Download Solutions - FloridaMAO

January Regional
Geometry Individual Solutions
1. Using the formula for the number of diagonals, we see that 23*20/2 is equal to 230. A
2. 3(90-x) = 3/5(180-x) – x/2.
B
90 – x = 36 – x/5 – x/6. 54 = 19x/30. x = 1620/19. A
3. You know that medians intersect at a point that is 2/3rds
the distance from each vertex to the midpoint of the opposite
side. So after solving BE to equal √13 (using Pythagorean
D
theorem), you know that DE is equal to √13/3. Also, using
AA, you know that ΔABE ≈ ΔFDE. Using ratios, you
find that EF is equal to 1. A
A
C
F
E
4. Definitions. B
5. If you reflect the point (12, 7) over the x-axis, you can make a large triangle by connecting the coordinates
(0, 5), (12, -7) and (0, -7). From that triangle, you can solve for the hypotenuse. 12√2. C
6. Set opposite angles equal to get 2y = 4x. y = 2x. Now, sum adjacent angles and set the value equal to 180.
8x + 2y = 180. Substitute 2x for y. 12x = 180. x = 15. y = 30. 2y = 60. D
7. The side opposite the 30° angle is half of the hypotenuse. Knowing this, we can solve for the hypotenuse to
equal 4. However, the answer must be converted to inches. 48. D
8. The question simplifies to asking for the inverse since repeats cancel out. B
9. Using the formula for the sum of all angles in a convex polygon, 21*180 = 3780. A
10. The points are collinear. This triangle does not exist. E
11. Definitions. Only IV) is not true. C
12. Definitions. B
13. If you set the angle measure of angle BCD equal to x, you get the equation, 2x + 40 = 180. x = 70. The
supplement of x is the value you’re looking for. 110. D
14. Definitions. A
15. All of the following are right triangles and can be checked using Pythagorean theorem. E
16. For simplicity, let’s call the side length of the hexagon s. Given that piece of
information, we can give the base of the zoomed in triangle the value (6-s)/2.
3
Since the triangle is a 30-60 right triangle (we know this because the upper
angle must be 30° since the adjacent angle is 90° and every angle in a
regular hexagon is 120°), we can solve for the hypotenuse, which turns
out to be (6-s). We can also reenact a similar process on the triangle to
the left of the square in which the side opposite the 60°angle measures
(6 – s) /2
3 (we know this because that side is half the length of the square).
The hypotenuse is then solved to be 2√3. Then, we can set s equal to the sum of the two hypotenuses and we
get the equation, 2√3 + 6 – s = s. s = 3 + √3. A
17. Definition. B
18. Each of the trisected pieces adjacent to z is equal to 5, because they are the midsegments of their
respective triangles which have base length 10. Knowing that we can write the equation for midsegments, (23
+ 10)/2 = 10 + z. z = 13/2. A
19. The exterior angle can be simply solved for by dividing 360 by 72, getting us 5. 180 – 5 = 175. D
20. Drawing a perpendicular line down from the angle measuring 75°, we see that this figure is essentially a
30-60 right triangle and a 45-45 right triangle. Thus, we solve to get the solution 23/2 + 23√3/2. A
21. The cosine of 60° is ½. A.
22. While you can solve this question using the Pythagorean theorem, you can also solve this by setting the
equations for are equal to each other. jk/2 = hi/2. i = jk/h. B
23. Using similar triangles, we get the ratio 5/3 = x/7. x = 35/3. The perimeter is 35/3 + 9 = 62/3. The semiperimeter is half of that, 31/3. B
24. As the figure to the right shows, the side of the square has been abbreviated to s.
Setting the base length equal to 2, we can solve for the side opposite the 30° angle
of the right triangle (we know this because of the AA postulate) in terms of s. We
can then solve the side opposite the 60° angle to be (2-s)√3/2, which is equal to s.
2√3 - s√3 = 2s. s = 2√3/(2 + √3) = 4√3 – 6. C
25. A 4, 5, 18 triangle is not possible. E
26. If 3√2 were to be the hypotenuse, the other side would be 3.
1 – s/2
s
January Regional
Geometry Individual Solutions
If 3√2 and 3 were to be the legs of the right triangle, we can use the Pythagorean theorem to solve the
hypotenuse to be 3√3. Thus, 3√3*3 = 9√3. B
27. If we map out the coordinates, we know that the distance between the coordinates will be the hypotenuse
of a triangle with legs measuring the differences in the x and y coordinates. Solving reveals the legs to be 20
and 21, which easily leads us to 29, which is the hypotenuse of this special right triangle. D
28. By using simple examples, we can solve for the formula of maximum sections to be x(x+1)/2 + 1. Using
this formula, we can solve. 23(24)/2 + 1 = 277. D
29. Because the triangle is isosceles, we know that lengths RA and RM are equal. RM is the length of one
side of the square. Knowing this, we can find the perimeter. 3 + 3 + 3 + 3 + 4 = 16. A
30. Knowing that ΔTOR is a right triangle with hypotenuse 5 and leg 3, we know that OR has length 4. We
know that ΔTGA ≈ ΔTRO by AA (vertical angles and right angles). Now, we can use ratios to solve for the
missing side. 3/5 = x/23. x = 92/3. B