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 Square L W s Triangle Parallelogram A C A hH 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 15 16 1) ________ cm 3 23 18 2) ________ 22 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 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. A 30 B F 16 15 E G H 3 23 A 18 J 22 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 of the trapezoid? (-2,3) (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 and . The white trapezoid’s area is of the square’s area. Express as a percent. 3x Answers 1. __________ R 2. __________ inches 3. __________ % x www.imlem.org Meet #2 December 2011 Solutions to Category 2 – Geometery Answers 1. 1. The area of a trapezoid is given by: 2. 3. ( ) 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 ). www.imlem.org 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 12 shaded area in the drawing? 2. In the kite below, 퐴퐶 = 18 inches, 퐵퐷 = 25 inches. 6 What is the length (in inches) of the perimeter of a square with the same area as the kite? A D B 3. In the diagram below, the trapezoid 퐴퐵퐶퐷 and the regular pentagon 퐴퐵푄푅푆 share a common edge 퐴퐵. 퐷퐶 = 9 inches. C R 퐴퐸 = 5 inches. Area of 퐴퐵퐶퐷 = 40 square inches. S Q What is the perimeter of ABQRS? (in inches). A B Answers 1. _______________ 2. _______________ D C 3. _______________ E www.imlem.org Solutions to Category 2 - Geometry Answers Meet #2, December 2009 1. 20 2. 60 3. 35 1. The shaded area is the difference between the triangle’s area and the square’s area. 푊푖푑푡 푕×퐻푒푖푔 푕푡 6×12 The triangle’s total area is = = 36 units squared. 2 2 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. 퐴퐶×퐵퐷 18×25 2. As the kite is made up of two triangles, its area is = = 225 squared 2 2 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. 퐻푒푖푔 푕푡×푆푢푚 표푓 푏푎푠푒푠 3. Recall that the area of a trapezoid is so in our case: 2 퐴퐸×(퐴퐵+퐶퐷) 5×(9+퐴퐵) = = 40 So we get 퐴퐵 = 7 inches. 2 2 The perimeter of the pentagon is 5 times the length AB, or 35 inches. www.imlem.org 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 27 10 figure? 11 6 9 M 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 in 2, how many inches long is side AT? Express your answer as a 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. The horizontal distance across the top is 8 + 29 = 37, so the 4 horizontal segments across the bottom also have 1. 140 a sum of 37. The vertical distance along the left is 27 + 6 = 33 and the vertical distance along the right will then 2. 1.5 also be 33. Since the segments can all be moved around to form a rectangle as shown, the perimeter is just 3. 576 37+33+37+33 = 140 . 8 29 29 27 10 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 =3 x. The formula for the area (bbh+ ) ( MHAT+ )5 (3 xx + )5 20 x A ==15 1 2 = = ===10x 15 of a Trapezoid is trap 2 2 22 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 24 2 = 576 . 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 28 mm 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 1 1 2 2 inches and the upper base is 1 4 inches.