Deductive geometry. This chapter at a glance Stage 5.1/5.2/5.3 After completing this chapter, you should be able to:

Transcription

1 eductive geometry eductive geometry This chapter at a glance tage 5.1/5.2/5.3 fter completing this chapter, you should be able to: apply the properties of complementary, supplementary and vertically opposite angles and angles at a point to find unknown angles, giving reasons find unknown angles on parallel lines, giving reasons use the angle sum of a triangle to find unknown angles, giving reasons use the exterior angle property of triangles to find unknown angles, giving reasons use the angle sum of a quadrilateral to find unknown sides and angles, giving reasons use the properties of the special quadrilaterals to find unknown sides and angles, giving reasons name a polygon according to the number of sides classify a polygon as either convex or non-convex, regular or irregular find the interior angle sum of a polygon find the size of the interior and exterior angles of a regular polygon use geometric properties to find unknown angles in diagrams involving more than one step prove unfamiliar geometric results by using formal reasoning apply the congruence tests to justify that two triangles are congruent apply the congruence tests to establish properties of triangles and quadrilaterals write formal deductive proofs involving the tests for congruent triangles use ythagoras theorem to prove unfamiliar results use the converse of ythagoras theorem to prove that a triangle is right-angled 118

2 hapter 4 : eductive geometry imple numerical exercises The questions in this exercise will review the basic geometrical properties of angles, triangles, quadrilaterals and general polygons that have been covered in previous years. The emphasis in this exercise is on the giving of correct reasons to justify each answer. ormal proofs of results and definitions of the special quadrilaterals are left to later exercises. djacent angles Two angles are adjacent if they: have a common vertex, and have a common ray, and lie on opposite sides of this common ray. β G H or example, G is adjacent to GH because: 1 is a common vertex, and 2 G is a common ray, and 3 the angles lie on opposite sides of G. omplementary and supplementary angles omplementary angles have a sum of 90. upplementary angles have a sum of 180. or example: β β and are complementary and are supplementary angles: + β = 90 angles: + β = 180 NOT: hen giving reasons, the terms complementary angles and supplementary angles must not be used. These terms simply mean that the angles have a sum of 90 or 180. In geometric reasoning, you need to explain why the angles have that sum (for example angles in a right angle, or angles on a straight line).

3 120 Mathscape 10 xtension ngles at a point ngles at a point are two or more angles that have a common vertex and whose sum is 360, or one complete revolution. or example: ngles at a point have a sum of 360. γ β + β + γ = 360. Vertically opposite angles Vertically opposite angles are formed by the intersection of two straight lines. Vertically opposite angles must be equal because they are adjacent and supplementary to a common angle. Vertically opposite angles are equal. or example: Y * T * TZ and YTX are vertically opposite angles. TY and ZTX are vertically opposite angles. Z X arallel lines arallel lines are two or more lines that have been drawn in the same plane and never meet. The notation means is parallel to. line that cuts two or more parallel lines is called a transversal. hen a pair of parallel lines is cut by a transversal, 8 angles are formed. These angles can be classified into 3 special pairs of angles: alternate angles, corresponding angles and co-interior angles. alternate angles corresponding angles co-interior angles

4 hapter 4 : eductive geometry 121 The angle sum of a triangle The angle sum of a triangle is 180. β That is, + β + γ = 180. γ The exterior angle of a triangle The exterior angle of a triangle is equal to the sum of the two interior opposite angles. β That is, γ = + β. γ ome other properties of triangles In an equilateral triangle, In an isosceles triangle, In any triangle, the longest all angles are 60. the equal angles are side is opposite the largest opposite the equal sides. angle and the shortest side is opposite the smallest angle The angle sum of a quadrilateral l The angle sum of a quadrilateral is 360. β γ δ That is, + β + γ + δ = 360.

5 122 Mathscape 10 xtension roperties of the special quadrilaterals Trapezium arallelogram hombus ectangle quare Kite roperties of a parallelogram: roperties of a rhombus opposite sides are parallel all properties of a parallelogram opposite sides are equal all sides are equal opposite angles are equal diagonals are perpendicular diagonals bisect each other diagonals bisect the angles at the vertices roperties of a rectangle: roperties of a square: all properties of a parallelogram all properties of a rectangle all angles are right angles all sides are equal diagonals are equal diagonals bisect the angles at the vertices diagonals are perpendicular roperties of a trapezium: roperties of a kite: one pair of opposite sides are two pairs of adjacent sides are equal parallel xercise ind the value of each pronumeral, giving reasons. a b c d 40 y 70 t 64 u

6 hapter 4 : eductive geometry 123 e f 25 g h 45 k 28 p m a a a i j k l 139 n z z z r 72 z 70 r w w n 63 w w w 2 ind the value of each pronumeral, giving reasons. a b c r 35 d e f (y 25) 2k f 2f 39 3 ind the value of each pronumeral, giving reasons. a b p c d e 28 f c h a t

7 124 Mathscape 10 xtension g h i q k s 50 4 ind the value of all pronumerals, giving reasons. ll lengths are in cm. a b c b 92 z 5n n d e f x a 8 u v 70 5 y 25 g h v i p 7 7 k 7 w 10 3 j k l d e 46 r s k j 12 m 18 n o f b 15 4 y z a

8 hapter 4 : eductive geometry ind the value of all pronumerals, giving reasons. a b c y p q 3a b 2a d e f 3g 76 g k 2k 4k 3k 4n 15 11n g h i 4b 60 2c 97 (12v 47) (5v + 16) 3b ind the value of all pronumerals, giving reasons. a b c a b u v d c 33 d 135 e 75 f 117 3m 2n k 13 g 152 j h i 7 ind the value of each pronumeral. (You do not need to give reasons.) a b c t n 75 r d e f 144 h c k

9 126 Mathscape 10 xtension 8 tate whether TU V in each of these. If the lines are parallel, give a reason. a b c T V T 81 U T 109 V U V U 9 ind the value of all pronumerals. o not give reasons. a b c a 72 b 3 4z 33 (p + 12) (4q + 27) c 2y 108 3s 2 10 ind the value of the pronumeral in each of these, giving reasons. a b c 3b (e + 9) 4g 2g 3g e 123 d e f 60 5n (7x 20) (3x + 30) (3t 18) 75 g (11s 20) (4s + 29) h i 44 (9y 12) 3y 76 8w

10 hapter 4 : eductive geometry ind the value of all pronumerals, giving reasons. a b n m 71 c b 71 a p q d e f c 48 d 59 g y h g h i 21 u r s v j k j k l f e 66 p q n 50 m ind the value of all pronumerals, giving reasons. a b c f 120 e 115 a b p q

11 128 Mathscape 10 xtension d 108 e 54 f g h v 36 u w v g h i c j k d t s 13 ind the value of all pronumerals, giving reasons. a b c m n 38 b c a b c d d e f z 76 a y b 73 y c z urther applications 14 ind the value of x. (You do not need to give reasons.) is a rhombus and =. ind the value of x. (You do not have to give reasons.) 72

12 hapter 4 : eductive geometry olygons ommon polygons polygon is a closed figure bounded by only straight sides. The name of a polygon is based on the number of sides that make up the boundary of the figure. The names of the first 10 polygons are shown below. ides olygon ides olygon 3 Triangle 8 Octagon 4 uadrilateral 9 Nonagon 5 entagon 10 ecagon 6 Hexagon 11 Undecagon 7 Heptagon 12 odecagon onvex and non-convex polygons polygon can be convex or non-convex. convex polygon is a polygon in which all of the diagonals lie within the figure. ll interior angles are less than 180. non-convex polygon is a polygon in which at least one diagonal does not lie onvex polygon completely within the figure. One or more Non-convex polygon interior angles is greater than 180. The interior angle sum of a polygon The sum of the interior angles of any n-sided polygon is given by = (n 2) 180. roof: Let n be a convex polygon with n sides. hoose any point O inside the polygon and join it to each of the vertices, forming n triangles. The angle sum of each triangle is 180, therefore, the sum of the angles in n triangles is 180n. However, this includes the angles around O whose sum is 360. These angles must be subtracted from the angles around the boundary of the polygon to give the interior angle sum. Hence, = 180n 360 = 180 (n 2), on factorising 1 n 2 O

13 130 Mathscape 10 xtension The exterior angle sum of a polygon hen one side of a polygon is produced, the angle between this produced side and an adjacent side of the polygon is called an exterior angle of the polygon. The sum of the exterior angles of any convex polygon is 360. roof: Let n be a convex polygon with n sides. If each side of the polygon is produced as shown, then the sum of the interior and exterior angles at each vertex is 180. s there are n interior angles and n exterior angles, the total sum of these angles is 180n. e know that the sum of the interior angles is 180 (n 2). The interior angles must be subtracted from the total angle sum to find, the sum of the exterior angles of the polygon. Hence, = 180n 180 (n 2) = 180n 180n = 360 n egular polygons regular polygon is a polygon in which all of the sides are equal and all of the angles are equal. The size of the interior and exterior angles in a regular polygon can be found by dividing the sum of these angles by the number of angles. In any regular n-sided convex polygon: 180 ( n 2) each interior angle measures n 360 each exterior angle measures n xercise How many sides has each polygon? a quadrilateral b hexagon c octagon d decagon e pentagon f heptagon g nonagon h dodecagon i undecagon 2 tate whether each polygon is convex or non-convex. a b c d

14 hapter 4 : eductive geometry tate whether each polygon is regular or irregular. a b c d 4 a Name a quadrilateral in which: i the angles are equal but the sides are not. ii the sides are equal but the angles are not. b re the quadrilaterals in a regular? xplain. 5 ind the angle sum of each pentagon by dividing the figure into triangles, as shown. a b onsolidation 6 ind the size of the interior angles in a regular: a pentagon b hexagon c octagon d decagon 7 ind the size of the interior angles in a regular polygon with 20 sides. 8 orm an equation and solve it to find the value of c. 2c c Use the formula θ = to find the size of each exterior angle in a regular: n a hexagon b pentagon c dodecagon 10 ind the size of the exterior angles of an equilateral triangle. 11 ind the size of the exterior angles in a regular polygon with 24 sides. 12 How many sides has a regular polygon whose exterior angles measure: a 45 b 36 c 20 d how that a regular polygon cannot have interior angles measuring onstruct a regular pentagon in a circle using a ruler and compasses. Measure the angles in the pentagon and verify that each angle is 108. urther applications 15 ind the interior angle sum of a regular polygon that has: a exterior angles measuring 72 b interior angles measuring 156

15 132 Mathscape 10 xtension TY THI atio of exterior angles If the exterior angles (a, b, c ) are in the ratio 4:5:6, find the ratio of the interior angles a : b : c. c' c b b' a a' 4.3 Harder numerical problems The logical order of the steps in these questions needs to be thought through before a written solution is attempted. ull reasons must be given for each step in the argument. G + xample 1 In the diagram, =, and. ind the value of a, giving reasons. a olution = 62 (vertically opposite s) = 62 (base s of an isosceles, = ) = 62 (alternate s, ) = 28 (adjacent s in a right angle) a = G G + xample 2 In the diagram,, G is a parallelogram and G bisects. ind the value of y, giving reasons. olution = 64 ( sum of is 180 ) = 116 (co-interior s, ) G = 58 (G bisects ) G = 58 (opposite s of a parallelogram) y = G y

16 hapter 4 : eductive geometry 133 xercise ind the value of x in each of the following, giving reasons. a b K N c X 107 Y 64 T 44 J L M V KL bisects JKM Z U d H J e I J f 67 G g h i 58 H G HG bisects G onsolidation I K M L N 155 U 23 T T 77 V U V bisects T 2 ind the value of x in each of these, giving reasons. a b c T X U Z Y U bisects d H e J L f 33 X V V K M N G OK bisects LN O 29

17 134 Mathscape 10 xtension g h i G T U 109 V U X 34 Y 81 H V Z j Y k l N V O K L 27 X J U M O is the centre of the circle I Z IJKLM is a regular pentagon 3 In each of the following, is a rectangle. ind the value of x, giving reasons. a b c 53 T 67 T T 21 urther applications 4 In each of the following, is a rhombus. ind the value of m, giving reasons. a b c m m m 17 d e f m 128 m m G =

18 hapter 4 : eductive geometry eductive proofs involving angles The questions in this exercise involve the proof of general results in figures by the use of angle relationships. In some questions it is necessary to begin by choosing one particular angle and writing a statement such as, let =. In other questions it may be necessary to choose two particular angles and write a statement such as let = and = β. hen the size of an angle is referred to by a Greek letter, the degrees symbol is not written, by convention. ull and correct reasons need to be given for each step in the argument. G + xample 1 In the diagram, U V T and V bisects. rove that U and T are supplementary angles. U olution Let U = V = U (corresponding s, U V) V = V = V (V bisects ) V = V + T = 180 (co-interior s, V T) + T = 180 T = 180 Now, U + T = = 180 U and T are supplementary angles 180 V T U V T

19 136 Mathscape 10 xtension G + xample 2 In the diagram, and. rove that =. G olution Let = and = β G + G = 180 (co-interior s, ) β + G = 180 G G = β G + G = 180 (co-interior s, G) β + G = 180 G = 180 β G = G (vertically opposite s) 180 = 180 β = β = xercise and are straight lines which intersect at. rove that =. 2 a rove that =. b rove that + = is a straight line through O. O bisects O. rove that O = O. O

20 hapter 4 : eductive geometry 137 onsolidation 4 is a straight line through. bisects. bisects. rove that. 5 T X UY ZV UY bisects TUV X rove that TX = UVZ. U Y Z V 6 rove that = + 7. β rove that = β + γ. γ 8 GH J rove that HG = JI H G urther applications 9 bisects bisects rove that. J I

21 138 Mathscape 10 xtension TY THI ngles in a rhombus is a rhombus. = 30. = 36. ind the value of x. 4.5 eductive proofs involving triangles The questions in this exercise involve the proof of general results in figures by the use of geometrical properties of triangles and angle relationships. s with the questions in the previous exercise, you should begin each question by choosing one or two angles and labelling them as or β. ongruent triangles are not required in this exercise. G + xample 1 In the diagram, = and =. rove that. olution Let = = (base s of an isosceles, = ) = = (vert. opp. s) = = (base s of an isosceles, = ) = Now, = (both equal to ) (alternate s are equal)

22 hapter 4 : eductive geometry 139 G + xample 2 In the diagram, = =. rove that. olution Let = and = β. = (base s of isosceles, = ) = = (base s of isosceles, = ) = β + + = 180 ( sum of is 180 ) + ( + β) + β = β = β = 90 = 90. β β xercise is an any triangle. rove that the angle sum of the triangle is 180. [Hint: onstruct through, parallel to.] 2 In, is produced to. rove that = +. [Hint: onstruct parallel to.] 3 In, = +. rove that the triangle is right-angled. 4 In, and are points on and respectively such that and. rove that =.

23 140 Mathscape 10 xtension 5 In the diagram, and =. rove that is isosceles. 6 In the diagram, = and. rove that bisects. 7 In, =. is a point on such that bisects. rove that = 3. 8 In, and = 30. is drawn to such that =. rove that is equilateral. 30 onsolidation 9 In XYZ, XY = XZ. The bisector of X meets the base YZ at. Let YX = and XY = β. a xplain why XZ = + β. b how that XY = + β. c Hence, prove that X YZ. 10 In,. is a point on such that =. a rove that is isosceles. b Hence, prove that is the midpoint of. 11 is a point on the circumference of a circle with centre O and diameter. a xplain why O = O = O. b Hence, prove that = 90. Y β X Z O

24 hapter 4 : eductive geometry In, is produced to. is a point on such that bisects. Let = and = β. a ind expressions for and, giving reasons. β b Hence, prove that + = In the diagram,, bisects and bisects. and meet at G. rove that. 14 G U V Z X Y In the diagram, V = VY and UX Y. rove that UVZ is isosceles. 15 In, is a point on such that bisects. is a point on such that =. Let = and = β. a xplain why = + β. b Hence, prove that =. β 16 In the diagram, =, and G bisects. rove that G. G urther applications 17 In the diagram, = and. bisects and bisects. rove that. [Hint: Let = and = β.] β

25 142 Mathscape 10 xtension 4.6 ongruent triangles review Two triangles are said to be congruent if they have exactly the same size and shape. ach triangle can be obtained from the other by performing one or more of the following transformations translation, rotation or reflection. The sides and angles that are in the same positions relative to other sides and angles are called matching sides and matching angles. The symbols and are used to mean is congruent to. If two triangles are congruent, then: the matching sides are equal in length the matching angles are equal in size the figures are equal in area. There are four standard tests that can be used to determine whether two triangles are congruent. 1 If the three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent (). 2 If two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the two triangles are congruent (). 3 If two angles and one side of one triangle are equal to two angles and the matching side of another triangle, then the two triangles are congruent ().

26 hapter 4 : eductive geometry If the hypotenuse and a second side of one right-angled triangle are equal to the hypotenuse and a second side of another right-angled triangle, then the two triangles are congruent (H). NOT: 1 If the three angles of one triangle are equal to the three angles of another triangle, then the triangles are not necessarily congruent. That is, is not a test for congruent triangles. 2 hen naming congruent figures, the vertices must be given in matching order. Thus, for the triangles below, we would write XYZ. Y X The standard congruence proof for triangles has five steps. Z To prove that two triangles are congruent: identify the triangles that are being used in the proof name the three pairs of equal sides or angles name the congruent triangles, giving the vertices of the triangles in matching order, and state the congruence test used. NOT: y convention, the sides or angles on the LH of the proof should belong to one triangle and the sides or angles on the H should belong to the other triangle. G + xample 1 and bisect each other at. a b rove that. Hence, show that. olutions a In and b = (matching s of congruent s) = ( bisects ) (alternate s are equal). = (vert. opp. s) = ( bisects ) ()

27 144 Mathscape 10 xtension G + xample 2 bisects. a rove that. b Hence, show that is isosceles. olutions a In and b = (matching sides of congruent s) = ( bisects ) is isosceles. = = 90 ( ) is a common side () G + xample 3 O is the centre of the circle OM. a rove that OM OM. b Hence, show that OM bisects O. olutions a In OM and OM OM = OM = 90 (OM ) O = O (equal radii) OM is a common side OM OM (H) b OM = OM (matching s of congruent s) OM bisects O. O M G + xample 4 XY = YZ Y bisects XZ. a b rove that YX YZ. Hence, show that Y XZ. X Z olutions a In YX and YZ b YX = YZ (matching s of congruent s) XY = YZ (given) ut, YX + YZ = 180 X = Z (Y bisects XZ) (adjacent s on a straight line) Y is a common side YX = YZ = 90 YX YZ () Y XZ. Y

28 hapter 4 : eductive geometry 145 xercise tate the test that could be used to prove that each pair of triangles are congruent. a b c d tate whether each pair of triangles are congruent. If they are congruent, state the test used. a 19 b c d L Is LMN? xplain. N M 4 a b T rove that T. G H rove that HG.

29 146 Mathscape 10 xtension c Y d M X Z rove that XYZ XZ. L K N rove that MLK MNK. onsolidation 5 a b I J K O and bisect each other at. rove that =. O is the centre of the circle and OJ IK. rove that OJ bisects IOK. c Y d O X Z XY = ZY and Y bisects XYZ. rove that XZ is isosceles. O is the centre of the circle and =. rove that O = O. e f I L J T, T and bisects T. rove that =. K IJ LK and LI KJ. rove that LJ bisects IJK.

30 hapter 4 : eductive geometry 147 g h = and. rove that bisects. T = T and T. rove that T. i j X O M bisects, and =. rove that. Y O is the centre of the circle and OM bisects XY. rove that OM XY. 6 In the isosceles triangle, =. X = Y. a rove that X Y. b Hence, show that XY is isosceles. 7 X Y In the isosceles triangle, =. and are the midpoints of and respectively. a rove that =. b Hence, prove that =. 8 In the isosceles triangle, =. L and M. a rove that L M. b rove that LN MN. c Hence show that LN = MN. L N M 9 a In, =. rove that =. b In, =. rove that =. (This is the converse of a.)

31 148 Mathscape 10 xtension 10 is equilateral. a onstruct, the bisector of. Hence, prove that =. b onstruct, the bisector of. Hence, prove that =. c Hence show that each angle in an equilateral triangle is 60. urther applications 11 is a triangle with =. is a point inside the triangle so that =. rove that bisects. 12 In a triangle LMN the sides LM and LN are equal and greater than MN. is any point on MN. On ML cut off M = N, and on NL cut off N = M. rove that =. 13 is an equilateral triangle. The perpendicular to at meets produced at, and the perpendicular to at meets produced at. rove that =. TY THI Intersecting parallelograms HIJK and GHI are both parallelograms. L is the midpoint of HK. rove that GJ = 3 HI. I L H J K G 4.7 eductive proofs involving quadrilaterals efinitions of the special quadrilaterals trapezium is a quadrilateral with at least one pair of opposite sides parallel. parallelogram is a quadrilateral with both pairs of opposite sides parallel. rhombus is a parallelogram with two adjacent sides equal in length. rectangle is a parallelogram in which one angle is a right angle. square is a rectangle with two adjacent sides equal in length. kite is a quadrilateral with two pairs of adjacent sides equal in length.

32 hapter 4 : eductive geometry 149 Tests for the special quadrilaterals arallelogram quadrilateral is a parallelogram if: the opposite sides are equal, or the opposite angles are equal, or one pair of opposite sides are equal and parallel, or the diagonals bisect each other. hombus quadrilateral is a rhombus if: all sides are equal, or the diagonals bisect each other at right angles. ectangle quadrilateral is a rectangle if: all angles are equal the diagonals are equal and bisect each other. quare quadrilateral is a square if: all sides are equal and one angle is a right angle, or all angles are right angles and two adjacent sides are equal, or the diagonals are equal and bisect each other at right angles. G + xample 1 is a rhombus. a rove that. b Hence, show that the diagonals of a rhombus are perpendicular. olutions a In and = (sides of a rhombus are equal) = (diagonals of a rhombus bisect the angles at the vertices) is a common side () b = (matching s of congruent s) ut, + = 180 (adj. s on a st. line) = = 90 the diagonals of a rhombus are perpendicular.

33 150 Mathscape 10 xtension xercise is any quadrilateral. rove that the angle sum of a quadrilateral is 360. [Hint: onstruct the diagonal.] 2 is a parallelogram. a rove that. b Hence show that = and =. c how that =. d hat property of a parallelogram have you proven? 3 is a parallelogram. The diagonals and meet at. a rove that. b Hence show that = and =. c hat property of a parallelogram have you proven? 4 is a rhombus. The diagonals and meet at. Let = and = β. a xplain why = and =. b imilarly, explain why = β and = β. c hat property of a rhombus have you proven? 5 is a rhombus. The diagonals and meet β at. Let = and = β. a xplain why = and = β. b ind + β. c Hence, explain why. d hat property of a rhombus have you proven? 6 is a rectangle. a rove that. b Hence show that =. c hat property of a rectangle have you proven?

34 hapter 4 : eductive geometry The proofs in this question verify the four standard tests for a parallelogram. a is a quadrilateral in which the opposite β angles are equal. Let = = and = = β. i ind the value of + β. β ii Hence show that and. b is a quadrilateral in which the opposite sides are equal. i rove that. ii Hence show that and. c is a quadrilateral in which = and. i rove that. ii Hence show that. d is a quadrilateral in which the diagonals and bisect each other at. i rove that. ii Hence show that = and. 8 is a quadrilateral in which the diagonals bisect each other at right angles at. a rove that. b Hence prove that =. c xplain why is a rhombus. 9 is a quadrilateral in which all angles are equal. rove that is a rectangle. 10 is a quadrilateral in which the diagonals are equal β and bisect each other. Let = and = β. a how that = and = β. b ind the value of + β. Hence prove that = 90. c xplain why is a rectangle.

35 152 Mathscape 10 xtension 11 is a parallelogram. The diagonals and meet at. line is drawn through, where lies on and lies on. a rove that. b Hence show that = and =. 12 is a parallelogram. The diagonal bisects. Let =. a rove that =. b xplain why is a rhombus. urther applications 13 is a parallelogram. is produced to and is produced to such that =. a how that =. b rove that. c Hence prove that is a parallelogram. 4.8 ythagoras theorem ythagoras theorem has been proven in more ways than any other theorem in geometry. In this exercise we will consider both numerical and geometric applications of the theorem. In any right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. That is, c 2 = a 2 + b 2 c 2 a 2 a c b b 2

36 hapter 4 : eductive geometry 153 OO: ata: is right-angled at, and squares are drawn on all sides. im: To prove that the square on side = sum of the squares on the other 2 sides. onstruction: Join, U and V as shown. roof: In U and, = (sides of a square are equal) U = (sides of a square are equal) U = (right + common ) U () Now = (= 90 ) is a straight line U 1 U = -- rectangle VU (same base and height) 2 1 = -- square (same base and height) 2 rectangle VU = square (double equal triangles) imilarly, rectangle VT = square square on = sum of squares on and V T The converse of ythagoras theorem If the square on one side of a triangle is equal to the sum of the squares on the other two sides, then the angle formed between these two sides is a right angle. OO: ata: such that 2 = im: To prove that is a right angle. onstruction: ssume is 90 and draw = so that = 90. roof: = 2 (ythagorean theorem) = 2 (since = ) ut = 2 (given) 2 = 2 = lso, is common and = (by construction) () = = 90

37 154 Mathscape 10 xtension G + xample 1 ind the value of x in each triangle, correct to 1 decimal place. a b 4 cm x cm 6 cm 11 cm x cm G + 9 cm olutions a y ythagoras theorem, b y ythagoras theorem, x 2 = x = 11 2 x 2 = 117 x = 121 x = 117 x 2 = 105 = 10.8 (correct to 1 decimal place) x = 105 = 10.2 (correct to 1 decimal place) xample 2 how that a triangle with sides 35 cm, 84 cm, 91 cm is right-angled. olution = = 8281 = 91 2 the triangle is right-angled (converse of ythagoras theorem). xercise ind the value of the pronumeral in each triangle. nswer in simplest surd form, where necessary. a b 7 c t 15 x a d e f u m e hich of the following could be the sides of a right-angled triangle? ll lengths are in mm. a 4, 5, 6 b 9, 12, 15 c 7, 10, 13 d 20, 21, 29

38 hapter 4 : eductive geometry In, is a point on such that. a ind and. 20 cm b Hence show that is right-angled. 16 cm 9 cm 4 n isosceles triangle has a base of 70 cm and congruent sides of 37 cm. ind the altitude of the triangle. 5 ind the length of. 12 m 13 m 15 m 6 In the diagram, T, = T, = 9 cm, = 16 cm and = 60 cm. ind T. 9 cm 16 cm 60 cm 7 ind the value of x. T 26 7 x 17 8 In a right-angled triangle, the hypotenuse is 8 cm longer than the shortest side and 1 cm longer than the third side. ind the length of the sides. 9 a a 2 how that is right-angled. a 10 how that is isosceles. 2xy x y

39 11 The diagonals of a rhombus are 32 cm and 60 cm. ind the perimeter. 12 In the diagram = and =. how that = rove that each set of expressions is a ythagorean triad. a 2x, x 2 1, x b 2xy, x 2 y 2, x 2 + y 2 14 In, is an altitude and = Let = x and = y. a how that = xy. b ind expressions for 2 and 2. x y c Hence, show that is right-angled. 15 In, is a point on such that. a ind an expression for 2 in. b ind an expression for 2 in. c Hence, show that = L In the quadrilateral KLMN, the diagonals KM and LN meet at. a how that KL 2 KN 2 = L 2 N 2. K b how that LM 2 MN 2 = L 2 N 2. c Hence show that KL 2 + MN 2 = KN 2 + LM 2. d hat geometric property of quadrilaterals have you proven? N M urther applications 17 In,. is any point on and is any point on. a rove that = b If and are the midpoints of and respectively, prove that = 5 2. [Hint: Let = 2x and = 2y.] 18 is a rectangle and O is any point in the interior. rove that O 2 + O 2 = O 2 + O 2. [Hint: onstruct O the altitudes OX, OY to, respectively.] 19 In the quadrilateral, and. a rove that 2 2 = 2 2. b Hence, prove that 2 2 = ( ) 2.

40 Mathscape 10 ext. - h04 age 157 Thursday, October 13, :17 M hapter eductive geometry 157 n unusual proof of ythagoras theorem ithin a large square LMN of side 7 units, a smaller square is drawn. L M a an you use this figure to show that = 5 2? b an you generalise from this argument to prove that 2 = M 2 + M 2? 0 N M XLOTION O N UILTL TINGL In this chapter you have been developing your skills in deductive reasoning. In working mathematically our aim is to fine tune these skills, by giving you an opportunity to make, refine and test your conjectures. You will also have the opportunity to make a generalisation from some specific cases and to prove a result. In these activities dynamic geometry software such as abri or Geometer s ketchpad are highly recommended. However if they are not available, geometrical instruments can be used. OU Introduction ON OKING N MTHMT ILLY I L LL LYY O U O N 0 K I N G M TTHHMMTTI G N I K O OU ON OU ON OKING MTHMTILLY TY THI 4:

41 The equilateral triangle Our starting point is the equilateral triangle. raw it in your book and write down as many geometrical facts about it as you can. OU ON OKING MTHMTILLY OU ON OKING MTHMTILLY If you are using a software program like Geometer s ketchpad or abri think carefully about how you will construct it. art of a ketchpad diagram using the circle by centre and radius tool is shown opposite. You can select and hide the circles. 2 L NING TIVITI 1 On your diagram construct and join the midpoints, and of the sides. 2 rove that the triangles, and are congruent. xplain why triangle is equilateral. 3 xplain why triangle is -- of the area of the 4 original equilateral triangle. 4 Now draw another equilateral triangle. onstruct the points, and such that 1 = = = --. rove that the triangles 3, and are again congruent. hat do you notice about the sizes of the angles of these triangles? 5 xplain why triangle is again equilateral. 6 hat fraction of triangle is triangle? 1 7 Make a conjecture about triangle as a fraction of triangle when 1 1 = = = --. ill it be greater or less than --? 4 3

42 % HLLNG TIVITI 1 e will now consider the general case where, and are all equal to some length x. x raw the diagram opposite and prove that the triangles, and are again congruent and that triangle is equilateral. In the usual notation, the side length a = b = c and the length of = = = a x. 2 Now consider the size of triangle as a fraction x of triangle. a rite down an expression for the area of triangle x b Use the cosine rule to find an expression for the length of 2. c rite down an expression for the area of triangle. d how that the ratio of the areas is a xa ( x) = a 2 a 1 e onfirm that the value of the ratio when x = -- is a 1 f onfirm that the value of the ratio when x = -- is a g alculate the value when x = -- and compare with your conjecture in question 7 above. 4 3 ere you surprised about the result? isappointed it was not so simple? Given that x is a fraction of a, will the ratio always be an exact rational number? hy? L T OMMUNIT iscuss in class what you have learned from this activity about using special cases to help clarify a general rule. % LTING Think over the power of algebra to generalise when we are tackling problems in geometry. OU ON OKING MTHMTILLY OU ON OKING MTHMTILLY

43 160 Mathscape 10 xtension 1 In a short sentence explain a quilateral triangles b ongruent triangles c conjecture d eductive reasoning e The specific as opposed to the general case. 2 The Macquarie Learners ictionary defines the word justify as follows: justify verb (justified, justifying) to show (an action, argument, etc.) to be right or reasonable: he can justify her decision to leave. How does the meaning of justify as given here to show to be right or reasonable differ from the mathematical use of the word? HT VI 1 ind the value of each pronumeral, giving reasons. a 115 d e X 63 Z Y c b H e G c 84 G p a T U V T bisects V f g I T J m X K U 52 M s 25 Y H L 10 N Z V HT VI

44 hapter 4 : eductive geometry 161 h 2 a how that. b how that,, are collinear points. 3 a Is KL MN? xplain. b c H I 112 J 85 Is X YZ? xplain. Is T UV? xplain. G K M 47 4 How many sides has: a a heptagon? b an undecagon? b I 105 J X 120 T 108 U 120 Y L N Z V 5 ind the angle sum and interior angles of a regular: a hexagon b decagon c pentagon d nonagon e octagon f dodecagon 6 ind the exterior angles of a regular: a pentagon b octagon c hexagon 7 How many sides are there in a regular polygon whose: a exterior angles measure 36? b interior angles measure 165? c angle sum is 2880? 8 an a regular polygon have interior angles of 124? xplain. 9 a b c G T U rove that and TU are supplementary. I H M K L IJ KL MN HM MN IM bisects KIJ rove that IKL = 2 IMH. G = rove that bisects G. T J N U HT VI HT VI

45 162 Mathscape 10 xtension HT VI d b e f 10 a = rove that is isosceles. G G G bisects bisects rove that. bisects = rove that is isosceles. O O is the centre = i rove that O O. ii Hence show that O bisects O. c d = i ii rove that. Hence show that is a parallelogram. Z Y XZ = ZXY XZ bisects ZY i rove that XZ YXZ. ii Hence show that YXZ is isosceles. G = bisects G i rove that G. ii Hence show that G. X HT VI

46 hapter 4 : eductive geometry T is a parallelogram. a rove that T T. b Hence show that the diagonals bisect each other. K N M KLMN is a rectangle. a rove that KNM LMN. b Hence show that the diagonals of a rectangle are equal. β Z Y XYZ is a rhombus. Let XY = and YX = β. a rove that ZX bisects XY, without the use of congruent triangles. b rove that XZ bisects Y at right angles, without the use of congruent triangles. V U TUV is a quadrilateral in which the diagonals bisect each other at right angles. L X T a b c xplain why TUV is a parallelogram. rove that V T. Hence show that TUV is a rhombus. = and are midpoints of and respectively. a how that =. b rove that. c d T Hence show that =. If T = T, show that T is isosceles., = = 2 rove that = 3. L M a b N L MN a M L = L N how that L = ab. b ind expressions for LM 2 and LN 2. c Hence show that LMN is right-angled. HT VI HT VI