Meet #2

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 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, 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. 30

ANSWERS

16

15

1) ________ cm 2) ________

3 22

23

18

3) ________ 4 www.imlem.org

Solutions to Category 2 Geometry Meet #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 each of the shorter sides).

1) 13 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. A

B

30

F

15

16

H

G

E

3

J

22

A

23 18

I

D

C 4 www.imlem.org

Meet #2

December 2011

Category 2 – Geometry 1. Given the coordinates in the diagram, how many units are in the area (-2,3) of the trapezoid?

(6,3)

(4, -2)

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 The white trapezoid’s area is Express

and

.

of the square’s area.

as a percent.

3x Answers 1. __________ 2. __________ inches

R

3. __________ %

x

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Meet #2

December 2011 Answers

Solutions to Category 2 – Geometery 1. 2. 3.

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

2. The square’s perimeter is is

(

)

units.

inches, so its side measures

, and so the triangle’s area is

If its height is

then

3. The triangles’ combined area is

.

and so

(

inches. Its area then

inches.

)

and so the

remaining trapezoid’s area is So we know that (Here

, or

of course, so we could divide by ).

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, or

.

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. 12

How many square units are in the shaded area in the drawing?

2.

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

6

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

square with the same area as the kite?

D

3.

B

In the diagram below, the trapezoid 𝐴𝐵𝐶𝐷 and the regular pentagon 𝐴𝐵𝑄𝑅𝑆 share a common edge 𝐴𝐵. C

𝐷𝐶 = 9 inches.

R

𝐴𝐸 = 5 inches. Area of 𝐴𝐵𝐶𝐷 = 40 square inches.

S

What is the perimeter of ABQRS? (in inches). Answers 1. _______________ 2. _______________ 3. _______________

A

D

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Q

B

C E

Solutions to Category 2 - Geometry

Answers 20 60 35

1. 2. 3.

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.

16

The square’s area is ( )2 = 16 units squared, so the difference is 20 square units. 4

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 𝐴𝐸×(𝐴𝐵+𝐶𝐷) 2

=

5×(9+𝐴𝐵) 2

𝐻𝑒𝑖𝑔 𝑡×𝑆𝑢𝑚 𝑜𝑓 𝑏𝑎𝑠𝑒𝑠 2

so in our case:

= 40 So we get 𝐴𝐵 = 7 inches.

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

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Category 2 Geometry Meet #2, November 2007

8 29

1. In the figure to the right, all angles are right angles. What is the perimeter of the figure?

10

27

11 6

9

M

A

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.

H

T

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. _______________

Solutions to Category 2 Geometry Meet #2, November 2007 Answers 1. 140 2. 1.5 3. 576

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. 29

8

29

10

27

11 6

6

9

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 (b + b ) h ( MH + AT )5 (3 x + x )5 20 x Atrap = 15 = 1 2 = = = = 10 x = 15 of a Trapezoid is 2 2 2 2 x = 1.5 = 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.

Category 2 Geometry Meet #2, December 2005

28 mm

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 12 inches and the upper base is 1 41 inches. The height of the trapezoid is 21 inch. The base of the triangle is 1 41 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.

1 12 in. 1 41 in.

1 2

in.

2 12 in.

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. 1.5 inches 0.71 inches

0.81 inches x inches

1.23 inches

0.55 inches 0.31 inches

Answers 1. _______________ 2. _______________ 3. _______________

x inches 3.15 inches

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Solutions to Category 2 Geometry Meet #2, December 2005 Answers 1. 84 2. 1

7 8

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.

3. 11

28 mm

2. The area of a trapezoid is one half the height times the sum of the bases, or 12 h (B1 + B2 ). Our trapezoid has an area of 12 ⋅ 12 ⋅ (2 12 + 1 41 ) = 41 ⋅ (3 43 ) = 41 ⋅ 154 = 15 16 square inches. The area of a triangle is one half the height times the base, or 12 hb . Our triangle has an area of 1 1 1 1 5 3 15 2 ⋅ 1 4 ⋅ 1 2 = 2 ⋅ 4 ⋅ 2 = 16 square inches. The total area of 15 15 30 15 7 + = = = 1 square inches. the figure is thus 16 16 16 8 8 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.

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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. 8 6

11 3

3 7

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. G

A

B

Answers 1. _______________ 2. _______________ 3. _______________ D www.Imlem.org

E

F

C

Solutions to Category 2 Geometry Meet #2, November 2003 Answers

1. 62 2. 38 3. 3

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. 8 6 12

11 3

3

12

7

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. 1 Thus the area of triangle EFG is ×1 ×1.2 = 0.6 square inches. Subtracting the 2 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.

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