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Category 2
Geometry
Meet #4 February 2007 – Practice #3
D
C
1) In the circle to the right angle ADC is 66 degrees and angle BAD
is 110 degrees. How many degrees are in the measure of angle CBA?
A
B
F
2)
Circle O at the left has diameter EG. The measure of angle FEG is 17
degrees and the measure of angle HGE is 56 degrees. What is the sum
of the measures of the angles FGE and GEH?
O
G
E
M
H
3) In circle B at the right triangle AMC is an isosceles triangle
with AM = MC. How many degrees are in the minor Arc AM?
A
C
B
4)
The two concentric circles to the left are designed such that the
area of the shaded region is equal to half the area of the
unshaded region. If radius AB is 4 cm, what is the length
of the radius(AC) of the larger circle? Give your answer as a
decimal to the nearest thousandth.
Category 2 - Geometry - Meet #4 February 2007 – Practice #3
D
1) * Since AD = AC as they are both radii of the same circle making ADC
an isosceles triangle. If ADC = 66 then ACD = 66 degrees as well. Since
the angles in triangle ADC must add up to 180 degrees and two of the
angles already add up to 66 + 66 = 132, the third angle DAC must equal
180 – 132 = 48 degrees. If angle DAC = 48 degrees then minor arc DC
also equals 48 degrees since DAC is a central angle. Angle CBD is called
an inscribed angle and is therefore equal to half the measure of the arc
it intersects. Since minor arc DC is 48 degrees, inscribed
angle CBD = half of 48 degrees or 24 degrees.
F
2)
O
G
E
H
A
B
*Any triangle inscribed in a circle that uses the circles diameter as one
of its sides is a right triangle. In this case both triangles would be right
triangles with angles EFG and angle EHG both right angles. Since the
sum of the angle is every triangle must be 180 degrees,
mFEG  mEFG  mFGE  180  and
mGEH  mHGE  mGHE  180  . Since we know four of the
angles already we can substitute what we know.
17  90  mFGE  180  and mHEG  56  90  180  . So
M
mFGE 73 and mGEH  34  .
3) * Since AC is the diameter of the circle and one side of triangle AMC,
Angle AMC must be a right angle. Since triangle AMC is isosceles,
angles MAC and MCA must be equal and both equal to 45 degrees.
A
Since angle MAC is an inscribed angle it is equal to half of the arc
in intersects, in this case minor arc AM. So minor arc AM must be
double angle MCA or 2(45) = 90 degrees
4)
C
C
B
*Since AB = 4cm, the inner circle has an area of 16  cm2.
That makes the shaded region equal to half that or 8  cm2. The
area of the larger circle then is 16  + 8  = 24  . If the area is
24  which is equal to  r2, so r2 = 24 and r = 24  4.8989
which rounds to 4.899