Download Geometry

Park Forest Math Team
Meet #3
Geometry
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 about GEOMETRY…
Properties of Polygons, Pythagorean Theorem
Formulas for Polygons where n means the number of sides:
• Exterior Angle Measurement of a Regular Polygon: 360÷n
• Sum of Interior Angles: 180(n – 2)
• Interior Angle Measurement of a regular polygon:
• An interior angle and an exterior angle of a regular polygon
always add up to 180°
Interior
angle
Exterior
angle
Diagonals of a Polygon where n stands for the number of vertices (which is equal
to the number of sides):
•
• A diagonal is a segment that connects one vertex of a polygon to
another vertex that is not directly next to it. The dashed lines represent
some of the diagonals of this pentagon.
Pythagorean Theorem
• a2 + b2 = c2
• a and b are the legs of the triangle and c is the hypotenuse (the side
opposite the right angle)
c
a
b
•
Common Right triangles are ones with sides 3, 4, 5, with sides 5, 12, 13,
with sides 7, 24, 25, and multiples thereof—Memorize these!
Category 2
50th
Geometry
Meet #3 - January, 2014
anniversary edition
26
Y
W
6
1) How many cm long is segment
XY ? All measurements are in
centimeters (cm).
Z
X
8
C
2) Angle ABC is a right
angle. Triangle BCD is
an isosceles triangle
such that DB = BC.
Find the value of X
if it is the measure
in degrees of angle
EDC and X < 180.
X
E
120
150
D
70
B
A
3) Moe and Larry race from point B to point S at a rectangular field.
Moe runs from B to A to S at an average rate of 5 feet every second.
Larry runs diagonally across the field from B to S at an average rate
of 10 feet every 3 seconds. If they both leave point B at the same
time, then who wins the race? Also, by how many seconds does the
winner finish ahead of the runner-up? (You must answer both
questions correctly to receive credit.)
S
E
ANSWERS
50 feet
1) __________ cm
2) __________
B
120 feet
A
3) __________ winner
_________ seconds
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Solutions to Category 2
Geometry
Meet #3 - January, 2014
Answers
1)
1) Use the Pythagorean Theorem twice - first to find
the length of WX and then XY.
24
2) 155
3) Moe
5
Use this result to find XY:
2) The measure of angle DBA is 20 degrees, because the sum of the
angles of a quadrilateral is 360 degrees.
The measure of angle DBC is 70 degrees, because angle ABC is a
right angle (90 degrees).
Since two sides (DB and BC) of triangle DBC are congruent, the
angles opposite those sides are congruent. The vertex angle, DBC,
measures 70 degrees, so the base angles are 55 degrees each,
including angle BDC.
angle X + 150 + 55 = 360, so X = 155.
3) Both answers must be answered correctly in order for students to
receive credit.
Moe:
runs 120 + 50, or 170 feet. At a rate of 5 feet per second,
it takes him 170 / 5, or 34 seconds to reach point S.
Larry: Use the Pythagorean Theorem to find that he has run
130 feet. At a rate of 10 feet every 3 seconds, it takes him
(130 / 10) x 3, or 39 seconds to reach point S.
Moe, therefore, reaches point S ahead of Larry by 39 - 34, or by 5
seconds, so Moe wins the race.
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Meet #3
January 2012
Category 2 – Geometry
(11, 4)
1. Given the coordinates in the diagram,
what is the distance between the
two points?
(-1, -1)
2. How many diagonals are there in a regular polygon with
sides (a
Hexadecagon)?
3. The sum of interior angles in a regular polygon is
measure of each of its exterior angles.
How many sides does the polygon have?
Answers
1. __________ Units
2. _________ Diagonals
3. __________ Sides
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times as great as the
Meet #3
January 2012
Solutions to Category 2 – Geometery
1. The horizontal distance is
Answers
1.
2.
3.
units, and the vertical distance is
units, so the total distance is √
units.
2. The formula for the number of diagonals in a polygon with N sides is:
so in our case we’ll have
diagonals.
3. The exterior angles of a polygon all add up to 360 degrees, so if there are
sides to the polygon, then each exterior angle measures
Every interior angle measures
degrees.
degrees, and their sum is therefore
degrees. So in our case we’re told that:
whice we can rewrite as:
.
Though this is technically a quadratic equation, we know that
number and can easily find that
[The other solution,
is a natural
is a solution (an Octagon).
, is clearly not an answer to our problem].
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Category 2 - Geometry
Meet #3, January 2010
1. The number of diagonals in a polygon is four times the number of its vertices.
How many vertices does it have?
(A diagonal is a line segment that connects two non-adjacent vertices of a polygon).
2. The exterior angle to a regular polygon N (with N sides) is half that of a regular
polygon M (with M sides). Polygon N has 7 times as many diagonals as polygon M.
What is the value of 𝑀 ∙ 𝑁?
3. Tom stands exactly 2 miles west of Jerry.
At 10:00am Tom starts walking east at 5 mph (miles per hour).
At 10:20am Jerry starts heading north at 9 mph.
How many miles between them at 12:00pm (noon)?
Answers
1. _______________
2. _______________
3. _______________
Remember: You do not have to specify units. Specifying
the wrong units will disqualify your answer.
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Solutions to Category 2 - Geometry
Meet #3, January 2010
1.
2.
3.
Answers
11
50
17
1. If we call the number of vertices V, then the number of diagonals is given by the
formula:
𝑉×(𝑉−3)
2
and for this to equal 4 × 𝑉 we get
2. The exterior angle is
Polygon M will have
360
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑖𝑑𝑒𝑠
𝑀∙(𝑀−3)
2
𝑉−3
2
= 4 or 𝑉 = 11.
so it should be clear that 𝑁 = 2𝑀.
diagonals, and polygon N will have
𝑁∙(𝑁−3)
2
=
2𝑀∙(2𝑀−3)
2
diagonals.
So we need to solve: 2𝑀 ∙ 2 ∙ 𝑀 − 3 = 7 ∙ 𝑀 ∙ 𝑀 − 3
which we can simplify to 4 ∙ 𝑀 − 6 = 7𝑀 − 21 and then to 3 ∙ 𝑀 = 15
and the solution is 𝑀 = 5, 𝑁 = 10 and 𝑀 ∙ 𝑁 = 50.
3. At 12:00pm, having walked for two hours, Tom is 10 miles to the east of his original
location, which means he’s 8 miles east of Jerry’s original location.
Jerry, after running at 9 mph for 1 hour and 40 minutes is 15 miles north of his (own)
original location.
The distance between them is 82 + 152 = 289 = 17 𝑚𝑖𝑙𝑒𝑠.
Note that we have to compare their locations to some ‘fixed’ or agreed-upon point.
You could have used any other point-of-reference and get the same answer.
www.imlem.org
Category 2
Geometry
Meet #3, January 2008
1.
If the exterior angle of a regular polygon is 24 degrees, how many sides does
the polygon have?
2.
In the figure to the right, four of the sides of
a regular polygon are shown. Also drawn
are all of the diagonals that use A as one of
the endpoints of the diagonal; however the
full lengths of the diagonals are not always
shown. How many diagonals does this
polygon have?
3.
A
In the figure below, triangle ABC is a right triangle with right angle at A and
quadrilateral BCDE is a square. If AB = 8 and AC = 18 , what is the area
of the pentagon ABEDC?
B
C
A
E
Answers
1. _______________
2. _______________
3. _______________
D
Solutions to Category 2
Geometry
Meet #3, January 2008
Answers
1. 15
1. Since the sum of the exterior angles is always 360 degrees, if
you divide 360 degrees by 24 degrees you get 15 which is the
number of equal exterior angles and therefore 15 vertices and
sides to the polygon.
2. 44
3. 20
2. If there are 8 diagonals and only 2 of the diagonals’ other
endpoints are shown there must be 6 vertices that are not
shown. Since 5 vertices are shown there are a total of 11
vertices and 11 sides. If there are 11 sides there are
diagonals total.
3. Using the Pythagorean Theorem:
→ √8 √18 → 8 18 26
is the area of the square, the area of square EBCD is 26. The pentagon
Since we are looking for though is equal to the area of the square minus the area of the
right triangle ABC. The area of triangle ABC =
of pentagon ABEDC = 26 – 6 = 20.
√√
√
6. The area
B
C
A
E
D
Category 2
Geometry
Meet #3, January 2006
1. How many degrees are in the measure of an interior angle of a regular 15-gon?
2. A certain polygon has 2 12 times as many diagonals as sides. How many sides
are there on this polygon? Note: A diagonal is a line segment that connects two
non-adjacent vertices of a polygon.
3. Three right triangles are joined together to form the concave pentagon shown
below. The measure of GA is 15 units, AS is 20 units, SM is 21 units, and ME is
35 units. How many units are in the perimeter of pentagon GAMES?
M
A
Answers
1. _______________
2. _______________
3. _______________
G
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S
E
Solutions to Category 2
Geometry
Meet #3, January 2006
Answers
1. 156
2. 8
3. 132
1. If we draw line segments from one vertex to all the
non-adjacent vertices, we can subdivide the 15-gon into
13 triangles. Each triangle has an angle sum of 180
degrees, so the sum of the interior angles of the 15-gon
must be 13 × 180 = 2340 degrees. Since the 15-gon is
regular, this total is shared equally among the 15 interior
angles. Each interior angle must have a measure of 2340
÷ 15 = 156 degrees.
2. The polygon in question must have an even number of sides, since we are
multiplying by 2 12 and the number of diagonals must be a whole number. A
square has only 2 diagonals, so let’s try a hexagon. From each vertex of a hexagon
we can draw 3 diagonals. If we simply multiply 6 × 3, we will have counted each
diagonal twice, so there must be only 6 × 3 ÷ 2 = 9 diagonals in a hexagon. This is
only 1 12 times the number of sides. Now let’s try an octagon. From each vertex of
an octagon we can draw 5 diagonals. That’s 8 × 5 ÷ 2 = 20 diagonals in all.
Twenty is exactly 2 12 times 8, so our polygon has 8 sides.
3. We can use the Pythagorean Theorem to find
the lengths of sides AM, GS, and SE. Let the the
measure of AM be x. Then 20 2 + 212 = x 2 . This
means x 2 = 400 + 441 = 841. Since 29 2 = 841,
x must be 29 units. Let the measure of GS be y.
Then 15 2 + 20 2 = y 2 . (Here we might recognize
a multiple of the 3-4-5 Pythagorean Triple.) This
means y 2 = 225 + 400 = 625. Since 25 2 = 625,
y must be 25 units. Finally, let the measure of
SE be z. Then 212 + z 2 = 35 2 . This means
z 2 = 35 2 − 212 = 1225 − 441 = 784 . Since 28 2 = 784 ,
z must be 28 units. The perimeter of polygon
GAMES is thus 15 + 29 + 35 + 28 + 25 = 132 units.
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M
29
A
15
G
21
35
20
S
25
28
E
Category 2
Geometry
Meet #3, January 2004
1. How many diagonals can be drawn in a decagon?
Note: A decagon is a polygon with 10 sides and a diagonal of a polygon is a
segment which connects any two non-consecutive vertices.
2. A regular polygon has an interior angle measure that is greater than 144
degrees and less than 150 degrees. How many sides does the polygon have?
3. In the figure at right, every
quadrilateral is a square and every
triangle is a right triangle. The area of
square ABCD is 144 square units and
the area of square EFGH is 16 square
units. Length LK is 5 units. How many
units are in the perimeter of the figure
(polygon ABCEFGIJKLMN)?
C
F
B
Answers
1. _______________
2. _______________
3. _______________
E
D
A
N
www.Imlem.org
L
M
G
H
K
I
J
Solutions to Category 2
Geometry
Meet #3, January 2004
1. From each of the ten vertices of a decagon, one can
draw seven diagonals. If we multiply 7 by 10 to get 70,
we will have counted every diagonal twice. Therefore
there must be 70 ÷ 2 = 35 diagonals in a decagon.
Answers
1. 35
2. 11
2. A polygon with n sides can be subdivided into n – 2
triangles, each with an angle sum of 180. This gives the
n-gon a total angle sum of 180(n − 2) or 180n − 360 . If
the n-gon is regular, then all n of its interior angles will
have the angle measure (180n − 360) n . We need to
solve the double inequality 144 < (180n − 360) n < 150
for a whole-number value of n.
3. 100
144 < (180n − 360) n
(180n − 360) n < 150
144n < 180n − 360
180n − 360 < 150n
360 < 180n −144n
180n −150n < 360
360 < 36n
30n < 360
10 < n
n < 12
The only whole number value between 10 and 12 is 11, so the polygon must have
11 sides.
3. Square ABCD has an area of 144 square units, so
its side length is 12. Length DL is 5, so triangle ADL
is a 5-12-13 triangle as the Pythagorean Theorem
confirms: 5 2 + 12 2 = 25 + 144 = 169 and 13 2 = 169 .
We are also given that square EFGH has an area of 16
square units, so its side length is 4 units. Triangle
EHK is a 3-4-5 triangle, since 5 2 − 4 2 = 25 −16 = 9
and 3 2 = 9 . We can now label all the side lengths in
the figure as shown at right. The perimeter is
12 + 12 + 13 + 4 + 4 + 5 +3 + 3 + 5 + 13 + 13 + 13 = 100
units.
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12
C
13
B
4 F4
E
D
12
G
I
5
3
K
L5 3J
A
13
13
N
H
13
M