PARCC Practice Test Question 32 (Day 175)

Question 32 of the PARCC Practice Exam is yet another question on inscribed angles:

32. The figure shows a circle with center P and inscribed isosceles triangle ABC.

If AC has the same length as the radius of the circle, what is the measure of angle ABC?

This question is not difficult if we remember the construction of a regular hexagon. We found out that a chord congruent to a radius subtends an arc of measure 60 degrees -- a sixth of the circumference, which explains why it leads to a regular hexagon. Then angle ABC, as an inscribed angle, would measure half of the subtended arc, or 30 degrees.

Notice that we made an assumption here -- in isosceles triangle ABC, AC is the base rather than one of the congruent legs. Of course, we made this assumption because of the way the triangle is drawn -- but in Geometry, we're not supposed to make conclusions based on mere drawings. Okay then -- suppose AC were a leg rather than a hypotenuse. That doesn't change the fact arc AC is still 60 degrees, and so inscribed angle ABC is still half of this measure, or 30 degrees.

Other than that, this should be a simple problem. I'm not sure what the most common student error will be -- 60, maybe, or otherwise the student doesn't know how to solve this problem at all.

PARCC Practice EOY Question 32
U of Chicago Correspondence: Lesson 15-3, The Inscribed Angle Theorem

Key Theorem: Inscribed Angle Theorem

In a circle, the measure of an inscribed arc is one-half the measure of its intercepted arc.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.C.A.2
Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.

Commentary: Today is our final activity day. I just couldn't resist having another construction activity, to construct the figure in the PARCC problem, which is a 30-75-75 triangle inscribed in a circle. To construct the figure, we start out just as we would for a regular hexagon -- draw a circle, mark a point A, then without changing the radius, mark another point C. The easiest way to locate B is to construct the perpendicular bisector of AC. We can take advantage of the fact that we already have a point on the perpendicular bisector of AC, namely P. Indeed, we can use the U of Chicago method and just mark arcs below AC, still without changing the radius, and then connect this point through P to the other side of the circle, B.