpendicular bisector property), if \(E\) is the mid - point of \(AB\) (because the line from the circum - center to the side is a perpendicular bisector), \(AE = EB\). But for \(AD\), since \(G\) is the circum - center and \(CD\) is the perpendicular bisector of \(AB\), \(AD=\frac{1}{2}AB\). Wait, no, actually, if \(G\) is the circum - center, the perpendicular bisectors of the sides of the triangle. Let's use the property:
- For \(AD\): Since \(G\) is the circum - center and \(CD\) is the perpendicular bisector of \(AB\), \(AD=\frac{1}{2}AB\). But wait, no, if \(E\) is the mid - point of \(AB\) (because the line from \(G\) to \(AB\) is a perpendicular bisector), \(AE=EB = 9\) (wait no, \(AB = 18\), so \(AE=EB=\frac{AB}{2}=9\)). Wait, no, hold on. Wait, the circum - center is the intersection of the perpendicular bisectors. So \(CD\) is the perpendicular bisector of \(AB\), so \(AD = BD\). But if \(AB = 18\), \(AD=\frac{AB}{2}=9\). Wait, no, no, wait, the length \(AC = 30\) is a red - herring. The key is that the perpendicular bisector of \(AB\) (passing through \(G\)) bisects \(AB\). So \(AD=\frac{AB}{2}\). But \(AB = 18\), so \(AD = 9\).

Step3: Find \(FC\)

Since \(G\) is the circum - center and \(BF\) is the perpendicular bisector of \(AC\) (by the property of circum - center), \(FC=\frac{1}{2}AC\). Given \(AC = 30\), so \(FC=\frac{30}{2}=15\)

Step4: Find \(EB\)

Since \(G\) is the circum - center and \(GE\) is the perpendicular bisector of \(AB\) (by the property of circum - center), \(EB=\frac{1}{2}AB\). Given \(AB = 18\), so \(EB = 9\)

Step5: Find \(AG\)

Since \(G\) is the circum - center, \(AG=BG = CG\). Using the Pythagorean theorem in \(\triangle BFG\) (where \(BF\) is the perpendicular bisector of \(AC\), \(BF = 22\), \(FG = 9\)). First, find \(BG\) (since \(AG = BG\)). In right - triangle \(BFG\), \(BG=\sqrt{BF^{2}+FG^{2}}\) (by the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(a = 22\), \(b = 9\), \(c = BG\)). So \(BG=\sqrt{22^{2}+9^{2}}=\sqrt{484 + 81}=\sqrt{565}\approx23.77\). But wait, no, wait, if \(G\) is the circum - center and \(BF\) is the perpendicular bisector of \(AC\), \(FC = 15\) (from step 3). Wait, no, hold on, we can also use the fact that \(AG = BG\). In \(\triangle BEG\), \(BE = 9\) (from step 4), \(BG\) (let's calculate using \(BG\) in \(\triangle BEG\) where \(BE = 9\), \(BG\) is the hypotenuse. Wait, no, in \(\triangle BFG\), \(BF\) is the length of the perpendicular bisector segment from \(B\) to \(AC\) (length \(22\)), \(FG\) is the distance from \(G\) to \(AC\) (length \(9\)). Then \(BG=\sqrt{22^{2}+9^{2}}=\sqrt{484 + 81}=\sqrt{565}\approx23.77\). But wait, another approach: since \(G\) is the circum - center, \(AG = BG\). In right - triangle \(BEG\) (where \(BE=\frac{AB}{2}=9\), and \(BG\) can be calculated from \(\triangle BFG\) (where \(BF = 22\), \(FG = 9\)). Wait, no, \(BF\) is not related to \(BE\) directly. Wait, correct formula: \(AG = BG\). In right - triangle \(BFG\) ( \(BF\) is the length of the perpendicular from \(B\) to \(AC\)'s bisector, \(FG\) is the distance from \(G\) to \(AC\)'s bisector). \(BG=\sqrt{22^{2}+9^{2}}=\sqrt{484+81}=\sqrt{565}\approx23.77\). But wait, if we consider the fact that \(AG = BG\) and using the Pythagorean theorem in \(\triangle BEG\) ( \(BE = 9\), \(EG\) (we will find \(EG\) next)). Wait, no, let's do it properly.
\[AG=\sqrt{AE^{2}+EG^{2}}\]
First, find \(EG\). In \(\triangle BFG\), \(BG^{2}=BF^{2}+FG^{2}\), \(BG^{2}=22^{2}+9^{2}=484 + 81=565\). Also, in \(\triangle BEG\), \(BG^{2}=BE^{2}+EG^{2}\). Since \(BE = 9\), \(EG^{2}=BG^{2}-BE^{2}=565 - 81 = 484\), \(EG = 22\). Then \(AG=\sqrt{AE^{2}+EG^{2}}=\sqrt{9^{2}+22^{2}}=\sqrt{81 + 484}=\sqrt{565}\approx23.77\)

Step6: Find \(EG\)

In \(\triangle BFG\), \(BG^{2}=BF^{2}+FG^{2}\) ( \(BG^{2}=565\)). In \(\triangle BEG\), \(BG^{2}=BE^{2}+EG^{2}\). Since \(BE = 9\) (from step 4), \(EG^{2}=BG^{2}-BE^{2}\). \(EG^{2}=565-81 = 484\), so \(EG = 22\)

Answer:

a) \(AD = 9\)
b) \(FC = 15\)
c) \(EB = 9\)
d) \(AG=\sqrt{565}\approx23.8\)
e) \(EG = 22\)