Question

find ab if f is the circumcenter of △aec, bf = 5.2, and fc = 12.7. round to the nearest tenth.

Explanation:

Step1: Recall Circumcenter Property

The circumcenter \( F \) of a triangle is equidistant from all vertices, and \( FB \perp AC \) (since \( B \) is the foot of the perpendicular from \( F \) to \( AC \)). So \( \triangle FBC \) is a right triangle with \( \angle FBC = 90^\circ \), \( FB = 5.2 \), \( FC = 12.7 \), and we need to find \( BC \) (wait, no, the question is to find \( AB \)? Wait, maybe \( AC \)? Wait, no, the diagram: \( F \) is circumcenter, \( FB \perp AC \), so \( B \) is midpoint? Wait, no, circumcenter is intersection of perpendicular bisectors. So \( FB \) is perpendicular bisector of \( AC \), so \( AB = BC \)? Wait, no, \( FB \perp AC \), so \( B \) is the midpoint? Wait, no, perpendicular bisector: so \( AB = BC \)? Wait, no, \( FB \) is perpendicular to \( AC \) and \( B \) is on \( AC \), so \( FB \) is the perpendicular bisector, so \( AB = BC \)? Wait, no, \( F \) is circumcenter, so \( FA = FC = FE \) (circumradius). Wait, the problem says "Find \( AB \) if \( F \) is the circumcenter of \( \triangle AEC \), \( BF = 5.2 \), and \( FC = 12.7 \)". Wait, \( FB \perp AC \) (since \( B \) has a right angle), so \( \triangle FBC \) is right-angled at \( B \). Wait, no, \( \angle FBC = 90^\circ \), so by Pythagoras, \( BC = \sqrt{FC^2 - FB^2} \)? Wait, no, \( FC \) is hypotenuse? Wait, \( FC = 12.7 \), \( FB = 5.2 \), right angle at \( B \), so \( BC = \sqrt{FC^2 - FB^2} \)? Wait, no, \( FB \) is one leg, \( BC \) is another leg, \( FC \) is hypotenuse. So \( BC = \sqrt{FC^2 - FB^2} \)? Wait, \( FC^2 = FB^2 + BC^2 \), so \( BC = \sqrt{FC^2 - FB^2} \). Wait, but the question is to find \( AB \). Wait, maybe \( AB = BC \)? Wait, no, maybe \( AC \) is the side, and \( B \) is the midpoint? Wait, no, circumcenter: perpendicular bisectors. So \( FB \) is perpendicular bisector of \( AC \), so \( AB = BC \)? Wait, no, \( B \) is on \( AC \), so \( AB = BC \) would mean \( B \) is midpoint, but also \( FB \perp AC \), so \( FB \) is perpendicular bisector, so \( AB = BC \). Wait, but the problem says "Find \( AB \)". Wait, maybe I misread: \( F \) is circumcenter of \( \triangle AEC \), so \( FA = FC = FE \). \( FB \perp AC \), so \( B \) is midpoint of \( AC \), so \( AB = BC \). Wait, but we need to find \( AB \). Wait, \( FC = 12.7 \) (circumradius), \( FB = 5.2 \), right triangle \( FBC \): \( BC = \sqrt{FC^2 - FB^2} \). Let's calculate that.

Step2: Apply Pythagorean Theorem

In right triangle \( \triangle FBC \), \( \angle FBC = 90^\circ \), so by Pythagoras:
\( BC = \sqrt{FC^2 - FB^2} \)
Substitute \( FC = 12.7 \) and \( FB = 5.2 \):
\( BC = \sqrt{12.7^2 - 5.2^2} \)
Calculate \( 12.7^2 = 161.29 \), \( 5.2^2 = 27.04 \)
Subtract: \( 161.29 - 27.04 = 134.25 \)
Take square root: \( \sqrt{134.25} \approx 11.6 \) (wait, no, \( \sqrt{134.25} = 11.59 \approx 11.6 \)? Wait, no, wait: \( 12.7^2 = 161.29 \), \( 5.2^2 = 27.04 \), \( 161.29 - 27.04 = 134.25 \), \( \sqrt{134.25} = 11.59 \approx 11.6 \). But the question is to find \( AB \). Wait, maybe \( AB = BC \)? So \( AB \approx 11.6 \)? Wait, no, maybe I made a mistake. Wait, the circumcenter: \( FA = FC = 12.7 \) (circumradius). Wait, \( FB = 5.2 \), \( \triangle FAB \) is right-angled at \( B \)? Wait, \( \angle FBA = 90^\circ \), so \( AB = \sqrt{FA^2 - FB^2} \). Since \( FA = FC = 12.7 \) (circumradius), so \( AB = \sqrt{12.7^2 - 5.2^2} \). Ah! That's the mistake. \( FA = FC = 12.7 \) (circumradius), \( FB \perp AC \), so \( \triangle FAB \) is right-angled at \( B \), so \( AB = \sqrt{FA^2 - FB^2} \).

So \( AB = \sqrt{12.7^2 - 5.2^2} \)
Calculate \( 12.7^2 = 161.29 \), \( 5.2^2 = 27.04 \)
\( 161.29 - 27.04 = 134.25 \)
\( \sqrt{134.25} \approx 11.6 \) (wait, \( \sqrt{134.25} = 11.59 \), which rounds to 11.6)

Step3: Round to Nearest Tenth

The value \( \sqrt{134.25} \approx 11.6 \) when rounded to the nearest tenth.

Answer:

\( 11.6 \)