a B. 3a D. 0 E. NOTA m RVS a. DB is parallel to EC and AB=3, DB=5, and

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1 Geometry Individual Test NO LULTOR!!! Middleton Invitational February 18, 006 The abbreviation "NOT" denotes "None of These nswers." iagrams are not necessarily drawn to scale. 1. Which set does not always define a plane?. three noncollinear points.. two distinct intersecting lines.. two parallel lines. a line and a point. NOT 6. The area of a regular rhombus is half of the area of a triangle. One side of the rhombus is 15. If one side of that triangle has length 1, then find the height to that side NOT 7. regular pentagon has one interior angle which has measure x degrees. Give the value of x NOT. Parallelogram has m 70. If m ( x10) then x 8. In the circle shown, mrvs a. R S NOT 3. n equilateral triangle has perimeter 9. What is the height of the triangle? V Find an expression for mrvsm RUSm RTS. U T NOT.. a. o a 3a. 0. NOT 4. quadrilateral RSTU is inscribed in a circle. If mr 30 and ms 40 then give the mt NOT In right, is the hypotenuse and is cm more than the longest leg. The other leg is half of. Give the length of the shortest leg, in cm NOT is parallel to and =3, =5, and =6. Find the length to the nearest tenth place NOT

2 Geometry Individual Test NO LULTOR!!! Middleton Invitational February 18, Given rectangle with = 15 cm and = 8 cm, circles with centers at and F are inscribed in and. Find F in cm.. 85 cm. 6 cm. 8 cm. 5 3 cm. NOT 11. is tangent to the circle at. The measure of arc is 11. Find m NOT 1. circle passes through two vertices and the orthocenter of an equilateral triangle with sides of length 3 cm. Find the area of this circle.. 4cm. 3cm. 3 cm. 3cm. NOT 13. n ornithologist stands 80 ft from the face of a vertical cliff and observes an eagle's nest on a ledge. She measures the tangent of the angle of elevation to be 3/7. If the height of her eye is 5 ft when she measures the tangent, how high is the nest? 14. The incenter of is located at point. If m = 3, find m NOT 15. Triangle contains sides = 8 cm, = 1 cm, = 16 cm. M is the midpoint of. Find the length of M.. 10 cm. 8 cm. 4 3 cm. 6 cm. NOT 16. If x y, m1 = a and then find m m4 = 4a - 4, NOT 17. right rectangular prism with dimensions (in cm) of 1, 15 and 16 is inscribed in a sphere. Find the surface area of the sphere.. 65cm. 440cm cm cm. NOT 18. In the diagram m = 108 and m = 64. Find m. 7 x y. 10 ft. 15 ft. 137 ft. 140 ft. NOT NOT

3 Geometry Individual Test NO LULTOR!!! Middleton Invitational February 18, n altitude of a right triangle intersects the hypotenuse at a point that divides it into a ratio of 4 to 9. If the hypotenuse measures 7, find the length of the altitude NOT SecantPintersects a circle at points and so that is 19 cm and P is 8 cm. Find the length of tangent segment PT. T. is a cyclic quadrilateral whose diagonals intersect at P. If P is 6, P is 8 and is 15, find length P NOT is externally tangent to circles and as shown in the figure below. If the radius of is 8 cm, the radius of is 15 cm and the distance between the centers is 5 cm, find the area of quadrilateral. P cm. 9 3 cm. 11 cm. 6 6 cm. NOT 1. regular octagon has a side of length 1 cm. Find its area cm. 55 cm. 575 cm. 76 cm. NOT 4. In the diagram below,,,, and are on the circle, m70and m 80. Find m P.. ( 8888 ) cm. ( ) cm. ( ) cm. ( ) cm. NOT P NOT

4 Geometry Individual Test NO LULTOR!!! Middleton Invitational February 18, In triangle, = = =. If the measure of is 100, what is the measure of? NOT 6. In the figure below,. Find the sum of x and y NOT x y 8. The quadrilateral formed by connecting the midpoints of the consecutive sides of a concave quadrilateral, none of which are congruent, is always a:. rectangle. rhombus. parallelogram. trapezoid. NOT 9. Floyd hates nglish class, so he began to watch the (analog) clock at 11:50 am. What is the fewest number of minutes that will elapse before the hands of the clock are at right angles? NOT plane contains a number of straight lines, no two of which are parallel and no three of which are concurrent. If these lines intersect in 153 distinct points, how many lines are in the plane? NOT 7. In right, sin = cos = 3. If = 8, find NOT

5 Geometry Individual Test NO LULTOR!!! Middleton Invitational February 18, 006 Solutions 1.. The point could be on the line... ngles and are supplementary. x + 10 = x = Perimeter 9 means side is 3. ltitude divides the triangle into a with short leg 3/. Long leg is (3/)( 3 ) 4.. Opposite angles of an inscribed quadrilateral are supplementary. mt = Let x = short leg. The hypotenuse is x (the "other" leg is half the hypotenuse ) and the long leg is x - ( less than the hypotenuse). Pythagorean Theorem x ( x) ( x ) simplifies to the quadratic x 8 x 4 0. Quadratic Formula yields x = 4 3. While 4-3 is also a solution, it is not a choice. 34 is a choice. 6.. regular rhombus is a square. Side length is 15. rea 5. The area of the triangle is 450 (twice the area of the square.) Using the triangle side of 1 as the base, h as the length of the altitude to that side, and the formula = (1/)bh leads to (1/)(1)h = 450. h = One interior angle of a regular pentagon is 108. x = 108. x = 54. x + 8 = SinceRVS is an inscribed angle, the arc it intercepts is a. RUS andrts intercept the same arc. They also measure a. a + a - a = a Let x = m.. 5x + 15 = 18. x = 0.6 3x iagonal divides the rectangle into congruent right triangles with sides of 15 and 8. is the hypotenuse of the Pythagorean triple. Solving, m is 17. The radius of a circle inscribed in a right triangle is half of (the sum of the legs minus the hypotenuse). r = (1/)( ) = 3 F is the hypotenuse of right triangle GF. See diagram. mg = 8 - r = 8-6 =. mgf = 15 - r = 15-6 = 9. y Pythagorean Theorem, mf = is perpendicular at the point of tangency. It must pass through the center making it a diameter and = 180. Subtracting m, = 68 = m. m is half of the arc that it intercepts, making it See diagram. m is given as 3, so m = 3 and the length of the altitude is 3. Since the orthocenter is also the centroid of an equilateral triangle, is and is also. The perpendicular bisector, F, of chord passes through the center of the circle at. ecause F is 60, F is a and is equilateral with sides =. This is also the radius of the circle. = r = Tangent is opposite (height of the nest) over adjacent (distance G F F to the cliff). Using a proportion, 3 h. h = 10 plus the height of eye The incenter is formed using the angle bisectors. Let the angle at measure a and the angle at measure b. Since measures 3, a + b = a + b = 74. Looking at the triangle formed by the vertices, and, m + a + b = 180. m =106.

6 Geometry Individual Test NO LULTOR!!! Middleton Invitational February 18, 006 a b c 15.. M is a median. y formula, the length of a median is where a and b are the sides 4 that form the vertex from which the median is connected to the midpoint of the opposite side, c mm = and4 are supplementary a + 4a - 4 = 180. a = is vertical with1, so they are congruent. m5 = = The space diagonal, d, of the prism is the diameter of the sphere. d = = 5. The radius is 5/. Surface area of a sphere is 4r = and are inscribed and equal to half the arcs they intercept. m = 64/ = 3. m = 108/ = 54. These two angles are the remote interior angles for exterior = = 86 x Let x = short part of the hypotenuse. Then,. x = 8 63 and 7 - x =. Since the 7x height, h, is equal to the square root of the product of x and x - 7, h = = = mpt = ( mpt ) = ( mp)( mp ) = 8 (8 + 19). 87 = See diagram. onstructing a square around the octagon, the square has sides of length The area of the octagon is the area of the square minus the 1 four triangles at the corners. (11 ) 4[ (6 ) ] rea = = The intersection of the diagonals divides each diagonal into two parts. The product of these parts is always equal. 3 P (P) = P ( P). P = 6(15-6)/8 = 7/4 = and are both perpendicular to the same segment, so they are parallel and is a trapezoid. Put a point on so that is parallel to. is a rectangle. = 15-8 = 7. The right triangle has a hypotenuse of 5 and one leg of 7. y the Pythagorean Theorem, the other leg 1 is 4 which is the height of the trapezoid. rea of the trapezoid is (8 15)(4) m = = 30. intercepts the same arc, so it is also 30. P is supplementary with the 70, so it is 110. mp = = Let m = a and m = c, Then a + c = 180. a + c = 80. m = a and m = c because and are isosceles. m = (a + c) = 100. m = m = ( )/ = 40. m = a because it is an exterior angle with both remote interior angles = a. a = 40. a = orresponding sides of similar triangles are proportional. and x 6 y x = 36/8 = 4.5 and y = 7/8 = 9. x + y = 13.5

7 Geometry Individual Test NO LULTOR!!! Middleton Invitational February 18, The right angle is at. is similar to a triangle with hypotenuse 1 and long leg 3 /. The short leg is 1/. is a with the 60 at. Short leg is 8, Long leg is onsider concave quadrilateral (a dart) with the diagonal that is contained within the quadrilateral. (iagonal is outside.) The segment from the midpoint of to the midpoint of is a midsegment of, paralel to and half its length. The segment from the midpoint of to the midpoint of is a midsegment of, paralel to and half its length. Segments parallel to the same segment are parallel to each other. oth are half of thus congruent. Segments of a quadrilateral that are both parallel and congruent define a parallelogram. 9.. The minute hand starts out 5/6 of the way around the clock face, thus it is pointing at the 10. The hour hand begins 5/6 of the way from the 11 to the 1. onsidering the face to be 360, the minute hand begins at 300 and the hour hand at 355. The 55 initial angle is decreasing as time continues. To reach a point where the hands are perpendicular, the minute hand must pass the hour hand and continue increasing the angle until the degree measure of the minute hand minus the degree measure of the hour 4 hand is 90. Solve. ( m) - ( m) = m = 145. m = 1450/55 = One line = 0 intersections; lines = 1 intersection; 3 lines = 3 intersections. Triangular numbers. ouble and factor to get n(n-1)/ = 153. Solve. n(n-1) = 306. Factors of 306 which are different by 1 are 17 and 18. The larger factor is n. n = 18.

8 Geometry Team Middleton Invitational February 18, 006 NO LULTOR 1. Let the points of concurrency for a triangle be represented by these letters: for incenter, for orthocenter, for centroid and for circumcenter. Record these letters in order to match the descriptions. First: intersection of the medians Second: intersection of the angle bisectors Third: intersection of the perpendicular bisectors of the sides Fourth: intersection of the altitudes Geometry Team Middleton Invitational February 18, 006 NO LULTOR. = The number of vertices in a polyhedron with 1 faces and 4 edges = The number of diagonals in a dodecagon = The measure of one exterior angle of a regular nonagon = The sum of the first six triangular numbers Find Geometry Team Middleton Invitational February 18, 006 NO LULTOR 3. = the height in cm of a cylinder with a diameter of 1 cm and a volume of 180cm 3 = the radius in cm of the base of a cone with a slant height of 9 cm and a surface area of 90cm = the side, in cm, of the square base of a pyramid with a height of 1 cm and a volume of 400 cm 3. = the width in cm of a right rectangular prism with a length of 1 cm, height of 7 cm and surface area of 47 cm. Find Geometry Team Middleton Invitational February 18, 006 NO LULTOR 4. ngle and angle are right angles. If = 10, = 1 and = 6, find the area of triangle.

9 Geometry Team Middleton Invitational February 18, 006 NO LULTOR 5. ircle M is circumscribed about trapezoid with = 10 and = and a height of 1. Find the area of the circle. M Geometry Team Middleton Invitational February 18, 006 NO LULTOR 6. The equations of two circles are given below. Let the point (, ) represent the center of the first circle and equal the radius. For the second circle, let the point (, ) represent the center and F equal the radius. Find F. y - x + 13 = 4y - x + 9x x + 6y - 14 = 8x - y - 3 Geometry Team Middleton Invitational February 18, 006 NO LULTOR 7. ircles and of radius 10 cm intersect in such a way that their centers are 10 cm apart. What is the area of the shaded region inside of and outside of both circles?

10 Geometry Team Middleton Invitational February 18, 006 NO LULTOR 8. onsider the following six statements. If the statement is true, add the value in parentheses. If it is false, subtract the value. Find the total.. iagonals of a parallelogram bisect each other (5). onsecutive angles of a kite are supplementary (13). Opposite angles of a rhombus are congruent (3). iagonals of a rectangle are perpendicular (37). n isosceles trapezoid has exactly one line of symmetry (41) F. The exterior angle of a regular dodecagon measures 30 (59) Geometry Team Middleton Invitational February 18, 006 NO LULTOR 9. 3 Find the area of a triangle formed by the x-axis, y-axis and the line y x 9. Geometry Team Middleton Invitational February 18, 006 NO LULTOR 10. The 90 vertex of isosceles right triangle LUV is the center of square MTH. If the square has a side of length 10 cm and point H is on the hypotenuse of the right triangle, what is the area of the overlapping (shaded) region? T V U M H L Geometry Team Middleton Invitational February 18, 006 NO LULTOR 11. n industrious (and very worried) squirrel begins storing nuts for the winter. On day 1 he stores 5 nuts. The next day he realizes that he must pick up the pace to have enough nuts to last the winter. He stores 13 nuts on day, 17 on day 3 and 1 on day 4. If he continues in this pattern, how many TOTL will he have stored after 30 days?

11 Geometry Team Middleton Invitational February 18, 006 NO LULTOR 1. kite has consecutive sides which measure 0 and 30 cm. If one vertex angle measures 90, Find the area of the kite. Geometry Team Middleton Invitational February 18, 006 NO LULTOR 13. Find the ordered pair for the centroid of a triangle whose vertices are located at (-13, 7), (, -14) and (-15, 13) Geometry Team Middleton Invitational February 18, 006 NO LULTOR 14. ~ MNP ( MNP which is not shown)if one side of MNP is 51 cm, and al sides are integers, then K is the length of the largest possible side of MNP. F~ XYZ (also, not shown) If YZ = 85 cm, then H is the largest possible altitude of XYZ. Find K + H F Geometry Team Middleton Invitational February 18, 006 NO LULTOR 15. The sides of squares and FG are 3 cm and 1 cm respectively. Find the area of triangle H expressed as a fraction. H F G

12 Solutions: Middleton Invitational Geometry Team SOLUTIONS 1. from definitions. uler's Formula for solids: vertices + faces = edges + (number of vertices) + 1 = 4 + = 14 For diagonals in an n-gon, use n (n - 3)/ For a dodecagon, n = 1. = 1(1-3)/ = 54 = 360/9 = 40 = = = = Volume of a cylinder = r h, where h is the height. 180= 1 h. = 5 Surface area of a cone = r + rl where r = radius of the base of the cone and l is the slant height. r + r (9) = 90 ivide pi out of both sides and solve the quadratic. r 9 r Factors are (r + 15) and (r - 6). For a positive r, r = 6. = 6 Volume of a square pyramid = height. 400 = (1) 3 s. 100 = s s = 10 = 10 s h where s = side length of the square base and h is the Surface area = (lw) + (lh) + (wh). 47 = (1)w + (1)(7) + (w)(7). w = 8 = = = 9 4. First find the height of triangle. 106 h h or 7 4 The area of the triangle is 5. To find the radius begin by constructing the height through the center of the circle and solving two right triangle problems to find x. r 5 (1 x ) and r 11 x so that x x 11 x and x =. r rea = r 15 1-x M x 5 r r ombine like terms. rrange in circle format. omplete the square. x 10 x y 4 y 13 x 8 x y 6 y 3 14 x 10 x 5 y 4 y x 8 x 16 y 6 y ( x 5) ( y ) 16 center (5, ) radius 4 ( x 4) ( y 3) 36center (4, -3) radius F = (-3) + 6 = 18

13 7. Label the point of intersection of the two circles inside of as. is equilateral with sides of 10. rea equals The area of each of the sectors and is The shaded area is the area of minus the area of the triangle and both of the sectors. rea = True +5 False -13 True +3 False -37 True +41 F True +59 Total is The x and y axis make this a right triangle with one leg equal to the y intercept of 9. To find the x intercept, solve the equation for y = 0. x = 6 rea =.5(9)(6) = GUI KUJ because both are 90 - mguj. lso, both are right triangles with a leg of 5. Their areas are equal. The shaded area is simply the area of square KUGH U I T G V 11. ay n Total Nuts Factors (n-1)(n+4) For n = 30 (30 1)(30 4) 006 M K J L H 1. The 90 angle has to be between the 0 cm sides because One diagonal is 0. To find the other, use the Pythagorean Theorem to find x = The area of the kite is half the product of the diagonals. 0 0/ 0/ x 30 1 (0 )( ) = ) 13. The centroid is located at the mean of the vertex coordinates , which is (-, ) 3 3

14 14. For ~ MNP, a scale factor of 3 times 17cm wil produce a side of 51 and integer side lengths for all three sides in the not shown triangle. The largest possible side is 3(4) = 7 = K. For F~ XYZ, a scale factor of 5 times the 17 cm side wil produce a side length of 85 and integer side lengths in the not shown triangle. lso, F is a right triangle. The largest possible altitude is the longest leg. 5(15) = 75 = H. K + H = = G is a right triangle with legs of 3 and 4. G is 5. G is similar to HG. To find H, solve the proportion 3 H 3 to find H = so that H is 9 and the area of 4 H is