Geometry Problems of the Day (Geometry Regents, August 2012)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Geometry Regents, August 2012

Part IV: A correct answer will receive 6 credits. Partial credit is available.


38. Chords AB and CD intersect at E in circle O, as shown in the diagram below. Secant FDA and tangent FB are drawn to circle O from external point F and chord AC is drawn. The mDA = 56, mDB = 112, and the ratio of mAC : mCD = 3:1.

Determine m∠CEB.

Determine m∠F.

Determine m∠DAC.

Answer:

Rules to remember: an inscribed angle is half the size of the arc it intercepts. Where two chords intersect, the vertical angles created will be equal to the average of the two arcs the chords intercept on either side. Finally, when two secants, or two tangents, or a tangent and a secant intercept a circle, the size of the angle of the two secants (or whatever) will be one-half of the difference between the size of the FAR (the bigger, more remote) arc minus the NEAR (the smaller, closer) arc.

In the circle given, you have measures for arcs AD and DB, 56 and 112, respectively, which total 168 degrees. There are 360 degrees total in the circle, so what remains, 360 - 168 = 192 is the sum of AC and CB. Note that arc ACB is a major arc, greater than a semicircle, so the number 192 looks reasonable.

We also know that AC:CB = 3:1, so AC = 3/4(192) = 144, and CB = 1/4(192) = 48.

m∠CEB = 1/2 (m CB + m DA) = 1/2(48 + 56) = 52 degrees.

m∠F = 1/2 (m ACD - m DB) = 1/2(192 - 112) = 40 degrees.

Angle DAC intercepts arc DBC, which in the sum of arcs DB and BC. Intercepted angles are half the size of the arc they intercept. So m∠DAC = 1/2 (112 + 48) = 80 degrees.

Can check your work here because the angles of triangle FAB must add up to 180. Angle F = 40, angle DAB is 1/2(112) = 56, and angle ABF is 1/2(168) = 84, and 40 + 56 + 84 = 180.

End of Exam.

More to come. Comments and questions welcome.

More Regents problems.

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