*The following problems were taken from the*

**GEOMETRY (COMMON CORE)**Regents Exam given on Thursday, January 26, 2017.Previous problems can be found here.

### Part 1

**21. In the diagram below of circle O, GO 8 and m∠GOJ = 60°. What is the area, in terms of π, of the shaded region?
**

(4) 160π / 3.

Since 60° is 1/6th of the 360° degree in the complete circle, then the unshaded region of the circle is (1/6) πr^{2} = (1/6) π8^{2}

and the shaded portion would be (5/6) πr^{2} = (5/6) π8^{2} = (5 * 64π) / 6 = (5 * 32π) / 3 = 160π/ 3.

**22. A circle whose center is the origin passes through the point (5,12).**

Which point also lies on this circle?

(3) (11, 2 sqrt(12))

The equation of the circle is *x*^{2} + *y*^{2} = *r*^{2}. We can find *r* using the *Distance Formula* or *Pythagorean Theorem*: *5*^{2} + *12*^{2} = *r*^{2}.

25 + 144 = 169 = r^{2}

r = 13 (which you really should have known. Look up **Pythagorean triples**.)

Which of the other points creates a right triangle with a hypotenuse of 13?

(10, 3) definitely do not -- they don't even create a triangle with a side of 13. (-12, 13) can't because the hypotenuse is longer than the legs (plus it would have to be -12 and either 5 or -5).

Check 11^{2} + (2sqrt(12))^{2} = 121 + 4(12) = 169 = 13^{2}. Looks good.

Check (-8)^{2} + (5sqrt(21))^{2} = 64 + 25(21) = way too much. No good.

Continue to the next problems.

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