Geometry: A Complete Course

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1 Geometry: omplete ourse with Trigonometry) Module Instructor's Guide with etailed Solutions for Progress Tests Written by: Larry. ollins RRT /010

2 Unit V, Part, Lessons 1, uiz Form ontinued. Match each statement in column I with a phrase in column II. olumn I olumn II Rectangle iagonal of a polygon Polygon onvex Polygon Square Parallelogram Trapezoid Vertex of a polygon uadrilateral Rhombus b i h f j g d e c a a) n equilateral parallelogram b) parallelogram that has one right angle c) closed path of four segments that does not cross itself d) quadrilateral that has exactly one pair of parallel sides e) n end point of a side of a polygon f) polygon in which any diagonal lies inside the polygon g) quadrilateral with opposite sides parallel h) simple closed curve made up entirely of line segments i) segment whose endpoints are two non-consecutive vertices of a polygon j) n equilangular equilateral quadrilateral. list of properties found in the group of seven special quadrilaterals is given below. Write the name of the special quadrilaterals) beside the given property for which that property is always present. a. oth pairs of opposite sides are parallel. Parallelogram, Rectangle, Rhombus, Square b. xactly one pair of opposite sides are parallel. Trapezoid, Isosceles Trapezoid c. oth pairs of opposite sides are congruent. Parallelogram, Rectangle, Rhombus, Square d. xactly one pair of oppsite sides are congruent. Isosceles Trapezoid e. ll sides are congruent. Square, Rhombus f. ll angles are congruent. Rectangle, Square 009 VideoTextInteractive Geometry: omplete ourse

3 Unit V, Part, Lessons 1, uiz Form ontinued. Indicate whether each of the following is true or false. a) very square is a rhombus. True b) very rhombus is a square. False c) very square is a kite. False d) very rhombus is a kite. False e) If a quadrilateral has three sides of equal length, then it is a kite. False f) very property of every square is a property of every rectangle. False g) very property of every trapezoid is a property of every parallelogram. False h) very property of a parallelogram is a property of every rhombus. True 009 VideoTextInteractive Geometry: omplete ourse

4 uiz Form lass ate Score Unit V - Other Polygons Part - Properties of Polygons Lesson - Midsegments Use the figure to the right for problems 1 and. 1. Point H is the midpoint of GJ. GH = Point L is the midpoint of GK. HJ = If GJ = 1, and HL = 9, Find GH, HJ, and KJ. KJ = 1 G GH = 1 GJ HJ = GH KJ = HL GH = 1 HJ = KJ = 9 1 KJ = 1 GH =. Point H is the midpoint of GJ. Point L is the midpoint of GK. 1 If m GLH is 1 degrees and KJ = 1, find m K and HL. L K J Problems 1 and m K = 1 O 7 1 HL = H m K = m GLH m K = 1 HL = 1 KJ HL = HL = 1 9 HL = 9 =7 1. Using the figure to the right, find,, m, m, and m. = 1 FG = FG m = = 1 = m = 0 = 1 = m = 90 = = 1 m = 0 O m = 90 O m = 0 O F 0 O G G 10 m = VideoTextInteractive Geometry: omplete ourse

5 Unit V, Part, Lessons, uiz Form ontinued. In the figure to the right, W, T, and S are midpoints of the sides of triangle F. If WT =, ST =, and SW = 7, What is the perimeter of F? = WT = = 10. Which of the following named quadrilaterals are parallelograms? a) b) W c) F = SW F = 7 F = 1 F = ST F = F = 1 Y Z Permimeter of F = 0 is not a parallelogram XYZW is a parallelogram GHIJ is a parallelogram Since point and are not midpoints. X J S G W I H T F. In the figure to the right, is a trapezoid with median MN as shown. a) If = 10t and MN = 1t, find. = 0t + MN = 10t + 1t = 0t = 10t + 0t = M N b) If = x and MN = x, find. = 1x + MN = +x x = x = +x 1x = 9 + c) If = 9 and =, find MN. MN = + MN = MN = VideoTextInteractive Geometry: omplete ourse

6 Unit V, Part, Lessons, uiz Form ontinued. In the figure to the right, point is the midpoint of, and point is the midpoint of. = x +, = y +, = x, and = y +. Find and. ) = = x += y+ = x - y+= x-) x+= y+ += y+ = ) - x-y=1 10 = y + = 10 - y+= x-10 -x +y = - =y = -x + y = -1 -x + y=-1) -7x = - =y = y + -x +y = - x= = + = 10. Which of the following named quadrilaterals are parallelograms? a) b) M c) 7 W 7 Z X N F Y P WXYZ is not a parallelogram MNP is a parallelogram F is not a parallelogram = 0 = 0. In the figure to the right, is a trapezoid with median MN as shown. M N a) If = x + and MN = 10x 1., find. b) If = and = 7, find MN. = 1x 7. MN = + MN = x x -1. = 0x -. = x + + 1x -7.= c) If =.7 and = 1., find MN. MN = MN = MN = MN = 1.1 MN = 10. MN = VideoTextInteractive Geometry: omplete ourse + MN = MN = +7 MN = 10 MN = MN =

7 uiz Form Unit V - Other Polygons Part - reas of Polygons Lesson 1 - Postulate 1 - rea Lesson - Triangles lass ate Score For problems 1, find the area of the given polygon using the appropriate Postulate, Theorem, or orollary from lesson 1 and. 1. = units. 7 = units 1 =b h = 0 O = 1 l l 0 O 1 = units ase: Side opposite 0 O angle x where x = 1 1 = = Height: Side opposite 0 O angle x where x = 1 1 = = 1 7 = units. = units. = units x + = x +9= x = x= =s s or s = = units Height is the long leg of the right triangle with short leg equal to and hypotenuse equal to. = 1 b h = 1. =. = units Height = 10 = 10 ase of triangle is 10. Height of triangle is the long leg of a right triangle with a hypotenuse of 10. So, the height is x where x = 10. = rea = 1 b h = = units 1 units = 1 10 = 10 = uni = ts O 9 ltitude of triangle is x where x =9 x= 9 x= 9 = 9 = 1 b h = = 9 9 = 1 = 1 u nits 009 VideoTextInteractive Geometry: omplete ourse

8 Unit V, Part, Lessons &, uiz Form ontinued For problems 7 and, find the area of each polygonal region = units. = o 10 o 1 1 ivide the region into a trapezoid with one base equal to 1, a second base equal to 1 and an altitude equal to ) and a x square. = 1 hb+b = = = 1 units 1 ) ) =s s= =9 units Total area of the polygonal region is 1 +9=1 + 1 = 1 units For Problems 9 and 10, find the area of the shaded region. 9. = 0 units 1 The area of the rectangle is b h or 1 10 or 10 units 1 rea of each trapezoid is hb+b or ) ) 10 = 1 = 9 = units rea of shaded triangles is the area of the rectangle minus the area of each trapezoid = 0 units o 17 ivide the region into a trapezoid with bases equal to 10 and and height equal to ), and a parallelogram with base equal to and height equal to 17 - ) ) = 1 hb+b 1 = 1 1+ ) = 1 = 9 =7 units =b h = 17- = 10-1 So, total area equals = ) 10. = units 0 o 9 0 o 009 VideoTextInteractive Geometry: omplete ourse Height of trapezoid and parallelogram is. O O O Use 0,0,90 right triangle properties) rea of shaded triangle is area of trapezoid minus area of parallelogram. 1 h b +b -b h= = -1 = ) ) 1 = - 1 = - 0 = 1 = units

9 Unit V, Part, Lessons &, uiz Form ontinued For Problems 7 and, find the area of each polygonal region. 7. =. = For the triangle on the right = 1 O O O Use,,90 right triangle properties) = = 9 For Problems 9 and 10, find the area of the shaded region ) units units For the triangle on the left = 1 = 9 Total rea = = 9+9 or 91+ ) or 9 1+ ) un its = units working from the top down.) rea = b h = = = 1= = 1= = 1= =7 1=7 = 1= =9 1=9 Total rea = units 0o 0o rea = 1 b h where h =, and b =. O O O Use 0,0,90 right triangle properties) 1 Total rea of the polygonal regions is b h = units = units rea of the shaded region is the area of the triangle minus 0 o VideoTextInteractive Geometry: omplete ourse the area of the trapezoid. Height of the trapezoid is or. = 1 b h- 1 ) hb+b 1 = 1 9 ) Use 0 O O O,0,90 right triangle properties) = =- = 7 - = 9 units

10 Unit V, Part, Lesson, uiz Form ontinued. Find the area of an equilateral triangle inscribed in a circle, with a radius of units. rea = units pothem = 1 = = = = 1 s a n = 1 1 s= = 1 = units s= s= s=1 9. Find the area of a square with an apothem of inches and a side of length 1 inches. rea = inches = 1 s a n or = s = 1 = inches = 1 1 = 1 = 1 1 = inches 10. Find the area of a regular hexagon with an apothem of 11 meters and a side of length meters. rea = 1 meters = 1 s a n = 1 11 = 1 meters VideoTextInteractive Geometry: omplete ourse

11 Unit V, Unit Test Form ontinued. Find the area of each of the following labeled polygonal regions using the appropriate postulate, theorem, or corollary. Note: figures which appear to be regular are regular) -) -) a) rea = units b) rea = units 11 rea of Triangle = 1 b h 7 Triangle) 9 Trapezoid) rea of Trapezoid = 1 h b +b ) 1 = 1 7 ) = 1 9+ = units = 1 1 = 17 = units -) -) c) rea = units d) 11 rea = units Rhombus) 0 o rea of Rhombus Parallelogram) = b h h = long leg of a = b h = x = = = = units Paralleloram) rea of Parallelogram = b h = 11 = units -1) -) e) rea = units f) rea = units rea of Rectangle = b h or l w -) 7 -) g) rea = units h) rea = units Rectangle) = = units Regular Pentagon) = 1 s a n = 1 7 = units 009 VideoTextInteractive Geometry: omplete ourse 10 Where x = + Regular Triangle) Square) = 1 s a n = 1 = 1 s a n = 1 10 raw line segments from the center of the triangle to the vertices, considering = the apothem to be. We now have two, right triangles. The long leg of either is 1 x 10 or. So, = = units or. = = =units

12 Unit VI, Part, Lessons 1,&, uiz Form ontinued. Use the figure to the right to complete the following statements. In the figure, JT is tangent to at point T. T a) If T = and J = 10, then JT = K Since T JT by orollary a, TJ is a right triangle. a +b =c T ) + JT ) = J) ) ) ) + JT =10 + JT =100 JT = JT = J b) If T = and JT = 1, then J = 17 Since T JT by orollary a, TJ is a right triangle. a +b =c T ) + JT ) = J) +1 = J) + = J) 9 = J) 17 = J c) If m JT = 0 and T =, then J = 1 Since T JT, TJ is a right triangle. If m JT = 0, TJ is a triangle, and T is the short leg of the triangle opposite the 0 degree angle. T = 1 / J in a triangle. T = 1 J = 1 J 1 = J d) If JK = 9 and K =, then JT = 1 Since T JT, TJ is a right triangle. a + b = c T ) + JT ) = J) + JT ) = JK + K) + JT ) = 9+ ) + JT ) = 17) JT 9 + ) = JT ) = 009 VideoTextInteractive Geometry: omplete ourse JT = 1

13 Unit VI, Part, Lessons 1,&, uiz Form ontinued. Find in, = 1. Find m in, m = O if = 10 and = 9. = from Theorem 7 if m = 9 O 7. Find in. =. Find and in. = 1 = 1 x x and, So bisects ) = ) + ) 9) = ) + ) 1 = + ) = ) ± = cannot be negative) 1 = 1 = = 1 = x x = x ± = x x cannot be negative) 1 = x = x so, = m = O from orollary 7a and, So bisects = call the radius " r") r r= r r r r = r r r = = + = + = 1 = + = + = 1 and are diameters) 9. Find in, given = 10. Find in. = that = 1, = 9 and = = 10 Theorem 7 - bisects, So = ) = ) + ) ) = ) + 9 ) ) = + ) ) = + 1 ) = 0 =± 0 cannot be negative) = 1 = 009 VideoTextInteractive Geometry: omplete ourse x = x = x = x ± = x x cannot be negative) = x = x = x = = )

14 Unit VI, Part, Lessons,,&7, uiz Form ontinued. Find x in. x =. Find m in. m = x -x 1 = 1 = x x 0 = x x 1 0 = x ) x+ ) 0 = x or 0 = x+ = xor = x x cannot be negative) x = 9 If we draw radii,,, and, we can prove triangles congruent. We can then use Theorem 7 and prove. m = 0 m m m m = m =. Find in. = 1. Find and in. = 1 = 9 7. Find and in. = 0.. and are m = 1 O = 1 tangents to, and bisects by = m = O. orollary 0a 0.x Find m x 1. = = x 1. = x 1. ) 1. )= 09. x) 0.x) 1. = 0. x 1. = x 0. = x ± = x x cannot be negative) = x = x = 0. ) x = 0. ) x = 0. ) = 0. ) = 0. = VideoTextInteractive Geometry: omplete ourse x+ x+ x+1 x 1 1 = x x+ x+ ) = x 1 ) x 1 )+ x ) x x+ )= x+ 1) x+ ) x + x = x + x+ x = x+ x = = x + x + = x + 1)+ x + = )+ )+ = 1 = + 1)+ = + = 9 + ) ) 1 m = m 1 = = 1

15 uiz Form lass ate Score Unit VI - ircles Part - ircle oncurrency Lesson 1 - Theorem - If you have a triangle, then that triangle is cyclic. Lesson - Theorem - If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic. 1. uadrilateral is cyclic. Find x and y. x = 100 x y y = 90 and are supplementary. orollary 7b) m + m = 10 m + m + m = 0 y + 90 = 10 x = 0 y = 90 x + 0 = 0 x = m = m Th eorem 7) 1 7 = m 10 = m or m ). uadrilateral Kite) is cyclic. Find m. m = 1 uadrilateral is a kite, so, Theorem 1), and m = 1. m + m = 10 is a semicircle) m + 1 = 10 m = 009 VideoTextInteractive Geometry: omplete ourse 11

16 UnitVI, Unit Test Form ontinued etermine whether each of the following is always, sometimes, or never true. sometimes always always never never sometimes always always always always sometimes sometimes never -7) 1. ongruent chords of different circles intercept congruent arcs. -) 1. n angle inscribed in a semicircle is a right angle. -1) 1. Two circles are congruent if their radii are congruent. -) 1. Two externally tangent circles have only two common tangents. -1) 17. radius is a segment that joins two points on a circle. -) 1. polygon inscribed in a circle is a regular polygon. -1) 19. secant is a line that lies in the plane of a circle, and contains a -) chord of the circle. 0. The opposite angles of an inscribed quadrilateral are supplementary. Postulate - p19) 1. If point X is on, then mx + mx = mx. -). The common tangent segments of two circles of unequal radii are congruent. -). Tangent segments from an external point to two different circles -) are congruent.. yclic quadrilaterals are congruent. -). If two circles are internally tangent, then the circles have three common tangents VideoTextInteractive Geometry: omplete ourse

17 UnitVI, Unit Test Form ontinued -). Find m m = 9 9. x = 10 -) m m 10 orollary 7b + m = 10 m = 9 + = ) = Theorem 7) + x = x x x = ) + ) x = 10 x =± 10 x cannot be negative) x = 10 x = Find m m = 0 1. x = 17-1) m + m = 10 is cyclic. orollary 7b) m = 10 = 0 x = 17; Theorem 0). Find =. Find. = 10 F 100 Given:, =, the radius of is. If =, then F = Theorem 7) ) + ) = ) -) -) -1) l F F F) + ) = ) F) + 1 = Given: l 1 and l are perpendicular F) = 9 bisectors of the sides of. F =± 9 = 0. F cannot be negative) F = 9 = 10, since = = orollary a) F = so, =. ongruent hords in the same circle are equidistant from the center of the circle. See Unit VI, Part, Lesson 7, xercise ) 009 VideoTextInteractive Geometry: omplete ourse 1 l1 x 17 x l

18 UnitVI, Unit Test Form ontinued -). Find m m = 9 9. x = 10 -) m m 10 orollary 7b + m = 10 m = 9 + = ) = Theorem 7) + x = x x x = ) + ) x = 10 x =± 10 x cannot be negative) x = 10 x = Find m m = 0 1. x = 17-1) m + m = 10 is cyclic. orollary 7b) m = 10 = 0 x = 17; Theorem 0). Find =. Find. = 10 F 100 Given:, =, the radius of is. If =, then F = Theorem 7) ) + ) = ) -) -) -1) l F F F) + ) = ) F) + 1 = Given: l 1 and l are perpendicular F) = 9 bisectors of the sides of. F =± 9 = 0. F cannot be negative) F = 9 = 10, since = = orollary a) F = so, =. ongruent hords in the same circle are equidistant from the center of the circle. See Unit VI, Part, Lesson 7, xercise ) 009 VideoTextInteractive Geometry: omplete ourse 1 l1 x 17 x l