32. The figure shows a circle with center

*P*and inscribed isosceles triangle

*ABC*.

If

*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*,

*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*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**~~AC~~, namely ~~AC~~, still without changing the radius, and then connect this point through

*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*P*. Indeed, we can use the U of Chicago method and just mark arcs below*P*to the other side of the circle,*B*.

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