Theta Applications Solutions
1. A
This is a form of the 7,24,25 right triangle so the side lengths are
14,48,50 and thus the perimeter is 112 yards. But since the answer is in feet
3X112=336 feet.
2. C
The sum of the nth row in Pascal’s Triangle is 2𝑛 . Thus the seventh
7
7
row would have a sum of 27 . log 8 27 = 7 log 8 2 = 3 log 2 2 = 3
3. D
The problem would look something like this:
Looking at it 2-dimensionally the circle can be represented by a line and the
sphere can be represented by a circle since the radius of the sphere, and the
circle, is 6 the distance from the center of the circle to where the line touches
the circle is 6 units long and since the vertical distance from the center to the
line is 3.
You can form a right triangle and find that half the length of the line, the
radius of the circle, is 𝑟 = √62 − 32 = √27. Thus finding the area gives us
27𝜋.
Theta Applications Solutions
4. A
Using tangent-secant theorem we get 82 = 4(4 + 𝑥) with simple
algebra we get x=12
5. C
Since we are looking for when y=5 we will plug in the value for h(x)
and solve for x (the horizontal distance). ℎ(𝑥) = 5 = 𝑥 2 − 10𝑥 + 5, hence
0 = 𝑥 2 − 10𝑥 = 𝑥(𝑥 − 10) and thus x=0 and x=10. Since we assumed
Brelbi isn’t at x=0, the answer is 10
6. D
7. A
2𝑧 + 2 = √(z + 1 )(5z − 5) 4𝑧 2 + 8𝑧 + 4 = 5𝑧 2 − 5
0 = 𝑧 2 − 8𝑧 − 9 0 = (𝑧 + 1)(𝑧 − 9) Hence 𝑧 = −1 and 𝑧 = 9
We are looking at something like this:
The radius of the inscribed circle is the apothem of the equilateral triangle
which is 2. Thus we get:
Where the base of the triangle is half the length of the base of the equilateral
triangle. Now using 30-60-90 triangle rules which states that the side length
across from the 30 degree angle is x, the side length across from the 60
degree angle is √3𝑥, and the side length across from the 90 degree angle is
2𝑥. Since we know x=2, the base of the triangle, half the base of the
equilateral triangle, is 2√3. Now that we know half the base we multiply it
by 2 to get the side length of the equilateral triangle, which is 4√3. Since the
equilateral triangle shares its side with the square, like seen in the first
picture, the square thus too has side lengths of 4√3. Now looking at the
square and the circumscribed circle we have:
Theta Applications Solutions
Now since we know all the sides of the square we can create a 45-45-90
triangle with the hypotenuse being the diameter of the circumscribed circle.
We know that the diameter,𝑑 2 = 𝑠 2 + 𝑠 2 where s is the side length of the
square. Thus 𝑑 2 = 2𝑠 2 which leads to 𝑑 = 𝑠√2 = (4√3)(√2) = 4√6. The
diameter of the circumscribed circle is 4√6 thus the radius, which is half the
2
diameter, is 2√6. The area of the circle is, 𝐴 = 𝜋𝑟 2 = (2√6) 𝜋 = 24𝜋
8. C
There are 8 yellow marbles and for the second pick there are 7, so
8
×
37
7
36
=
14
333
.
𝑑
𝑑
𝑟
300
30
Let 𝑑 = 𝑟𝑡 and thus 𝑡 = . The total time, 𝑡𝑡𝑜𝑡𝑎𝑙 =
9. B
7𝑑
150
. Thus we have 𝑟 =
10. A
78+68+88+118+𝑥
5
2𝑑
7𝑑
150
=
+
𝑑
2
50
+
𝑑
2
150
=
7
= 90; 78 + 68 + 88 + 118 + 𝑥 = 450 Thus x=98
11. B
The line crosses the x-axis at (12, 0) and the y-axis at (0, 6). The
shape formed is a cone with height 6 and radius 12. Volume = 288π.
12. D
Any right triangle inscribed in a circle shares its hypotenuse with the
diameter. Thus 𝑑 = √352 + 122 and since 2r=d, 𝑟 =
√1369
2
thus the area is
1369𝜋
4
13. C
You get
log 3
log 2
×
log 4
log 3
×
log 5
log 4
× …×
log 108
log 107
, noticing that they cancel out
you are left with log 2 108 which simplifies down to 2 + 3log 2 3
14. D
The sum of an arithmetic sum is
(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑟𝑚𝑠)(𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚+𝑙𝑎𝑠𝑡 𝑡𝑒𝑟𝑚)
2
15. B
16. A
1
3
1
1
10
7
𝑥
21
+ = ,
=
1
𝑥
=
(101)(201+1)
thus 𝑥 =
𝐹𝑎𝑐𝑒𝑠 + 𝑉𝑒𝑟𝑡𝑖𝑐𝑒𝑠 = 𝐸𝑑𝑔𝑒𝑠 + 2
2
= 1012 = 10201
21
10
9+V=17+2 Hence V=10
Theta Applications Solutions
17. C
18. C
{
𝑥 + 𝑦 = 3500
, 𝑥 = 3250
0.3(2𝑥 + 10𝑦) = 2700
Number of different handshakes =
𝑛(𝑛−1)
2
=
12(11)
2
= 66
19. A
The circumference of the circle is 𝜋𝑑. The first circle has a
circumference of 18𝜋 and the second circle has a circumference of 12𝜋.
Since the belt only goes around half the circumference of each circle we get
9𝜋 + 6𝜋 = 15𝜋. Thus the sum of the semi circumferences is 15𝜋. When
finding the internal common tangent you can make a right triangle with the
sum of the radii and the distance between the centers like this:
We see the right triangle with the sum of the radii, the internal tangent and
the distance between the centers. This gives a right triangle with a leg of 15
and a hypotenuse of 25. If we assign the tangent length x we have
252 = 𝑥 2 + 152 which gives us 252 − 152 = 𝑥 2 so we get
𝑥 = √625 − 225 = √400 = 20. Thus the internal tangent has a length of
20. But since there are two the length is 40.Thus the length of the belt is
15𝜋 + 40.
20.D
The circumference of the circle is 𝜋𝑑. The first circle has a
circumference of 18𝜋 and the second circle has a circumference of 12𝜋.
Since the belt only goes around half the circumference of each circle we get
9𝜋 + 6𝜋 = 15𝜋. Thus the sum of the semi circumferences is 15𝜋. This time
the right triangle is formed by the external common tangent:
We see that the right triangle is made by the external tangent, the distance
between the centers, and the positive difference between the radii. Thus we
have the 252 = 𝑥 2 + (9 − 6)2 which gives us 252 − 32 = 𝑥 2 so we get
Theta Applications Solutions
𝑥 = √625 − 9 = √616 = 2√154. Thus the length of the external tangent is
2√154. But again since there are two the sum of the tangents is 4√154.
Thus the length of the belt is 15𝜋 + is 4√154.
10
3
ℎ
10
3
3
𝑉𝑓𝑟𝑢𝑠𝑡𝑢𝑚 = (𝑅2 + 𝑟𝑅 + 𝑟 2 )𝜋 =
21. A
(64 + 40 + 25)𝜋 =
(129)𝜋 = 430𝜋
22. B
To be champion you need to win 4 matches after group stages.
256
Therefore Germany needs to win all 4 games: (0.8)4 = 0.4096 =
.
625
First the perimeter of the triangle is 18, its altitude is 3√3. The radius
23. C
of an inscribed circle is: 𝑟 =
2𝐴
𝑃
=
2(62 √3)
4(18)
= √3 Thus the circumference of
the circle is 2𝜋√3
|𝐴𝑥+𝐵𝑦+𝐶|
|5(5)+12(2)−23|
26
24. A
Shortest distance= 2 2 =
= =2
2
2
√𝐴 +𝐵
Let A= Bobbie’s rate. Let B = Tom’s rate. Let C= Ajay’s rate.
25. C
Because 𝐷 = 𝑅𝑇, we can establish three equations:
𝑑
𝐵
13
√5 +12
=
𝑑−8
𝐶
, 𝑎𝑛𝑑
𝑑
𝐴
=
equation, we get that
𝑑 − 31
𝐶
𝑑 − 25
𝐵
𝑑
𝐴
=
𝑑 − 25
𝐵
,
. Seeing a similarity between the first and third
=
𝑑 − 31
𝐶
. Subtracting the second equation from
this new equation, we find that 23B=25C. After substituting for B, 𝐵 =
25𝐶
23
into the second equation, we solve for d, which equals 100.
26. E
Rewriting the quadratic we get 𝑥 2 + (𝑘 + 2)𝑥 + 𝑘 + 3 = 0. Taking
the determinant and setting it less than 0 will determine when the quadratic
will have imaginary roots. Hence, (𝑘 + 2)2 − 4(1)(𝑘 + 3) < 0. Cleaning
this up, we get 𝑘 2 < 8 . Thus the answer −2√2 < 𝑘 < 2√2, not
−2√2 < 𝑥 < 2√2 .
,
Theta Applications Solutions
27. D
We are given the equation 𝑑 =
information, 20 =
𝑘(22 )
√16(5)
𝑘𝑡 2
√𝑔𝑅
. Solving for k with the given
, we get 𝑘 = 100. Now solving for d: 𝑑 =
100(32 )
√4(15)
and hence we get 𝑑 = 30 𝑚𝑒𝑡𝑒𝑟𝑠.
𝐶4 × 5𝐶4 × 2𝐶1
20
28. C.
Probability=8
29. A
Since 𝑓(𝑥)𝑎𝑛𝑑 𝑔(𝑥) are inverses of each other 𝑔(1) =
15𝐶9
=
143
𝑥 𝑎𝑛𝑑 𝑡ℎ𝑢𝑠 𝑓(𝑥) = 1. Hence, 𝑥 3 + 2𝑥 2 − 𝑥 − 2 = 𝑥 3 so we get 2𝑥 2 − 𝑥 −
2 = 0. Since the sum of the roots of a quadratic in the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 =
𝑏
−1
𝑎
2
0 is – = −
30. C.
𝑥=
=
1
2
Again, 𝑓(𝑥)𝑎𝑛𝑑 𝑔(𝑥) are inverses of each other. So letting 𝑔(𝑥) = 𝑦,
𝑦3
𝑦 3 +2𝑦 2 −𝑦−2
⇛𝑥=
𝑦3
(𝑦−1)(𝑦+1)(𝑦+2)
. Now the horizontal asymptotes are
𝑦 = −1, 𝑦 = 1, 𝑎𝑛𝑑 𝑦 = −2 (due to the undefined nature those play when
plugged in) and the vertical asymptote would be 𝑥 = 1 since the limit of x as
it approaches infinity is 1. What you may see is that if you find the
asymptotes of 𝑓(𝑥), which are 𝑦 = 1, 𝑥 = −1, 𝑥 = 1, 𝑎𝑛𝑑 𝑥 = −2 and
reflect it across the line 𝑦 = 𝑥, also a 90 degree rotation, will yield the
answer: ) 𝑦 = 1, 𝑦 = −1, 𝑦 = −2, 𝑎𝑛𝑑 𝑥 = 1.