Park Forest Math Team

Meet #2

GeometryGeometry

Self-study Packet

Problem Categories for this Meet: 1. Mystery: Problem solving

2. Geometry: Angle measures in plane figures including supplements and complements

3. Number Theory: Divisibility rules, factors, primes, composites

4. Arithmetic: Order of operations; mean, median, mode; rounding; statistics

5. Algebra: Simplifying and evaluating expressions; solving equations with 1 unknown including identities

Important Information you need to know for Meet 2, Category 2…

GEOMETRY: Area and Perimeter of Polygons

Shape Perimeter Area Rectangle 2L + 2W LW

Square 4s s2 Triangle A + B + C ! Bh

Parallelogram 2A + 2B Bh Trapezoid A + C + B + b !h(B + b)

Rectangle

L W

Triangle

Square

s Parallelogram

A C H

B

Trapezoid b A 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!

h

h

A h

B

Category 2 Geometry Meet #2 - November, 2015 1) What is the maximum (most) number of non-overlapping squares of perimeter 8 centimeters that can be cut from a square that has an area of 144 square centimeters? 2) The area of rectangle MATH is 72 square millimeters. Points W and C are the midpoints, respectively, of sides AT and TH. How many square millimeters are in the area of triangle MWC ?

3) For Kamel to walk a mile in his local rectangular park, he must either walk the width 132 times or the perimeter 40 times. How many square feet are in the area of the park? (one mile = 5280 feet)

Answers

1)

2)

3)

A

C

W

H T

M

Solutions to Category 2 Geometry Meet #2 - November, 2015 1) A square of perimeter 8 has a side length of 8 / 4, or 2, Answers and an area of 2x2, or 4. Divide the area of the large square by 4: 144 / 4 = 36 squares. 1) 36 2) The area of triangle MWC = the area of rectangle 2) 27 MATH minus the sum of the areas of the lettered triangles X, Y and Z. Triangle A is 1/4 of the area of MATH. 3) 1040 Triangle B is also 1/4 of the area of MATH. Triangle C is 1/8 of the area of MATH. MATH - (X + Y + Z) = 72 - [(0.25)(72) + (0.25)(72) + (0.125)(72)] = 72 - [18 + 18 + 9] = 72 - [45] = 27 3) Divide 5280 by 132 to get the width of the rectangle: 5280 / 132 = 40. Let X = the length of the rectangle. The perimeter is 40 + 40 + X + X, or 80 + 2X. The actual perimeter is 5280 divided by 40 (a different 40, which is the number of times Kamel had to traverse the perimeter to walk the mile). 80 + 2X = 5280 / 40 80 + 2X = 132 2X = 52 X = 26 So, the length of the rectangle is 26 and the width is 40, so the area is length x width, or (26)(40) = 1040 square feet. Check: 40 times the perimeter = 40 (132) = 5280 132 times the width = 132 (40) = 5280.

A

C

W

H T

M Y

X Z

30

22

3

15 16

18

4

23

Category 2GeometryMeet #2 - November, 2013

1) A pentadecagon is a polygon with 15 sides. ("penta" means 5, while "deca" means 10.) In a certain pentadecagon, eleven of the sides have the same length while each of the remaining sides measures 16 cm. If the perimeter of the pentadecagon is 207 cm, then how many cm long is one of the shorter sides?

2) The figure to the right is an example of a Domino - a rectangle consisting of two congruent squares that share a side. A giant model of a Domino measures 6 inches by 12 inches. How many Dominoes are used to completely fill the bottom of a square pan whose area is 81 square feet? (12 inches = 1 foot)

3) Find the perimeter of the figure below. All angles are right angles. Measurements are not to scale.

ANSWERS

1) ________ cm

2) ________

3) ________

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Solutions to Category 2GeometryMeet #2 - November, 2013

Answers 1) 207 - (4 x 16) = 207 - 64 = 143 (the total length of the 11 congruent sides. 143/11 = 13 (the length of 1) 13 each of the shorter sides).

2) 162 2) Each Domino contains 6 x 12, or 72 square inches

(or half of a square foot). A square pan of 81 3) 150 square feet contains 162 of these model Dominoes.

Also consider that two Dominoes combine to make one square foot.

Another approach is to convert the 81 square feet to square inches, where 1 square foot = 144 (12 x 12) square inches. 81 x 144 = 11,664 square inches. Dividing 11,664 by 72 yields 162 Dominoes. Of course, this technique involves much more arithmetic.

3) The width of the narrow horizontal bar at the top of the figure can be found by subtracting 18 from 23, (DE from BC) which is 5. The sum of this 5 and the 3 (HI) is the difference between the 15 (AJ) and the unlabelled vertical segment (FG), so the length of that vertical segment (FG) is 7. The adjacent horizontal unlabelled segment (HG) when added to the width of the leftmost large space (30 - 20 = 10) totals to 22 (IJ), so the length of that horizontal unlabelled segment (GH) is 22 - 10, or 12. (Lengths are not to scale.)

So, the perimeter:

AJ + AB + BC + DC + DE + FE + FG + GH + HI + IJ = 15 + 30 + 23 + 4 + 18 + 16 + 7 + 12 + 3 + 22 = 150.

30

22

3

15 16

18

4

23

A

C

B

D

EF

A

I

G H

J

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Meet #2 December 2011

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Category 2 – Geometry

1. Given the coordinates in the diagram,

how many units are in the area

of the trapezoid?

2. The shaded triangle and square in the diagram share one side.

The square’s perimeter is square inches, and its area is

the area

of the triangle. How many inches are in the triangle’s height?

3. The diagram shows a square whose side is , inside of which are

two right triangles whose short legs are and .

The white trapezoid’s area is of the square’s area.

Express

as a percent.

Answers

1. __________

2. __________ inches

3. __________ %

(4, -2)

(6,3) (-2,3)

R

3x

x

Meet #2 December 2011

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Solutions to Category 2 – Geometery

1. The area of a trapezoid is given by:

( )

From the diagram we can observe the height (the distance between the

horizontal bases) is 5 units, the small base measures 4 units, and the large base

measures 8 units. So the area is

( ) units.

2. The square’s perimeter is inches, so its side measures inches. Its area then

is , and so the triangle’s area is

.

If its height is then

and so inches.

3. The triangles’ combined area is

( ) and so the

remaining trapezoid’s area is

So we know that , or , or .

(Here of course, so we could divide by ).

Answers

1.

2.

3.

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Category 2 - Geometry

Meet #2, December 2009

1. The big triangle’s height is 12 units, its width is 6 units,

and the inscribed square’s perimeter is 16 units.

How many square units are in the

shaded area in the drawing?

2. In the kite below, 𝐴𝐶 = 18 inches, 𝐵𝐷 = 25 inches.

What is the length (in inches) of the perimeter of a

square with the same area as the kite?

3. In the diagram below, the trapezoid

𝐴𝐵𝐶𝐷 and the regular pentagon

𝐴𝐵𝑄𝑅𝑆 share a common edge 𝐴𝐵.

𝐷𝐶 = 9 inches.

𝐴𝐸 = 5 inches.

Area of 𝐴𝐵𝐶𝐷 = 40 square inches.

What is the perimeter of ABQRS? (in inches).

Answers

1. _______________

2. _______________

3. _______________

A

B

C

D

S

R

Q

D C

B A

E

6

12

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Solutions to Category 2 - Geometry

Meet #2, December 2009

1. The shaded area is the difference between the triangle’s area and the square’s area.

The triangle’s total area is 𝑊𝑖𝑑𝑡 𝑕×𝐻𝑒𝑖𝑔𝑕𝑡

2=

6×12

2= 36 units squared.

The square’s area is (16

4)2 = 16 units squared, so the difference is 20 square units.

Editor’s note: The original problem was height 6, width 5, inscribed square’s

perimeter 12. Unfortunately, this square wouldn’t fit in the triangle. The answers 6

and 7.5 were accepted.

2. As the kite is made up of two triangles, its area is 𝐴𝐶×𝐵𝐷

2=

18×25

2= 225 squared

inches. A square with the same area will have a side of length 225 = 15 inches, and

therefore a perimeter of 15 × 4 = 60 inches.

3. Recall that the area of a trapezoid is 𝐻𝑒𝑖𝑔𝑕𝑡×𝑆𝑢𝑚 𝑜𝑓 𝑏𝑎𝑠𝑒𝑠

2 so in our case:

𝐴𝐸×(𝐴𝐵+𝐶𝐷)

2=

5×(9+𝐴𝐵)

2= 40 So we get 𝐴𝐵 = 7 inches.

The perimeter of the pentagon is 5 times the length AB, or 35 inches.

Answers

1. 20

2. 60

3. 35

Category 2

Geometry Meet #2, November 2007 1. In the figure to the right, all angles are right angles. What is the perimeter of the figure?

2. Quadrilateral MATH to the left has sides MH and AT parallel to each other with MH being three times as long as AT. If the shortest distance between the parallel sides is 5 inches and the area of the quadrilateral is 15 in2, how many inches long is side AT? Express your answer as a decimal.

3. A regular hexagon and a regular octagon, both with whole number side lengths, have the same perimeter which is between 80cm and 100cm. What is the number of square centimeters in the area of a square that has the same perimeter as the octagon and the hexagon?

Answers

1. _______________

2. _______________

3. _______________

9

11

10

29

8

6

27

M

H

T

A

Solutions to Category 2

Geometry Meet #2, November 2007

1. The horizontal distance across the top is 8 + 29 = 37, so the 4 horizontal segments across the bottom also have a sum of 37. The vertical distance along the left is 27 + 6 = 33 and the vertical distance along the right will then also be 33. Since the segments can all be moved around to form a rectangle as shown, the perimeter is just 37+33+37+33 = 140.

2. Since MH and AT are parallel, MATH is a trapezoid and since MH is three times as long as AT, we could call AT =x, and MH =3x. The formula for the area

of a Trapezoid is 1 2( ) ( )5 (3 )5 20

15 10 152 2 2 2

1.5

trap

b b h MH AT x x xA x

x AT

+ + += = = = = = =

= =

3. An octagon with whole number side lengths could have perimeter 80, 88, or 96. A hexagon with whole number side lengths could have perimeter 84, 90, 96. So 96 must be the perimeter of both if their perimeters are equal. A square with

perimeter 96 would have side lengths of 96÷4 = 24, and an area of 242 = 576.

Answers 1. 140 2. 1.5

3. 576

9

11

10

29

8

6

27

29

6

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Category 2 Geometry Meet #2, December 2005

1. The long diagonal of a regular hexagon from vertex to opposite vertex measures 28 mm. How many millimeters are in the perimeter of the hexagon?

2. The figure at right is a triangle on top of a trapezoid. The lower base of the trapezoid is 2 1

2 inches and the upper base is 1 14 inches.

The height of the trapezoid is 12 inch. The

base of the triangle is 1 14 inches and its height

is 1 12 inches. How many square inches are in

the area of the figure? Express your answer as a mixed number in lowest terms. 3. Find the perimeter of the rectilinear figure shown below. Express your answer to the nearest hundredth of an inch. Note: Rectilinear means that the figure consists entirely of straight lines and right angles.

Answers 1. _______________ 2. _______________ 3. _______________

1.5 inches

0.81 inches

1.23 inches x inches

0.71 inches

0.31 inches

0.55 inches

x inches

3.15 inches

28 mm

2 12 in.

1 14 in.

1 12 in.

12 in.

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Solutions to Category 2 Geometry Meet #2, December 2005

1. A regular hexagon can be subdivided into 6 equilateral triangles as shown below left. The side lengths of the equilateral triangle are equal to the side length of the hexagon. The long diagonal consists of exactly 2 of these side lengths, and the perimeter consists of 6 of these side lengths, which is 3 times as much. The perimeter of the hexagon must be 3 × 28 = 84 mm. 2. The area of a trapezoid is one half the height times the sum of the bases, or 1

2 h B1 + B2( ). Our trapezoid has an area of 1

2 ⋅ 12 ⋅ 2 1

2 + 1 14( )= 1

4 ⋅ 3 34( )= 1

4 ⋅ 154 = 15

16 square inches. The area of a triangle is one half the height times the base, or 1

2 hb. Our triangle has an area of 12 ⋅1 1

4 ⋅1 12 = 1

2 ⋅ 54 ⋅ 3

2 = 1516 square inches. The total area of

the figure is thus 1516

+ 1516

= 3016

= 158

= 1 78

square inches.

3. The total height of the figure is the sum of the three heights given on the left, which is 0.71 + 0.55 + 0.31 = 1.57 inches. The four unlabeled heights on the right side of the figure must have the same sum, so there will be another 1.57 inches in the perimeter. The total width of the figure is the sum of the three measures given at the top, which is 1.5 + 0.81 + 1.23 = 3.54 inches. The width along the bottom is 3.15 + x inches. We can determine the unknown length x by sbtracting 3.15 from 3.54, which is 0.39 inches. The two horizontal lines where the figure pokes in on the left add extra perimeter equal to 2x. The total perimeter is thus 2 × 1.57 + 2 × 3.54 + 2 × 0.39 = 3.14 + 7.08 + 0.78 = 11 inches.

Answers 1. 84

2. 1 78

3. 11

28 mm

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Category 2 Geometry Meet #2, November 2003 1. Find the number of feet in the perimeter of the figure below. All angles are right angles and all lengths are in feet. 2. The two squares shown side by side in the figure below have a total area of 74 square inches. The side length of each square is a whole number of inches. How many inches are in the perimeter of the figure? (Note: The line where the two squares touch is not to be counted in the perimeter of the figure.) 3. In the figure below, ABCD is a rectangle. The points E and F trisect side DC, meaning they cut it in three equal pieces. G is a point on side AB . The length of side AB is 3 inches and the length of BC is 1.2 inches. Find the number of square inches in the shaded area of the figure.

Answers 1. _______________ 2. _______________ 3. _______________

A B

D C

G

E F

3

7

3

6 11

8

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Solutions to Category 2 Geometry Meet #2, November 2003

1. The unmarked height on the left side of the figure is the sum of the three vertical lengths on the right, namely 6 + 3 + 3 = 12 feet. Similarly, the other unmarked length must be 8 + 11 – 7 = 12 feet also. The perimeter of the figure is thus: 8 + 6 + 11 + 3 + 12 + 3 + 7 + 12 = 62 feet. Alternatively, one could reason that there must be two vertical totals of 12 feet and two horizontal totals of 19 feet, for a perimeter of: 2 ×12 + 2 ×19 = 24 + 38 = 62 feet.

2. The two perfect squares with a sum of 74 must be 49 and 25. This means the side length of the smaller square is 5 inches and the side length of the larger square must be 7 inches. Although there are two unknown lengths where the smaller square meets the larger square (call one of them x and the other y), the total height x + 5 + y must equal 7. Thus the perimeter of the figure is 4 × 7 + 2 × 5 = 28 + 10 = 38.

3. The area of rectangle ABCD is 1.2 × 3 = 3.6 square inches. Since E and F trisect side DC, the length of segment EF must be 3 ÷ 3 = 1 inch. Also, regardless of where point G is on AB , the height of triangle EFG is 1.2 inches.

Thus the area of triangle EFG is 12

×1×1.2 = 0.6 square inches. Subtracting the

area of triangle EFG from the area of rectangle ABCD, we get 3.6 – 0.6 = 3 square inches, which is the area of the shaded regions.

Answers 1. 62 2. 38 3. 3

12 3

7

3

6 11

8

12