Download Meet 2 - Category 2

Park Forest Math Team
Meet #2
Self-study Packet
Problem Categories for this Meet (in addition to topics of earlier meets):
1. Mystery: Problem solving
2. Geometry: Area and perimeter of polygons
3. Number Theory: Divisibility GCF, LCM, prime factorization
4. Arithmetic: Fractions, terminating and repeating decimals, percents
5. Algebra: Word problems with 1 unknown; working with formulas; reasoning in number
sentences
Important Information you need to know for Meet 2, Category 2…
GEOMETRY: Area and Perimeter of Polygons
Shape
Rectangle
Square
Triangle
Parallelogram
Trapezoid
Perimeter
2L + 2W
4s
A+B+C
2A + 2B
A+C+B+b
Rectangle
Area
LW
s2
! Bh
Bh
!h(B + b)
Square
L
W
s
Triangle
A
Parallelogram
C
A
H
h
h
B
B
Trapezoid
b
A
h
C
B
To find the area of a more complex polygon, break the area into smaller parts and
find the area of each part. Then add the areas together. If you memorize the
formula for area of a rectangle and a triangle, you can find the area of virtually
any polygon!
Category 2
Geometry
Meet #2 - November, 2014
Figures are not necessarily drawn to scale.
C
1) The area of square ABFE is
25 square feet. The area of
triangle BFC is 10 square feet.
The measure of angle D in
triangle CDE is 45 degrees.
How many feet are in the length
of segment AD ?
B
F
A
D
E
2) The perimeter of square A is 2/3 of the perimeter
of square B, and the perimeter of square B is
2/3 of the perimeter of square C. If
one side of square A is
4 centimeters, then how many
B
square centimeters are in the
A
area of square C ?
C
3) Triangle XYZ has a perimeter of 10 inches. The lengths of its
sides are all whole numbers. If n is the length of side YZ, then
what is the difference between the largest and smallest possible
values of n ?
Answers
1)
2)
3)
Solutions to Category 2
Geometry
Meet #2 - November, 2014
1)
Start with the square. If the area is 25, then one
side is the square root of 25, or 5. Then the base of
of the triangle, BF, is also 5. If the area of triangle
BFC is 10, then its altitude, CF, is 4. That makes
CE = 5 + 4 = 9. Because angle D is 45 degrees,
right triangle CED is isosceles. So, ED is also 9.
Finally, AD = AE + ED = 5 + 9 = 14.
Answers
1) 14
2) 81
3)
2
2) Since one side of A is 4, so its perimeter is 4(4), or 16. Then this 16 is
2/3 of the perimeter of B, so the perimeter of B is 16(3/2), or 24.
Then 24 is 2/3 of the perimeter of C, so the perimeter of C is 24(3/2),
or 36. So, one side of C is 36/4, or 9. The area of C is 9(9), or 81.
3) This problem utilizes the notion that the sum of any two sides of a
triangle must be larger than the third side. Since the perimeter is a
fixed value of 10, this chart shows the possibilities for the lengths of
the sides (and the ones that fail):
side 1
side 2
side 3
possible?
1
1
1
1
2
2
2
3
1
2
3
4
2
3
4
3
8
7
6
5
6
5
4
4
no
no
no
no
no
no
yes
yes
The first six options fail because the sum of the first two sides does
not exceed the third side. The bottom two succeed because the sum
of any two of the sides does exceed the third side.
So, the smallest possible value of n is 2 and the largest possible
value is 4, and their difference is 2.
Category 2
Geometry
Meet #2, November/December 2012
1. Find the number of inches in the
perimeter of the figure at right. All angles
are right angles and all lengths are in inches.
2. Five squares of gold all have the same thickness, but they have edge
lengths of 1 cm, 5 cm, 7 cm, 7 cm, and 11 cm. If the gold is melted down
and recast with the same thickness as before into five identical squares,
how many centimeters are there in the edge length of each square?
3. In trapezoid ABCD, side AD, which is 15 cm, is parallel to side BC,
which is 21 cm. The area of trapezoid ABCD is 234 square centimeters.
If point E is on side AD, how many square centimeters are there in the
area of triangle BEC? Express your answer to the nearest tenth of a
square centimeter.
Answers
1. ______________ inches
2. __________________ cm
3. _______________ sq. cm
Solutions to Category 2
Geometry
Meet #2, November/December 2012
1. The total vertical rise on the right side of the
figure must be equal to the 9 inches we see on the
left side of the figure, so a + 3 + b = 9. Similarly,
the total of the horizontal lengths would equal
the 12 inches we see on the bottom, but there is
an extra 3 + 3 = 6 inches because the figure
turns into itself for 3 inches and then must come
back 3 inches. The total perimeter is thus 9 + 12
+ 9 + 12 + 6 = 48 inches.
Answers
1. 48 inches
2. 7 cm
3. 136.5 sq. cm
2. The total surface area of the 5 squares is 12  52  72  72  112 
1 25  49  49 121 245 square cm. If the gold is to be recast in 5
equal squares of the same thickness as before, they must each have a
surface area of 245 ÷ 5 = 49 square cm. The side length of each square

would be√49 = 7 cm. Notice that we squared the side lengths, then
averaged these squared numbers, and finally took the square root of
this average. This square root of the mean of the squares is called the
“root mean square” or the “quadratic mean.”

1
3. Using the area formula for a trapezoid A  hb1  b2 , we substitute
2
in the known values and solve for the unknown height h as follows:
1
234  h21 15 234  17h  h  13.
2

Triangle BEC has the same heigth, so it’s

1
area is  2113  10.5 13
2
= 136.5 square centimeters.

Meet #2
December 2010
A
Category 2 – Geometry
1. The perimeter of rectangle
measures
How many centimeters in the measure of
B
cm.
?
E
C
D
2. In the drawing below, the area of trapezoid
rectangle
If
is four times the area of
. [The drawing is not to scale].
measures
A
B
E
C
inches, then how many inches are
there in the measure of
?
D
3. The rectangle below is divided into
The total area of all is
congruent (identical) rectangles.
square inches.
How many inches in the perimeter of each one?
Answers
1. _______________
2. _______________
3. _______________
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Meet #2
December 2010
Solutions to Category 2 – Geometery
1. For the perimeter to measure
The area of triangle
cm,
must measure
then is
equal the area of triangle
cm.
and this must
which can be expressed as
When we plug in the known values we get
2. If we call the trapezoid’s height
area
.
.
[Using similar triangles, you can also observe that
rectangle’s area is
Answers
1. 12
2. 10
3. 70
].
, then we know that the
, and the trapezoid’s area is larger by the triangle’s
. Knowing that
write:
, and naming
, we can
[Expressing the fact that the triangle’s area is 3
times the rectangle’s, and cancelling out
3. Let’s call a rectangle’s width
]. Solving, we get
, and its height
.
Each reactangle’s area is
.
In the drawing we see 4 rectangles in the top row and 3 in the bottom row, so
we can conclude that
or
first equation we get
or
. When we subtitute this in the
.
So we get
, and the perimeter is:
[Another way to solve is to notice that the whole area is
. Combined with
this leads to
www.imlem.org
].
Category 2
Geometry
Meet #2, December 2008
1.
The four squares below have areas of 16, 9, 4, and 1 respectively. The
squares are lined up one next to the other as shown below. What is the perimeter
of the overall shape below?
2. A rectangle with a perimeter of 60 inches is cut into
4 smaller congruent rectangles by cutting the rectangle
in half both horizontally and vertically as shown to the
right. How many inches are in the sum of the
perimeters of the four new rectangles?
3.
The trapezoid on the right has an area that is 16 more than the area of the
trapezoid on the left. What is the value of h (the height) in the trapezoid on
the right?
19
16
h
10
24
Answers
1. _______________
2. _______________
3. _______________
29
Solutions to Category 2
Geometry
Meet #2, December 2008
Answers
1.
1.
The total perimeter of the figure is:
4 + 1 + 3 + 1 +2 + 1 + 1 + 1 + 1 + 2 + 3 + 4 + 4 = 28
4
28
1
2. 120
3.
3
1 2
4
9
1
1
1
3
4
2
1
2. Labeling the rectangles’ length and width with L and W we can call the
perimeter of the original rectangle 2W + 2L. Each of the smaller rectangles would
!
#
have a length of and a width of giving them each a perimeter of W + L and the
"
"
four of them combined would be 4W + 4L. That’s exactly twice the perimeter of
the original, so the combined perimeter of the four rectangles must be 2(60) = 120
L
L
2
W
W
W
W
2
2
L
2
L
3. The area of the trapezoid on the left is $ %&'()*+,- .
9::
/01 203 456
"
.
/782"9457:
"
.
. ;<<= So the area of the trapezoid on the right is 200+16 = 216. Using the
formula for area of a trapezoid we can find the missing height given the area.
"
;>? .
/7@2"@456
9A6
"
;>? .
"
;>? . ;BC
DDDDDE . DC
Category 2
Geometry
Meet #2, December 2006
1. The square at the far left in the figure below is a unit square. As you move to
the right, the side length of each square increases by 1 unit. How many units are
in the perimeter of the entire figure?
2. In the figure below, sides TR and AP are parallel, but sides TP and RA are not
parallel. The length of side TR is 11 units and the length of side AP is 17 units.
If the area of TRAP is 84 square units, how many units apart are the two parallel
lines?
R
T
P
A
3. If the area of the inner-most square EFGH is 9 square units, how many units are
in the perimeter of the outer-most square ABCD?
A
Answers
1. _______________
2. _______________
3. _______________
D
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B
E
F
H
G
C
Solutions to Category 2
Geometry
Meet #2, December 2006
Answers
1. 70
2. 6
3. 48
Students with correct
answer in a cluster of
6 schools:
1. 28/36
2. 28/36
3. 20/36!
(Many got 144 for #3,
which is the AREA of
the outer square, not
the perimeter.)
1. The distance across the bottom of the figure is
1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 units. The sum of the
horizontal distances across the tops of the squares is also
28 units. The height of the square at the far right is 7
units. There are also seven 1 unit lengths on the left of
each square. The total perimeter of the figure is thus 28
+ 28 + 7 + 7 = 70 units.
2. Quadrilateral TRAP is a trapezoid. The formula for
1
the area of a trapezoid is A = h (b1 + b2 ), where h is the
2
height and b1 and b2 are the two parallel bases. We know
the area and the lengths of the two bases. Solving for h
in the area formula as shown below, we find that the two
parallel lines are 6 units apart.
1
84 = h (11 + 17)
2
1
84 = h (28)
2
84 = 14h
84
h=
=6
14
3. Each larger square in the figure is twice the area of
the one inside it. Since the area of square EFGH is 9
square units, the area of square ABCD must be 9 × 2 × 2
× 2 × 2 = 144 square units. The side length of square
ABCD must be 12 units, since 12 × 12 = 144. The
perimeter of square ABCD is thus 4 × 12 = 48 units.
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Category 2
Geometry
Meet #2, November 2004
1. How many different rectangles have a perimeter of 48 units, if both the length
and the width of the rectangle are positive whole numbers?
Note: A rectangle with length a and width b is considered the same as a rectangle
with length b and width a.
2. An irregular octagon is made on a grid as
shown at right. Notice that some sides are longer
than others. If the octagon has an area of 252
square inches, how many inches are in the length
of a short side?
3. Three right triangles and two squares were cut out of a rectangle with length 20
centimeters and width 12 centimeters as shown below. The dimensions of the
triangles are given in the picture and the squares are each 3.5 cm by 3.5 cm.
How many square centimeters are in the area of the shaded region?
8 cm
3 cm
Answers
1. _______________
2. _______________
3. _______________
10 cm
5 cm
3 cm
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7 cm
Solutions to Category 2 Average team got 13.74 points, or 1.1 questions correct
Geometry
Meet #2, November 2004
Answers
1. 12
2. 6
3. 171
1. Since the perimeter of a rectangle is twice the sum of
the length and the width, we will consider all whole
number lengths and widths with a sum of 24 units.
The 12 possible dimensions of the rectangle are: 1 × 23,
2 × 22, 3 × 21, 4 × 20, 5 × 19, 6 × 18, 7 × 17, 8 × 16,
9 × 15, 10 × 14, 11 × 13, and 12 × 12.
2. There are 5 full squares and 4 half squares in the
shaded area, for a total of 7 squares the size of the grid.
Since the area of the shaded region is 252 square inches,
each square in the grid must be 252 ÷ 7 = 36 square
inches. This means each grid square must be 6 inches on
a side (6 × 6 = 36), and there are also 6 inches in the
short side of the octagon.
3. Before the triangles and squares were cut out of the
rectangle, it had an area of 12 cm × 20 cm = 240 square
centimeters. The areas of the triangles are 10 × 3 ÷ 2 =
15 cm2, 8 × 3 ÷ 2 = 12 cm2, and 5 × 7 ÷ 2 = 17.5 cm2.
The area of the two squares is 2 × 3.5 × 3.5 = 24.5 cm2.
Subtracting these areas from the original 240 square
centimeters, we get 240 – 15 –12 –17.5 – 24.5 = 171
square centimeters.
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