Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – Circles & Polygons Topic Test – Theta Division
1. (A) Each side is 2. The area = 1.5(2) 2 3  6 3 .
2.
(B) The interior angle is 177 degrees; the exterior
angle is 3 degrees. Number of sides = 360/3 = 120.
3.
(C) The perimeter is unchanged.
14. (A) Forget the actual measurements – of the
hexagon. Let’s go general to save our sanity. Call
each side of the hexagon 2X. Then we are worried
about the shape shown below in quadrant I.
4. (A) The area is half the product of the diagonals.
5. (C) Corresponding angles of similar figures are
congruent.
6. (C) Any interior angle and exterior angle are
supplementary in a regular polygon.
7. C) The inscribed right angle forms a 180 arc, so
the hypotenuse of the triangle is the diameter of the
circle. Pythagoras takes care of the rest.
8. (B) The radius is half the diagonal. Each side is
4 2 , and the area is 32.
9. (C) mBGF  50 by vertical angles. An exterior
angle equals the sum of the non-adjacent interior
angles, so 110  mB  50 , and the answer is 60.
10. (B) Consecutive angles of a parallelogram are
supplementary. 3x  3 y  180 , then x  y  60 .
Rotating about the x-axis will yield a cone atop a
cylinder. The cone has volume


volume  X 3
12. (A) If each side of the square is x, so is the midline.
The area of the square is x 2 . The hypotenuse is x,
and each side is
x
. The area of the triangle is
2
2
1 x 
x2

, and the ratio of the areas is 1/4.
2  2 
4
 M  2
 M 
  r  48 and 
 (2 r )  8
 360 
 360 
M
Let’s solve both for
and set the results equal to
360
13. (B) 
each other.
48 4

 48r  4r 2  r  12 . Plug back
r2 r
into either formula to find M  120 .
 ( X )  3X  . The total volume
2
3
is then 4 X 3 .
Rotating about the y-axis yields a frustum. Its
volume is

 X 3
3
X
2
 X (2 X )  (2 X )
2


7 X 3 3
.
3
Squaring both and setting the first atop the second
11. (C) A hypotenuse of 30 would give a non-integral
leg of 10 5 by the Pyth. Theorem.

2
1
 X 3 ( X )  X 3 . The cylinder has
3
yields
16 X 6 2
48
1

49 2 X 6 (3) 49
9
15. (C) Let k be the constant of proportionality. The
area of the large circle is 25k 2 . The area of the
“ring” is 16k 2  9k 2  7 k 2 . The probability of
7k 2
7

hitting the shaded area is
.
2
25k  25
16. (B) Opposite angles of a convex trapezoid are
supplementary.  L is opposite J .
17. (A) n(n  3) / 2  9(6) / 2  27
18. (B) The sum of the angles of a pentagon is
180(5-2)=540. So x  (1/ 5)(540  x) ,and x = 90.
19. (B) The diameter of the small circle is 2, and so is
each side of the square. The diagonal of the square
Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – Circles & Polygons Topic Test – Theta Division
is 2 2 , and the radius of the large circle is half
that. Its area is 2 .
20. (B) The area of a regular polygon is half the product
of its apothem and perimeter.
0.5(2.2)(16)  17.6  18 .
21. (D) The slopes of y= x+2 and y=-x are negative
reciprocals, so the lines are perpendicular. The
triangle is isosceles, because it’s symmetric about
the line x = -1. The distance formula confirms this.
36 3 . The sector has an area of
120 2
(12 )  48 . x  y  84 .
360
28. (A) Note that CB || AD . ABC  130 by
IASST. Arc BDC  260 . BD is a diameter; its arc
is 180 degrees. So arc CD = 260-180 = 80.
22. (A) The square in the middle has an area of 4. Each
angle of a regular octagon is 135 , so the white
areas are 45-45-90 triangles. The 4 rectangles have
sides measuring 2 and 2 , and areas of 2 2 . The


sum = 4 2 2  4  4  8 2 .
23. (D) The ratio of the perimeters is the square root of
the area ratio.
29. (C) Let EF  x , and r = the radius of the circles.
APB is isosceles, so mPAE  mPBF .
AE  BF  (r  x) . So APE  BPF by SAS.
Now EPF is isosceles. Draw a line which
perpendicularly bisects EF at G. Now
APG  BPG  BQG  AQG by HA. Then
QG  PG , and the 4 small right triangles are
congruent by SSS. The corresponding angles at A
and B are all congruent, so the 4 obtuse triangles are
congruent by SAS. Finally, the four sides of the
quadrilateral in question are congruent.
24. (A) Note that the vertices of the polygon do not
necessarily have to be among the listed points.
A possible quadrilateral is shown above.
 ( s  a )( s  b)( s  c)

s
 (9  5)(9  6)(9  8) 15 5


9
9
3
25. (C)
26. (D) The circumference of the circle is 36 , and the
arc between each consecutive vertex of the 18-gon
measures 2 . We go across 12 such arcs, and the
sum is 24 .
27. (D) The triangle formed is isosceles, and can be split
into two 30-60-90 triangles, each with a short leg of
6 and a long leg of 6 3 . The sum of their areas is
30. (D) The dodecagon can be divided into 24 congruent
right triangles. In each, the angle at the center is
(360/24) = 15, and the angle at the vertex is 75. The
short side of the triangle is (x/2) and to find the
height the trig ratio tan(75)  h  ( x / 2) is solved.
h  ( x / 2) tan(75) . The area of the triangle is
( x 2 / 8) tan 75 . There are 24 of these triangles
comprising the dodecagon, so the total area is
3x 2 tan 75 .