Park Forest Math Team. Meet #3. Geometry. Self-study Packet

Transcription

1 Park Forest Math Team Meet #3 Self-study Packet Problem Categories for this Meet: 1. Mystery: Problem solving 2. : ngle measures in plane figures including supplements and complements 3. Number Theory: Divisibility rules, factors, primes, composites 4. rithmetic: Order of operations; mean, median, mode; rounding; statistics 5. lgebra: Simplifying and evaluating expressions; solving equations with 1 unknown including identities

2 Important Information you need to know about GEOMETRY Properties of Polygons, Pythagorean Theorem Formulas for Polygons where n means the number of sides: Exterior ngle Measurement of a Regular Polygon: 360 n Sum of Interior ngles: 180(n 2) Interior ngle Measurement of a regular polygon: n 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): 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 a 2 + b 2 = c 2 a and b are the legs of the triangle and c is the hypotenuse (the side opposite the right angle) c a 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! b

3 Category 2 Meet #3 - January, th anniversary edition W 26 Y 1) How many cm long is segment XY? ll measurements are in centimeters (cm). 6 Z 8 X C 2) ngle BC 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. E X D B 3) Moe and Larry race from point B to point S at a rectangular field. Moe runs from B to 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? lso, by how many seconds does the winner finish ahead of the runner-up? (You must answer both questions correctly to receive credit.) NSWERS 1) cm E S 50 feet 2) 3) winner B 120 feet seconds

4 Solutions to Category 2 Meet #3 - January, ) 24 1) Use the Pythagorean Theorem twice - first to find the length of WX and then XY. 2) 155 3) Moe Use this result to find XY: 5 2) The measure of angle DB 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 BC 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 = 360, so X = ) Both answers must be answered correctly in order for students to receive credit. Moe: runs , or 170 feet. t 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. t 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.

5 Meet #3 January 2012 Category 2 1. Given the coordinates in the diagram, what is the distance between the two points? (-1, -1) (11, 4) 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 times as great as the measure of each of its exterior angles. How many sides does the polygon have? 1. Units 2. Diagonals 3. Sides

6 Meet #3 January 2012 Solutions to Category 2 Geometery 1. The horizontal distance is units, and the vertical distance is units, so the total distance is units 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 degrees. Every interior angle measures 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 is a natural number and can easily find that is a solution (an Octagon). [The other solution,, is clearly not an answer to our problem].

7 Category 2 - Meet #3, January The number of diagonals in a polygon is four times the number of its vertices. How many vertices does it have? ( 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. t 10:00am Tom starts walking east at 5 mph (miles per hour). t 10:20am Jerry starts heading north at 9 mph. How many miles between them at 12:00pm (noon)? Remember: You do not have to specify units. Specifying the wrong units will disqualify your answer.

8 Solutions to Category 2 - Meet #3, January 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 V 3 = 4 or V = The exterior angle is 360 Number of sides Polygon M will have M (M 3) 2 diagonals. so it should be clear that N = 2M. diagonals, and polygon N will have N (N 3) 2 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. = 2M (2M 3) 2 3. t 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 = 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.

9 Category 2 Meet #3, January 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. lso drawn are all of the diagonals that use 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. In the figure below, triangle BC is a right triangle with right angle at and quadrilateral BCDE is a square. If B = 8 and C = 18, what is the area of the pentagon BEDC? B C E D

10 Solutions to Category 2 Meet #3, January 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. 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: Since is the area of the square, the area of square EBCD is 26. The pentagon we are looking for though is equal to the area of the square minus the area of the right triangle BC. The area of triangle BC = 6. The area of pentagon BEDC = 26 6 = 20. B C E D

11 Category 2 Meet #3, January How many degrees are in the measure of an interior angle of a regular 15-gon? 2. certain polygon has times as many diagonals as sides. How many sides are there on this polygon? Note: 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 G is 15 units, S is 20 units, SM is 21 units, and ME is 35 units. How many units are in the perimeter of pentagon GMES? M S E G

12 Solutions to Category 2 Meet #3, January 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 = 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 = 156 degrees. 2. The polygon in question must have an even number of sides, since we are multiplying by and the number of diagonals must be a whole number. 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 = 9 diagonals in a hexagon. This is only times the number of sides. Now let s try an octagon. From each vertex of an octagon we can draw 5 diagonals. That s = 20 diagonals in all. Twenty is exactly times 8, so our polygon has 8 sides. 3. We can use the Pythagorean Theorem to find the lengths of sides M, GS, and SE. Let the the measure of M be x. Then = x 2. This means x 2 = = 841. Since 29 2 = 841, x must be 29 units. Let the measure of GS be y. Then = y 2. (Here we might recognize a multiple of the Pythagorean Triple.) This means y 2 = = 625. Since 25 2 = 625, y must be 25 units. Finally, let the measure of SE be z. Then z 2 = This means z 2 = = = 784. Since 28 2 = 784, z must be 28 units. The perimeter of polygon GMES is thus = 132 units. 15 G M 21 S E

13 Category 2 Meet #3, January How many diagonals can be drawn in a decagon? Note: decagon is a polygon with 10 sides and a diagonal of a polygon is a segment which connects any two non-consecutive vertices. 2. 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 BCD 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 BCEFGIJKLMN)? B C D F E H K L J G I N M

14 Solutions to Category 2 Meet #3, January 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. 2. 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. 144 < ( 180n 360) n 144n < 180n < 180n 144n 360 < 36n ( 180n 360) n < n 360 < 150n 180n 150n < n < < n n < 12 The only whole number value between 10 and 12 is 11, so the polygon must have 11 sides. 3. Square BCD has an area of 144 square units, so its side length is 12. Length DL is 5, so triangle DL is a triangle as the Pythagorean Theorem confirms: = = 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 triangle, since = = 9 and 3 2 = 9. We can now label all the side lengths in the figure as shown at right. The perimeter is = 100 units. 12 B N C D F 4 E G H 5 I K 3 L J M