6
2012
High School Math Contest
Geometry
Exam
3
4
2
Lenoir-Rhyne University
Donald and Helen Schort School of
Mathematics and Computing Sciences
Geometry
Solutions
This exam has been prepared by the following faculty from Western Carolina University:
Risto Atanasov
Mark Budden
Andrew Chockla
Axelle Faughn
Nathan Borchelt
Geooff Goehle
John Wagaman
GEOMETRY - Solutions, March, 2012
1. Solution (C)
The circle of diameter 3 has area π
3 2
2 .
The circle of diameter 1 has area π
2
2
π 32 − π 12
purple-painted area to the gold-painted area is
= 8.
2
π 12
1 2
2 .
The ratio of the
2. Solution (B)
√ 2 √ 2
Let r be the radius of the semicircular. Then r2 = ( 28) + 228 = 28 + 7 = 35. The area of the
semicircle is 21 πr2 =
35π
2 .
3. Solution (D)
From x − 40◦ + 2x + 10◦ = 180◦ we get x = 70◦ . Then 30◦ = 3y − 18◦ , which gives y = 16◦ . Therefore,
x + y = 70◦ + 16◦ = 86◦ .
4. Solution (C)
2
The area of the square is (2r) = 4r2 and the area of the circle is πr2 . Hence, the shaded region has
2
2
area 4r − πr .
5. Solution (A)
√
3 3
2
3
The area of the triangle is
1
2
(3)
√ 3 3
2
=
√
9 3
4
cm2 .
6. Solution (B)
√
The hight of the triangle is h = 102 − 62 = 8 and its area is 21 (12)(8) = 48. Let L and W denote
48
the length and the width of the rectangle respectively. Then L = W
= 48
4 = 12. The perimeter of the
rectangle is 2(12 + 4) = 32.
7. Solution (B)
Let A and B be the centers of the circles. Then the Pythagorean theorem gives us that P Q2 =
2
AB 2 − (AP − BQ) = 25 − 9 = 16. Hence P Q = 4.
8. Solution (D)
The surface area is given by s2 + 2sl. Hence, 832 = 162 + 32l. Therefore, l = 18 m.
9. Solution (B)
The area is 12 (b1 + b2 ) · h = 12 (40 + 60) · 25 = 1250 ft2 . Since
of topsoil is 84.
1
1250
15
= 83.3, the minimum number of bags
10. Solution (B)
Since 152 + 202 = 252 , the triangle is a right triangle. The two legs AC and BC are altitudes with
= 150, we get DC = 12.
lengths 15 and 20 respectively. The area of 4ABC is 150. From 25·DC
2
B
20
25
D
15
A
C
11. Solution (C)
The area
of the region
that consists of the equilateral triangle AOB topped with the semicircle is
√
√
3
3
1
1
1 2
= 4 + 18 π. The area of the lune results from subtracting from this the area of the
2 · 1 · 2 + 2π 2
√
√
1
(6 3 − π).
sector of the larger semicircle which is π6 . The area of the lune is 43 + 81 π − 16 π = 24
12. Solution (C)
C
r
A
O
r
2
6
E
F
B
D
4CF D is a right triangle becasue CD is a diameter of the circle. Then 4CF D ∼ 4DOE, so
i.e. r2 = 24. The area of the circle is 24π.
13. Solution (C)
√
2
2
√
Area = 4
1
2
2
√ 2 √ √ √
√
√ 2
+ 2 2 2 + 2 2 + 2 2 = 4 + 4 2 + 4 + 4 2 = 8 1 + 2 ft2
14. Solution (B)
Since 6 CAT = 25◦ = 6 EBF , we have 6 BF E = 180◦ − 25◦ − 6 F EB = 155◦ − 95◦ = 60◦ .
2
2r
6+2
= 6r ,
15. Solution
(B)
2
2
4
30◦
⇒ Diameter = 8
16. Solution (A)
2
3
45◦
√
45◦
2
√3
2
2
5
D=
q
√
(2 +
2 2
2 )
+ (5 −
√
5 2 2
2 )
=
p
√
42 − 23 2 ft.
17. Solution (D)
The lateral area is πr` where r is the radius and ` is the lateral surface length. Since l = 13, we get
that the lateral area is 65πcm2 .
` = 13
12
5
18. Solution (A)
Of the 10 possible combinations only (2, 3, 4), (2, 4, 5), and (3, 4, 5) satisfy the Triangle Inequality.
19. Solution (B)
The total area of the triangles P AD and P CB is half that of the parallelogram. Hence
the area of the parallelogram is 6. The area of the parallelogram is 36.
20. Solution (C)
Denote the height by h. Then 7 · 12 · 15 = 7 · 18 · h or h =
3
12·15
18
= 10 in.
1
2
−
1
3
=
1
6
of
21. Solution (C)
Drop perpendiculars AH
√ and DK to BC from√A and D, respectively. Then triangle HAB is half a
square with diagonal 6, so that HA = HB
√ = 3. Also, triangle KCD is half an equilateral triangle
with side 6, so that CK = q
3 and DK = 3 3. Hence HK = 8 while the difference of AH and DK is
√
√
√
2 3. The length of AD is 82 + (2 3)2 = 2 19.
22. Solution (E)
Pythagorean Triple (8, 15, 17) satisfies 17 − 8 = 9. The area = of the triangle is
1
2
· 8 · 15 = 60.
23. Solution (D)
E
D 1
F I
G
3
A
2
C
3
H 5
B
Let H be on line AB such that EH ⊥ AB, and let I be the intersection of EH and DC. Then F G = 2.
FG
EI
EI
EI
Since 4ABE ∼ 4F GE we have AB
= EH
= EI+IH
. Hence 25 = EI+3
, i.e. EI = 2 and EH = 5.
1
25
A4ABE = 2 AB · EH = 2 .
24. Solution (A)
B
A
Oα
D
C
Let α be the central angle generated by the chord BC. Then 6 DBC = α2 (tangent chord theorem)
and 6 BAC = α2 (inscribed angle theorem). Hence 6 DBC = 6 BAC. Both 4BAC and 4BDC are
isosceles. Since 6 ABC = 26 BDC we have
π = 6 BAC + 6 ABC + 6 ACB = 6 BAC + 46 BDC
π = 6 BDC + 6 DBC + 6 DCB = 6 BDC + 26 BAC
Hence 6 BDC = π − 26 BAC and
π = 6 BAC + 46 BDC = 6 BAC + 4(π − 26 BAC) =
= 4π − 76 BAC
Therefore 6 BAC =
3π
7 .
4
25. Solution (A)
A
F
B
D
4
E
4
√
We have CE = 2 2,
have
C
p
√
√
DE = 2 2, AE = AD2 + DE 2 = 17. Since 4ADE is a right triangle, we
√
√
2 2·3
=3 2
2
1
A4ADE = AE · DF
2
√
√
√
6 34
1√
6 2
=
3 2=
17 · DE ⇒ DE =
2
17
17
A4ADE =
26. Solution (C)
√
Half the diagonal of the square base has length 3 2 2 . Since the pyramid is half of a regular octahedron,
√
its height has the same length. Hence the height of the top of the building from the ground is 2 + 3 2 2 .
27. Solution (A)
The sum of the interior angles of an n-sided polygon is (n − 2) · 180◦ , so this 2n-sided polygon has
interior angle sum given by
(2n − 2) · 180◦ = 6 A1 + 6 A2 + . . . + 6 An + (360◦ − 6 B1 ) + . . . + (360◦ − 6 Bn ).
Since 6 A1 = . . . = 6 An , 6 A1 + 10◦ = 6 B1 = 6 B2 = . . . = 6 Bn , we have
2(n − 1) · 180◦ = n6 A1 + 360◦ n − n6 A1 − 10◦ n
360◦ n − 360◦ = 350◦ n
n = 36
5
28. Solution (D)
2
Sector area = 61 π(4) = 83 π
√
√
Triangle area = 21 (4)2 3 = 4 3
√ Shaded area = 2 38 π + 2 83 π − 4 3 =
32
3 π
√
− 8 3 cm2 .
29. Solution (D)
The diagonal crosses 32 horizontal grid lines and 74 vertical grid lines. However, the greatest common
divisor of 33 and 75 is 3 so that it crosses exactly one corner of a square twice (at 13 and 23 distance).
Thus it has 32 + 74 − 3 = 103 points of intersections on it, and they divide it into 102 segments of
positive length.
30. Solution (E)
The volume of a sphere is proportional to the cube of its radius. Hence n ≥
of n is 38 and it exceeds 20.
6
1000
27 .
The minimum value