(1)
A garden is laid out in the fashion shown in the diagram. If only the shaded
isosceles triangles are used for planting, what is the total area, in square feet, that is to be
used for planting?
20 ft.
16 ft.
20 ft.
(2)
A rectangle is formed. Its length is increased by 50% and its width is cut in
half in forming a second rectangle. Then each of the two new dimensions is decreased by
20% to form a third rectangle. What percent of the original rectangle’s area is the area of
the third rectangle?
(3)
A square with sides 6 inches is shown. If P is a point such that the segment
P A, P B, P C are equal in length, and segment P C is perpendicular to segment F D, what
is the area, in square inches, of triangle AP B?
F
C
D
P
A
6′′
B
(4)
What is the ratio of the shaded area to the area of triangle ABC if all the
triangles shown are equilateral?
A
B
C
(5)
A clock has struck 4 o’clock. In exactly how many minutes will the two
hands first be at right angles? Express your answer as a mixed number.
(6)
An equilateral triangle is stacked on top of a square, which in turn is stacked
on top of a regular hexagon. If each of these figures has side 8 and touching sides coincide
exactly, what is the number of units in the perimeter of the resulting figure?
(7)
A line segment 40 cm long is to be divided into three segments whose
lengths are integers. How many centimeters long is the longest piece if they for the sides of
a right triangle?
(8)
A cylindrical beaker is 8 cm high and has a radius of 3 cm. How many such
beakers of water will it take to fill a spherical tank of radius 6 cm?
(9)
How many square units are in the area of the enclosed region in the
diagram? All angles pictured are right angles.
10
2
6
2
4
10
2
4
2
6
2
10
(10)
Express the area of the shaded region in circle O in terms of π.
O
6
8
(11)
The following ordered pairs are graphed on the coordinate plane and
connected by line segments: A(0, 1), B(3, −2), C(6, 1), D(3, 4) to form a closed figure
ABCD. What is the most specific mathematical term describing figure ABCD?
(12)
A wrecker’s iron ball eight inches in diameter weighs 120 pounds. How many
pounds would a similar iron wrecking ball twelves in diameter weigh?
(13)
The length of a side of a cube is 8 units. Express, as a common fraction,
the ratio of the number of square units in the surface area of the cube to the number of
cubic units in its volume.
(14)
Parallelogram ABCD has its vertices at A(2, −3), B(4, 5), C(−5, 5), and
D(−7, −3). What is the area of the parallelogram?
(15)
Terry bicycled due north for 7 miles and due east for 24 miles. He then
returned straight back to his starting point. How many miles did he bicycle on the trip?
(16)
(17)
√
The diagonal of a square is 3 2. Find the perimeter.
Three cubes are stacked as shown. If the cubes have edge lengths 1, 2, and
3 as shown, what is the length of the portion of the segment AB that is contained in the
center cube?
A
1
2
3
(18)
B
The area of rectangle P QRS is 48 sq cm. P Q = 31 QR. U is a point on QR
such that QU = UR. T is a point on P S such that P T = 2T S. Find the perimeter of the
quadrilateral RST U.
(19)
A person walks completely around the outer circle of the two concentric
circles with centers E beginning at C and ending at C. She continues her walk from C to
D and then around the inner circle beginning and ending at D. The distance from C to D
is 50 yards and the distance from A to B is 200 yards. What is the total number of yards
that she walked?
C
D
A
E
B
(20)
An 8 inch by 10 inch rectangular pictures is surrounded by a border that is 2
inches wide on all sides. Find the ratio of the area of the border to the area of the picture.
Express your answer as a common fraction.
(21)
The sides of the base of a triangular pyramid are 3, 4 and 5 feet and the
altitude is 6 feet. What is the number of cubic feet in the volume of the pyramid?
(22)
Two ants are crawling on a 4-inch-by-6-inch-by-8-inch brick. What is the
farthest, in inches, the two ants could get from one another? Round your answer to 3
significant digits.
(23)
In a right triangle, the sum of the lengths of the legs is 17 mm. The area of
the right triangle is 30 square millimeters. Find the number of millimeters in the length of
the hypotenuse.
(24)
Parallelogram ABCD has vertices A(3, 3), B(−3, −3), C(−9, −3), and
D(−3, 3). If a point is selected at random from the region determined by the
parallelogram, what is the probability that the point is not above the x -axis? Express your
answer as a common fraction.
(25)
Given quadrilateral ABCD with A(3, 7), B(3, 5), C(−1, 1), and D(−1, 3),
find the number of square units in the area of the quadrilateral.
(26)
Suppose the points A, B, C, D, E, and F are the vertices of a regular
hexagon with sides of length 1 unit. What is AC?
(27)
Point Z is the midpoint of MN. Point P is the midpoint of ZN. Point Q is
the midpoint of P N and Point R is the midpoint of QN. If P R is 24 inches long, how
many feet long is MN. Express your answer as a mixed number.
(28)
A circle is divided into twenty equal sectors. How many adjacent sectors
combined would form an angle of 72 degrees?
(29)
A regular hexagon is 4 cm on each side. How many equilateral triangles, 2
cm on each side, would be required to tile the hexagon?
(30)
A maple tree casts a shadow of 9 feet on the ground. From the top of the
tree to the tip of the shadow of the tree measures 41 feet. How many feet tall is the tree?
Copyright MATHCOUNTS Inc. All rights reserved
Answer Sheet
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Answer
256
48 %
27/4
9/16
5
5 11
72 units
17
4
72 square units
12.5π − 24
square
405 pounds
3/4
72
56 ml
12
√
6 √
14 + 2 5 cm
300π + 50
11
10
12 cubic feet
10.8 in
13
1
2
8
√square units
3
10 2/3 feet
4 sectors
24
40
Problem ID
3DAA1
C42B
1BC51
135A1
DB3A1
DCCA
A02C
DB4D1
05C51
CAAA1
123A1
0DA41
1AAA1
A2BB
A10C
A3012
035D1
CC3B
4C1C
AC541
4C0C
432B
AC1C
C10D1
AAC41
B14A1
5DA41
C4032
D2BB
322B
Copyright MATHCOUNTS Inc. All rights reserved
Solutions
(1) 256
ID: [3DAA1]
No solution is available at this time.
(2) 48 %
ID: [C42B]
No solution is available at this time.
(3) 27/4
ID: [1BC51]
We first extend line segment CP so that it intersects AB. We’ll call this point of
intersection point E, so CE is a perpendicular bisector to segment AB and AE = EB = 3.
We also let x = the lengths of segments P A, P B, and P C, so line segment P E will have
length 6 − x . Now we have that △AEP is a right triangle. Using the Pythagorean
Theorem and solving for x , we have that:
AE 2 + P E 2 = P A2
⇒
⇒
32 + (6 − x )2 = x 2
9 + 36 − 12x + x 2 = x 2
12x = 45
15
.
⇒
x=
4
Thus, △AP B has base 6 and a a height of 6 − x = 6 −
1
27
1
9
an area of bh = · 6 ·
square inches.
=
2
2
4
4
⇒
(4) 9/16
ID: [135A1]
No solution is available at this time.
5
(5) 5 11
ID: [DB3A1]
No solution is available at this time.
(6) 72 units
ID: [DCCA]
No solution is available at this time.
15
4
= 94 . It follows that △AP B has
(7) 17
ID: [A02C]
No solution is available at this time.
(8) 4
ID: [DB4D1]
We first have to remember our formulas for the volumes of 3 dimensional objects. The
volume of a cylinder with radius r and height h is r 2 hπ and the volume of a sphere with
radius r is 34 r 3 π. Since the cylindrical beaker has a height of 8 centimeters and a radius of
3 centimeters, that means that its volume is 32 · 8 · π = 72π cubic centimeters. Since the
sphere has a radius of 6 centimeters, its volume is 43 · 63 π = 288π cubic centimeters. The
number of beakers of what it will take to fill the spherical tank is just the ratio of the
288π
= 4.
volume of the tank to the volume of the cylinder, which is given by
72π
(9) 72 square units
ID: [05C51]
No solution is available at this time.
(10) 12.5π − 24
ID: [CAAA1]
No solution is available at this time.
(11) square
ID: [123A1]
No solution is available at this time.
(12) 405 pounds
ID: [0DA41]
No solution is available at this time.
(13) 3/4
ID: [1AAA1]
No solution is available at this time.
(14) 72
ID: [A2BB]
No solution is available at this time.
(15) 56 ml
ID: [A10C]
No solution is available at this time.
(16) 12
ID: [A3012]
Let s be the
the square. By 45-45-90 triangle relationships, the diagonal
√ of √
√ side length
length is s 2. So s 2 = 3 2, or s = 3. Thus, the perimeter is 4s = 4(3) = 12 .
(17)
√
6
ID: [035D1]
It turns out that the amount of segment passing through a given cube is proportional to
the side length of the cube. To see why this is, let the intersection of the smallest two
cubes be plane j, the intersection of the largest two cubes be plane k, and let AB intersect
j and k at X and Y respectively. Imagine dropping perpendiculars from A and X to planes j
and k – these perpendiculars, along with the portions of AB passing through the cubes,
form similar triangles, resulting in the aforementioned proportionality.
The center cube has 2/(1 + 2 + 3) = 1/3 of the three cubes’ total side length. To find
the length of AB, let the front
√ bottom left vertex of the largest cube be√C and look at
right triangle ACB. CB = 3 2 and AC = 6, so by Pythagoras, AB = 3 6. Thus, the
√
√
center cube has (1/3) · 3 6 = 6 of the total length of AB.
√
ID: [CC3B]
(18) 14 + 2 5 cm
No solution is available at this time.
(19) 300π + 50
ID: [4C1C]
No solution is available at this time.
(20)
11
10
ID: [AC541]
No solution is available at this time.
(21) 12 cubic feet
ID: [4C0C]
Because we have 32 + 42 = 52 , the base of the pyramid is a right triangle with legs of
lengths 3 and 4. Therefore, the area of the base is 3 · 4/2 = 6 square feet. The altitude of
the pyramid is 6 feet. The volume of the pyramid is one-third the product of the area of
the base and the altitude, which is 6 · 6/3 = 12 cubic feet.
(22) 10.8 in
ID: [432B]
No solution is available at this time.
(23) 13
ID: [AC1C]
No solution is available at this time.
(24)
1
2
ID: [C10D1]
Let us first call the point where the x -axis intersects side AB point E and where it
intersects CD point F .
D
A
F
C
E
B
Now, since the x -axis is parallel to bases AD and BC of the parallelogram, EF is parallel
to the two bases and splits parallelogram ABCD into two smaller parallelograms AEF D
and EBCF . Since the height of each of these parallelograms is 3 and the length of their
bases equals AD = BC = 6, both parallelograms must have the same area. Half of
1
parallelograms ABCD’s area is above the x -axis and half is below, so there is a
2
probability that the point selected is not above the x -axis.
(25) 8 square units
ID: [AAC41]
Quadrilateral ABCD is a parallelogram, but we don’t know the base or height. Instead of
directly finding the base and height of ABCD, notice that we can create a point E so that
quadrilateral AECD is a right trapezoid with a height of 4 units and bases of 2 and 6 units.
Now we can find the area of ABCD by subtracting the area of △BEC from the area of
trapezoid AECD. The area of △BEC is 21 bh = 21 (4)(4) = 8 and the area of trapezoid
2 )h
= (6+2)4
= 16. The area of quadrilateral ABCD is 16 − 8 = 8 square
AECD is (b1 +b
2
2
units.
A(3, 7)
B(3, 5)
6
D(−1, 3)
4
2
C(−1, 1)
(26)
4
E(3, 1)
√
3
ID: [B14A1]
No solution is available at this time.
(27) 10 2/3 feet
ID: [5DA41]
No solution is available at this time.
(28) 4 sectors
ID: [C4032]
No solution is available at this time.
(29) 24
ID: [D2BB]
No solution is available at this time.
(30) 40
ID: [322B]
No solution is available at this time.
Copyright MATHCOUNTS Inc. All rights reserved