Park Forest Math Team. Meet #4. Geometry. Self-study Packet

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1 Park Forest Math Team Meet #4 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 Circles rea of a Circle: =!r 2 Circumference of a Circle: C=!d or C=2!r *(Very Important!) e sure to use the test s given value for! and not the! button on your calculator!!! Other properties: If C is a diameter of a circle and point is any other point on the circle, angle C will be a right angle. The measure of an inscribed angle is half the measure of the arc it subtends. For example, if arc is 70, then the measure of angle D is 35

2) C is the diameter of circle O. D = DC. = 9 feet. O O C = 40 feet. C How many square feet are in the area of quadrilateral CD?

3 Category 2 Meet #4 - February, 2016 Calculator Meet 1) ngle O is a right angle. The radius of circle O is 8 centimeters. How many square centimeters are in the area of the shaded region? Use Round your final answer to the nearest tenth. 2) C is the diameter of circle O. D = DC. = 9 feet. O O C = 40 feet. C How many square feet are in the area of quadrilateral CD? Round your final answer to the nearest whole number. D 3) J = 12 decimeters. UJ = 8 decimeters. QU = 18 decimeters. How many units long is Q? Quadrilateral JQU is tangent to the circle. nswers Q 1) U 2) 3) J

4 Solutions to Category 2 Meet #4 - February, ) The shaded area is 1/4 the area of the circle nswers = = 1) 50.3 = = ) 600 = Rounding to the nearest tenth yields ) 22 2) n angle inscribed in a semi-circle is a right angle. Triangle C is a right triangle with right angle at vertex. Triangle DC is isosceles as well as right. To find the area of triangle DC, it is necessary to know the length of the diameter of the circle that is also the hypotenuse of the two right triangles. Let x = the length of D. So, the area of quadrilateral CD 3) The "around the world" strategy is very effective here, utilizing the property of circles that when two tangents are drawn to a circle from an exterior point, the tangent segments are congruent. Let x = the distance from point to the tangent point to its left. Then the following relationships exist, as we go around the circle clockwise: Then Q = x + {18 - [8 - (12 - x)]} = x + {18 - [ x]} = x + { x} = x x = 22. U = the area of triangle DC! = (area of C) + (area of DC) = (0.5)(9)(40) = = = 600 when rounded to the nearest whole number. J Q

Category 2 Meet #4 - February, 2014 50th anniversary edition Calculator Meet 1) ngle C is a right angle. is the diameter of a semi-circle.

5 Category 2 Meet #4 - February, th anniversary edition Calculator Meet 1) ngle C is a right angle. is the diameter of a semi-circle. How many square cm are in the area of the entire figure if C = 51 cm and C = 45 cm? Use. C 2) The circumference of the larger circle is 46π. The circumference of the smaller circle is 14π. What fractional part of the larger circle is shaded? Express your answer as a common fraction (lowest terms). 3) rc VWU is a quarter-circle with point Y at the center. XZ = 10 cm. The sum of the length and width of rectangle WXYZ is 13 cm. How many centimeters are in the perimeter of the figure bounded by the points VWUZXV? Use 3.14 π. NSWERS 1) sq. cm V X W 2) 3) cm Y Z U

Solutions to Category 2 Meet #4 - February, 2014 nswers 1) 763.2 1) Use the Pythagorean Theorem to find the length of diameter : 2) 480 529 3) 32.7 So, the radius of is half of 24, or 12.

6 Solutions to Category 2 Meet #4 - February, 2014 nswers 1) ) Use the Pythagorean Theorem to find the length of diameter : 2) ) 32.7 So, the radius of is half of 24, or 12. The total area of the entire figure = (area of semi-circle) + (area of triangle) 2) Using the formula for the circumference of a circle, C= 2π r, we find that the radius of the smaller circle is 7 and the radius of the larger circle is 23. Using the formula for the area of a circle, π r2, we find that the area of the smaller circle is 49π and the area of the larger circle is 529π. y subtracting the area of the smaller circle from the area of the larger circle, we get the area of the shaded region = 480π. Therefore, the fractional part of the larger circle that is shaded is 480π 529π = ) The key that unlocks this puzzle is the notion that the diagonals of a rectangle are congruent, so that XZ = YW = 10 cm = the radius of the circle = YU = VY. The length of the arc VWU, the quarter-circle, is 0.25(2)(3.14)(10), or 15.7 cm. VX + XY = radius = 10 and YZ + ZU = radius = 10. VX + XY + YZ + ZU = = 20 VX + (length of rectangle + width of rectangle) + ZU = 20 VX + (13) + ZU = 20, therefore VX + ZU = 7. (Tricky, huh!!) So, perimeter of VWUZXV = = 32.7 cm.

7 Meet #4 February 2012 Calculators allowed Category 2 Use 1. The area of a circle is square inches. How many inches are there in its circumference? Give an exact answer with no rounding. 2. The radius of the circle shown is centimeters. The radius points to o clock, and the radius points to o clock. How many centimeters are there in the arc? O 3. square is inscribed inside a circle. What percentage of the square s perimeter is the circle s circumference? Round your answer to the nearest whole percent. nswers 1. inches 2. cm 3. %

8 Meet #4 February 2012 Calculators allowed Solutions to Category 2 Geometery nswers 1. The area of a circle is so in our case the radius is inches The circumference is inches. 2. The hours on the clock divide the central angle to equal parts (each one being degrees then). The arc from to will measure of the whole circumference, or in our case cm 3. If we call the square s side, then its perimeter is. Its diagonal is, and that is the circle s diameter, so the circle s circumference equals

9 You may use a calculator today! Category 2 - Meet #4, February ssume that the Earth orbits the Sun along a perfect circular orbit with a radius of 150 million kilometers, and completes the orbit in 365 days and 6 hours. What is the Earth s average speed around the Sun? Express your answer in Kilometers per Hour (km/hour), rounded to the nearest integer. Use π = Kite CD is inscribed inside a circle whose center is point O. DC=25 degrees. How many degrees are in the measure of OD? [ Kite is made up of two isosceles triangles] O C D 3. The radii (plural of radius) of both circles in the diagram measure 10 inches. They intersect each other in such a way that the distance measures 10 inches. How many inches are in the perimeter of the resulting shape? Express your answer in inches, rounded to the nearest hundredth. Use π = nswers

10 You may use a calculator today! Solutions to Category 2 - Meet #4, February Remember that Speed = Distance / Time. nswers , The distance, in kilometers, is the perimeter of the orbit, namely: 2 π R = kilometers. The time, 365 days plus 6 hours, equals to hours. The average speed then is: ( ) = ,766 = 107, ,461 km/our 2. DC = 25 = DC Therefore CD = 130 degrees (to complete to 180 degrees). D is inscribed on the chord D, and so equals 180 CD = 50 degrees. Finally, OD = 2 D = 100 degrees, since it s the central angle on the same chord. Something to think about: Does CD have to be a kite? 3. The perimeter is the sum of the two circles perimeters, minus the two arcs. If we connect and to a circle s center O, we get an equilateral triangle, since we know that equals the radius of the circle. Therefore the central angle O measures 60 degrees, and so the arc represents one-sixth of the circle s perimeter. This of course holds for the second circle as well, as it has the same radius. So the anwer is 5 of the two perimeters, or π R = 10 3 π 10 = = inces. O

Category 2 Meet #4, February 2008 You may use a calculator today! E F D C G H 1. In the semi-circle to the right, Point O is the center, and Points and J are at opposite ends of a diameter.

11 Category 2 Meet #4, February 2008 You may use a calculator today! E F D C G H 1. In the semi-circle to the right, Point O is the center, and Points and J are at opposite ends of a diameter. The 8 points, C, D, E, F, G, H, and O I are equally spaced around the semicircle and each is connected to both ends of the diameter J. What is the sum of the degrees in the measures of the angles J, CJ, DJ, EJ, FJ, GJ, HJ, and IJ? I I D J 2. In the square IDF at the right two semicircles are drawn using I and DF as diameters and the two semicircles are tangent to each other. F = 8 cm. n ant crawls along the lines and arcs in a path that takes it from Point and then through, C, D, E, F, G, H, I, J, and back to, in that order. How many centimeters long was the path the ant travelled? Express your answer as a decimal to the nearest tenth of a centimeter. J H C G E F 3. circle is inscribed in the square to the left and the area between the two shapes is shaded. Using 3.14 as an estimation for Jimmy calculated that the area of the shaded region is cm 2. Using 3.14 as an estimation for again, how many centimeters are in the circumference of the circle? Express your answer as a decimal to the hundredths place. nswers

12 Solutions to Category 2 Meet #4, February 2008 nswers ll 8 of the triangles use the diameter as one side and have the 3 rd vertex on the circle, so all 8 of the triangles are right triangles. The sum of the angles is just The path the ant travels will be along the entire circumference of the circle once and along the diameter twice. So it will travel a total of I D H J C E Here is what the path looks like with arrows to guide you. G F 3. If we call the radius of the circle r, then the sides of the square are all 2r making the area of the square (2r) 2 = 4r 2. Since the area of the circle is 3.14r 2, the area of the shaded region must be 4r r 2 =.86r 2 which we know to be So and 22 Therefore the circumference of the circle is 3.14(22) = 69.08

13 Category 2 Meet #4, February 2006 You may use a calculator 1. toy car has wheels with a diameter of 1 inch. How many turns does each wheel make if the car rolls 12 feet across the floor? (There are 12 inches in 1 foot.) Use 3.14 for and round your answer to the nearest whole number of turns. 2. How many centimeters are there in the circumference of a circle with an area of 36 square centimeters? Use 3.14 for and express your answer as a decimal to the nearest tenth of a centimeter. 3. Three semi-circles of diameter 2 centimeters are cut from three sides of a 4-cm by 4-cm square to form the figure below. circle of radius 1-cm is placed above the square without overlap. How many square centimeters are in the area of the figure? Use 3.14 for and express your answer to the nearest tenth of a square centimeter. nswers

14 Solutions to Category 2 Meet #4, February 2006 nswers Wheels with a diameter of 1 inch have a circumference of 1 = inches, which is about 3.14 inches. If the car rolls 12 feet across the floor, then it rolls = 144 inches. The question now is how many turns of 3.14 inches there are in 144 inches. Dividing 144 by 3.14, we get about turns, which is 46 to the nearest whole number of turns. 2. The formula for the area of a circle is circle = πr 2. We can find the radius of the given circle by solving the equation 36π = πr 2. Since 6 2 = 36, the radius must be 6 centimeters. The formula for the circumference of a circle is C = πd or 2πr, so the circumference of our circle is 12. Using 3.14 as an approximation of, we get = or 37.7 to the nearest tenth of a centimeter. 3. If we cut the circle above the square in half, we can fill two of the voids on the sides of the square That would leave a square with just one semicircular region cut out of it. The area of the square is 4 cm 4 cm = 16 square centimeters. The area of a circle with radius 1 centimeter is 1 2 = square centimeters, so the area of a semicircle with radius 1 cm is 0.5. Thus the figure has an area of = = or 14.4 square centimeters to the nearest tenth.

Meet #4. Math League SCASD. Self-study Packet. Problem Categories for this Meet (in addition to topics of earlier meets):

Meet #4. Math League SCASD. Self-study Packet. Problem Categories for this Meet (in addition to topics of earlier meets): Math League SCASD Meet #4 Self-study Packet Problem Categories for this Meet (in addition to topics of earlier meets): 1. Mystery: Problem solving 2. : Properties of Circles 3. Number Theory: Modular Arithmetic,