# Rocket City Math League 2011-2012 Gemini Test Round 2 1. If the

```Rocket City Math League
2011-2012
Round 2
Gemini Test
1. If the prism has triangular bases, it will have 5 faces, 6 vertices, and 9 edges. 5+6+9=20
20
2. A square may be split into two 45-45-90 triangles that share a relationship with the diagonal and one of the sides:
(side ) = diagonal
side =
25/2
side =
Area= 25/2
3. The sum of an interior angle and its exterior angle is always 180 degrees. Since each interior angle is 150, each
exterior is 30 and there must be 360/30 or 12 sides. 12 x 180 = 2160.
4. The side opposite of the 30 degree angle would be seven divided by root three, which is seven root three over
three. Then to get the hypotenuse you multiply it by two and you get 14 3
14 3
3
.
3
5. If arc AC is 150 degrees, then arc CB would be 30 degrees. Sector COB is therefore 1/12 of the circle’s area. If
the arc length is 15π for 150&ordm; angle, the radius of the circle is 18. The area of the circle is 324π. The area of the
sector is 27π.
6. 42π = 2πr2 + 2πrh dividing by 2π gives
21= r2 + rh and 21 = r2 + r(7)
27
7  133
2
Complete the square and r  7  133
2
7. If the apothem of the regular hexagon is 3 3 , then one of the sides is 3. Thus, the hexagon may be split into 6
2
equilateral triangles with sides measuring 3. A  6
2160
s 2 3 27 3

4
2
27 3
2
8. Cyclic quadrilateral ABCD is a kite. If line segment BD cuts the kite into two congruent triangles, angles at A
and C are congruent and supplementary and therefore are right angles. BD = 4 3 and is a diameter of the circle.
So, the area of the circle is 12π.
12π
9. The sum of the supplementary angles is 180, so 12 x2  5x  77  0 must be solved. (4x – 11)(3x + 7) = 0 and
4
976
7 11
.
x   , . Both of these give possible solutions. The is 108 or
9
9
3 4
108
4
9
10. It can be derived, but the volume of a tetrahedron is given by: V  2 a3 , so the side of the tetrahedron, a, is
12
equal to 12. The radius of the inscribed sphere is r 
a
2 6

12
2 6
 6 . The volume of the sphere is
8 6
3
4
4
V   r 3   6  8 6 .
3
3
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```