1. LINE SEGMENTS. a and. Theorem 1: If ABC A 1 B 1 C 1, then. the ratio of the areas is as follows: Theorem 2: If DE//BC, then ABC ADE and 2 AD BD

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1 Chapter. Geometry Problems. LINE SEGMENTS Theorem : If ABC A B C, then the ratio of the areas is as follows: S ABC a b c ( ) ( ) ( ) S a b A BC c a a b c and b c Theorem : If DE//BC, then ABC ADE and AD AE DE AB AC BC AD AE AD DB DB EC AE EC Theorem 3: If ACB =ADC = 90, then AC AB AC AB AD AD AC () BC BD BC AB BD AB BC () CD BD CD AD BD AD CD (3) CD AB AC BC CD AB AC BC () CD BC AC (5) Theorem (Angle Bisector Theorem): The angle bisector of a triangle divides the opposite side into segments that are proportional to the adjacent sides. AB BD AB AC or. AC CD BD CD The length of the angle bisector is AD AB AC BD DC.

2 Chapter. Geometry Problems Theorem 5. Tangents to a circle from an outside point are congruent. PA = PB. Example. (008 NC Algebra II) An isosceles triangle ABC with point D on AB, has AC = BC = BD and AD = DC. If AB =, find the length of CD. A. B. C. 5 D. 0 3 E. 3 5 Solution: E. Let x = CD. Since ΔABC ~ ΔABC CD : AC = AC : AB x = (AC). AD + BD = AB x + AC = x = ( x) x 6x + = 0 x = 3 5. Example. (00 NC Algebra II) A small tree 5 feet from a lamp post casts a shadow feet long. If the lamp post were feet higher the shadow would only be 3 feet long. How tall is the tree? A. 7/ ft. B. /5 ft. C. 8/3 ft. D. /3 ft. E. 7/5 ft. Solution: B. AB DE a b ABC DEC, then BC EC 5 9b a () a b Similarly () b Substituting () into (): b b Example 3: (0 Illinois Algebra II) In Triangle ABC, AB =6, BC =5, and AC =9. AD bisects CAB, and CD bisects ACB. Expressed in simplest radical form, AD = k w, where k and w are positive integers. Find the value of (k + w).

3 Chapter. Geometry Problems Solution: 59. Applying the angle bisector theorem, we have AB AC CE CE 5 CE CE CE = 9/7. By the angle bisector length formula, we have AE AB AC CE EB AE 69 (5 ) AE Applying the angle bisector theorem again, we have 9 CE AC AE 7x x ED AD AE x x x AE k = and w = 57. So k + w = + 57 = 59. Example. (00 NC Algebra II) A circle is inscribed in a triangle with sides of lengths of 8, 7, and 9. Let the segments determined by the point of tangency on the side of length 8 be w and r, with w < r. Find the ratio of w to r. A. 3:5 B. 5:7 C. :3 D. :5 E. 7:9 Solution: A. r + w = 8 () 9 r = 7 w () Solving the system of equations: r = 5 and w = 3. The ratio is 3:5.. PYTHAGOREAN THEOREM (For right triangles only): a + b = c (a and b are two legs. c is the hypotenuse). Proof: In right triangle ABC, draw CD AB. We know that ABC ACD CBD and: AC BC AB AD AB BD () () 3

4 Chapter. Geometry Problems () + (): AC BC AB AD AB BD AB( AD BD) AB AB AB Example 5. (00 UNC- Charlotte Algebra II) For what positive value of x is there a right triangle with sides x +, 6x, and 6x +? A. 8 B. 6 C. D. 5 E. 9 Solution: C. The number x must satisfy (x + ) + (6x) = (6x + ) since 6x + is clearly the largest of the three numbers. Thus x + 8x x 36x x = x 6x = 0 from which it follows that x =. Example 6. (00 NC Algebra II) When the base of a ladder is 6 ft from the base of a wall, 3 ft of the ladder projects beyond the top of the wall. When the base of the ladder is 9 ft from the base of the wall, 8 ft of the ladder projects beyond the top of the wall. How long is the ladder? A. 5 ft B 6.5 ft C. 0 ft D. 3 ft E. 5.5 ft Solution: D. Let the length be x. b 6 ( x 3) () b 9 ( x 8) () () (): 6 9 ( x 3) ( x 8) x x QUADRATIC EQUATION/FUNCTION APPLICATION Example 7. (00 Tennessee Algebra II) A rectangular garden measures 8 feet wide and 6 feet long. The gardener wants to put a strip of gravel of uniform width around the garden. There is enough gravel or square feet. How wide will the strip be? A. feet B. foot C. 6 inches D. feet E. 3 feet. Solution: A. ( 6 x )(8 x) 86 x 8x 0 x x 8 0 ( x )( x ) 0. So x =. Example 8. (005 NC Algebra II) Each spring a meter meter rectangular garden has its length increased by meters but its width decreased by 50 centimeters. What will be the maximum attainable area of the garden? A. m B. 76 m C. 89 m D. 00 m E. 5 m

5 Chapter. Geometry Problems Solution: E. The area will be ( + y)( 0.5y)= + y 6y y = +8y y, where y is the number of years. This quadratic expression has a maximum at its vertex, which occurs when y = b/(a) = 8/( ) = 9. The area when y = 9 is +8(9) 9 = 5. Example 9. (0 Illinois Algebra II) A rectangular park with dimensions 5 feet by 30 feet is surrounded by a sidewalk of uniform width of x feet along the sides of the park. If the area of the sidewalk is between 900 square feet and 7900 square feet and if x is an integer, find the sum of all distinct possible values for x. Assume all edges of the sidewalk are composed of straight line segments. Solution: (5 x )(30 x) x 70x x (x 355) Since , we test x = : x ( x 355) 5. So x = is good. Since , we test x = : x ( x 355) 7. So x = is not good. We also see that when x = 0, x ( x 355) So x = 0 is not good. Thus the sum of all values for x is = 9.. AREAS AND VOLUMES.. Given a triangle with side a, b, and c and height h on the side a, the area of this triangle is A and A ah A a b sin C A s ( a b c) a b c abc R sin A sin B sin C A A A r s perimeter Surface area of a sphere: A r. Volume of a sphere: V r d. 3 6 Example 0. (00 Tennessee Algebra II) The area of a triangle is 600 square feet. If the height is three times the base, what is the height of the triangle? A. 0 feet B. 0 feet C. 60 feet D. 30 feet E. 0 6 /3 feet. Solution: C. s( s a)( s b)( s c) 5

6 Chapter. Geometry Problems A b h 600 h h 3 h h. Example. (007 NC Algebra II) In square ABCD, X lies on DC such that DX : XC = 5 : and Y lies on BC such that BY :YC = 3 :. The ratio of the area of ΔAXC to the area of ΔABY is A. :7. B. :3. C. 3:. D. :9. E. 9:6. Solution: B. Since 5 + = 7, and 3 + = 7, each side length is a multiple of 7, so 5 : can be represented as 5a : a and 3 : can be represented at 3a : a. Then the area AXC : areaaby = (a) 7 : (3a) 7 = : 3. Example. (00 Indiana Algebra II) The number of cubic centimeters in the volume of a sphere is the same as the number of square millimeters in its surface area. The radius of the sphere in millimeters is: A. 300 B. 30 C. 3 D E. none of these Solution: C. Surface area of a sphere: A r. Volume of a sphere: V r 3 3 So r = r r = r = Example 3. (999 NC Algebra II) The height of a triangle is 6 units less than the length of its base. If the area of this triangle is square units, how many units is the length of its base? A B C. 60 D. 0 E. 5. Solution: A. Let b be the length of the base. b ( b ) b 6b b 3 93, and b 3 93 (ignored). Copyright 03 mymathcounts.com 6

7 Chapter. Geometry Problems PROBLEMS Problem. (00 NC Algebra II) If the length and width of a rectangle were increased by, the area would be 8. The area would be 8 if the length and width were diminished by. Find the perimeter P of the original rectangle. A. 0 < P < 0 B. 0 < P < 30 C. 30 < P < 0 D. 0 < P < 50 E. none of the above Problem. (00 NC Algebra II) The altitudes of a triangle are three distinct integers, the larger two of which are and 66. Find the length of the shortest altitude. Which of the following intervals contains the integer solution? A. (0, 3) B. [3, 6) C. [6, 9) D. [9, ) E. none of the above Problem 3. (00 NC Algebra II) Suppose the lengths of three of the four lateral edges of a pyramid with a rectangular base are 8, 5, and in that order. Find the length of the fourth lateral edge. A. 79 B. 7 C. 0 D. E. 3. Problem. (003 NC Algebra II) An isosceles right triangle region of area 36 is cut from a corner of a rectangular region with sides of length 6 and 6( ). What is the perimeter of the resulting trapezoid? A. 36 B. 8 8 C. 30 D. E.. Problem 5. (003 NC Algebra II) Pictured are two semicircles. AB is tangent to the smaller semicircle and parallel to CD. If AB is 6, find the area of the shaded region. a. 6 b. 3 c. 8 d. e. 8 Problem 6. (00 NC Algebra II) A triangle has one side of length a and two sides of length b. The area of the triangle is A. a b a B. a b a C. 3 ab D. b a b E. Problem 7. (00 NC Algebra II) A 5 foot tall ladder is placed along the vertical wall of a house. The foot of the ladder is 0 feet from the bottom of the house. If the top of the ladder slips 8 feet, then the foot of the ladder will slide how many feet? A. 3 ft. B. 5 ft. C. 8 ft. D. ft. E. 7 ft. b a b 7

8 Chapter. Geometry Problems Problem 8. (00 NC Algebra II) The perimeter of a rectangle is 00 cm, and its diagonal has length x. Express the area of the rectangle as a function of x. (50 x) A. cm (50 x) B. cm 500 x C. cm D. x 500cm E. x 500cm. Problem 9. (00 NC Algebra II) The volume of wood V in a tree varies jointly as the height h and the square of the girth g (girth is the distance around). If the volume of a redwood tree is 6m3when the height is 30 m and the girth is.5 m, what is the height of a tree whose volume is 960m3 and girth is m? A.. 8 m. B. 5 m. C. 50 m. D. 75 m. E. 95 m. Problem 0. (005 NC Algebra II) A street sign 0 feet from a lamp post casts a shadow 5 feet long. If the lamp post were foot taller the shadow would only be feet 8 inches long. How high is the lamp on the post? A. 5 ft. B. 7 / ft. C. 5 /3 ft. D. 35 ft. E. 70 ft. Problem. (005 NC Algebra II) A rectangular computer image in enclosed with a two pixel wide frame composed of 50 pixels. If the width of the image is increased by 0% then the frame needs 5 pixels. How many pixels does the image have? A. 3,600 pixels B. 3,7 pixels C. 3,800 pixels D. 5,00 pixels E. None of these. Problem. (007 NC Algebra II) If the sum of the squares of the lengths of all the sides of a rectangle is 00, then the length of a diagonal of the rectangle is A. 5 B. 3 C. 3 D. 5 E. 0. Problem 3. (007 NC Algebra II) A circle is centered at the vertex of the right angle of an isosceles right triangle. The circle passes through the trisection points of the hypotenuse of the triangle. If the length of the radius of the circle is 0, find the area of the triangle. A. 5. B. 60. C. 75. D. 90. E. 05. Problem. (007 NC Algebra II) An open box is to be created from a nine-inch by twelve-inch piece of posterboard by cutting congruent squares from each corner and folding up the sides. The goal is to cut squares of the size that will produce the box with maximum volume. Determine that volume to the nearest cubic inch. A. 8. B. 98. C. 8. D. 50. E. 6. 8

9 Chapter. Geometry Problems Problem 5. (008 NC Algebra II) The perimeter of a right triangle is six times longer than its shortest side. What is the value of the ratio of the longer leg to the shorter one? A..083 B.. C.. D..6 E.. 6. Problem 6. (008 Tennessee Algebra II) The lengths of the three sides of a right triangle are consecutive multiples of three. What is the area of the triangle? A. 08 B. 5 C. 36 D. 90 E. 5 Problem 7: (0 Illinois Algebra II) Alicia is 5 feet tall and walks on a perfectly horizontal flat surface at the rate of 5 feet per second away from a streetlight with its lamp 6 feet above horizontal flat surface. Alicia is now directly under the lamp and walk away from the lamp for seconds. Find the number of feet in the length of Alicia s shadow directly after those seconds. Problem 8. (03 Alabama Algebra II) A square is inscribed in another square, such that each vertex divides a side of the outside square into intervals of length x and y, where x > y. What is x/y, if the area of the inscribed square is /5 of the area of the outside square? A. 5 B. 6 C. 5 D. 5 E. 6. Problem 9. (009 Indiana Algebra II) When a right triangle of area 3 square units is rotated 360 º about its shortest leg, the solid that results has a volume of 30 cubic units. What is the volume, in cubic units, of the solid that results when the same right triangle is rotated about its longer leg? Problem 0. (00 Tennessee Algebra II) A rectangular garden measuring 80 by 60 meters has its area doubled by adding a border of uniform width along both shorter sides and one longer side. Find the width of the border. A. 0 B. 5 C. 0 D. 5 E. 30. Problem. (03 Illinois Algebra II) In a regular pentagon, each diagonal has a length of k w p. The perimeter of the pentagon can be written in reduced simplest radical form, f where k, w, p, and f represent positive integers. Find the smallest possible value of (k + w + p + f ). 9

The Theorem of Pythagoras

CONDENSED LESSON 9.1 The Theorem of Pythagoras In this lesson you will Learn about the Pythagorean Theorem, which states the relationship between the lengths of the legs and the length of the hypotenuse