Chapter 19 Exercise 19.1

Transcription

1 hapter 9 xercise 9... (i) n axiom is a statement that is accepted but cannot be proven, e.g. x + 0 = x. (ii) statement that can be proven logically: for example, ythagoras Theorem. (iii) The logical steps used to prove a theorem, e.g. when proving two triangles are similar, we show they have equal angles. (iv) statement derived (found) from a theorem, e.g. it can be derived from ythagoras Theorem that + >, where, and are the sides of a right-angled triangle and is the hypotenuse. (v) theorem in reverse order, e.g. the angles in a quadrilateral sum to 60. The converse is: a polygon whose angles sum to 60 is a quadrilateral... (i) tatement: a = b onverse: b = a (true) (ii) tatement: when you stick two equal squares together you get a rectangle onverse: rectangle is made up of two equal squares (false) quare quare ectangle.. tatement onverse onverse is True/False If the largest angle in a triangle is obtuse, then the two smaller angles are acute. If two triangles are congruent, then they are similar. If a triangle is equilateral, then it is also isosceles. If a number is divisible by 6, then it is also divisible by. If the two smaller angles in a triangle are acute, then the largest angle is obtuse. If two triangles are similar then they are congruent If a triangle is isosceles then it is also equilateral If a number is divisible by, then it is divisible by False False False False.. Given: Triangles and with parallel lines l and m as shown below =. I m To prove: Δ Δ onstruction: Label angles,, and roof: = alternate = given = alternate Δ Δ ctive Maths (trands ): h 9 olutions

2 .. (i) Given: ircle with chords [] and [T] as shown below. To prove: Δ Δ T onstruction: Label angles,,, roof: = vertically opposite = radii = alternate Δ ΔT T () (ii) T = diameters [ = T = = ] Note: is the centre of the circle.. 6. (i) Given: Triangles F and LM = 90 = 0 M = 90 = 0 F = L To prove: ΔF = ΔLM roof: F = L given = given F = 80 ( ) = 0 angle is a Δ L = 80 ( ) = 0 angle is a Δ F = L ΔF ΔLM () (ii) Given: Triangles and = = 0 = = To prove: Δ Δ roof: = given = = 0 given is a common side Δ Δ () (iii) Yes, by (iv) Yes, by. 7. (i) Yes = common angle =. corresponding =. corresponding. (ii) Yes =. both 90 xercise 9. = because triangle is similar to Δ and therefore =.. Given: quare as shown. To prove: = onstruction: Label angles, and ctive Maths (trands ): h 9 olutions

3 roof: = 90 square = square = isosceles Δ = 80 angle is a Δ = 90 =.... Given: iagram as shown W X.. Given: Triangle as shown. + To prove: Δ is isosceles onstruction: Label angle roof: = + Δ is isosceles... Given: Triangle with extension and as shown =. Y To prove: YZW = 90 roof: XY = XZ = XW o we can draw a circle around Δ WYZ with X as the centre with XY, XZ and WX as radii YZW = 90 angle in a semicircle... Given: iagram as shown. Z To prove: = onstruction: Label angles,, and roof: = 80 straight angle = 80 straight angle but = given = 80 =. 6. Given: iagram as shown. To prove: > onstruction: Label angles,, roof: = + > To prove: = + onstruction: Label angles,,, roof: = + exterior angle = alternate = + ctive Maths (trands ): h 9 olutions

4 . 7. Given: Triangle as shown.. 9. Given: iagram as shown. 7 8 X To prove: X > onstruction: Label angles, and roof: = + >. 8. Given: Isosceles Δ as shown. To prove: = 90 + onstruction: Label angles,,, roof: = isosceles Δ = 80 straight angle = 80 ( + ) angles in a Δ = 80 ( ) = 80 = 80 + = To prove: Δ Δ onstruction: Label angles,,,,, 6, 7, 8 roof: = given = given = isosceles Δ = isosceles Δ = 80 = 80 = 6 7 = 8 equal angles in a Δ implies the rd angles are equal Δ Δ (). 0. (i) M = M (common) M = M (given) = (given) ΔM ΔM (side, side, side) (ii) M in ΔM corresponds to M in ΔM, Therefore M = M (iii) M = M = 80 (traight line) but M = M M = M = 90 Therefore M ctive Maths (trands ): h 9 olutions

5 .. Given: Δ XYZ as shown. X 6 Y Z To prove: + + = 60 onstruction: Label angels, and 6 roof: = + 6 = + 6 = = ( ) but = = (80 ) = 60 angle in a Δ = = 80 angles in a Δ = 90 = Δ XY is isosceles (ii) ut Δ XY may be equilateral 6 = + 6 = If Δ XY is equilateral 6 = = = 90 = 90 = 0 o when = 0 Δ XY is equilateral.. Given: iagram as shown... (i) Given: Δ as shown below = Y X To prove: Δ XY is isosceles onstruction: Label angles 6 roof: = isosceles Δ = 80 angles in a Δ 6 = 90 = 90 = vertically opposite To prove: = onstruction: Label angles,,,,, 6 roof: = isosceles Δ = isosceles Δ = 80 angles in a Δ + + = = 80 angles in a Δ + = = + = ctive Maths (trands ): h 9 olutions

6 .. Given: Triangles and as shown below X = Z orresponding W = Z Therefore [ bisects 6 To prove: + = 90 onstruction: Label angles 6 as shown roof: = isosceles Δ = + = = 6 isosceles Δ = + 6 = + = 80 straight angle + = 80 ( + ) = 80 + = 90.. (a) Given: (b) No, the result in (a) would not still apply. ngle Y would not be equal to angle X. xercise 9... Given: arallelogram as shown. 6 Y X To prove: Δ Y similar to Δ XY onstruction: Label angles,,,,, 6 roof: = alternate = vertically opposite = 6 alternate Δ Y and Δ XY are similar.. Given: hombus as shown. z w T To prove: [] bisects onstruction: Label angles X Y To prove: [ bisects. roof: Y = W lternate Y = X isosceles roof: = alternate = isosceles Δ = [] bisects 6 ctive Maths (trands ): h 9 olutions

7 .. Given: arallelogram and as shown. To prove: Δ onstruction: Δ Δ roof: = parallelogram = parallelogram = parallelogram = parallelogram = Δ Δ.. Given: ectangle FG as shown To prove: F = G () onstruction: Label angles and roof: = GF opposite sides = right angles F is a common side in Δ F and Δ FG Δ F Δ FG F = G orresponding sides.. Given: arallelogram LM as shown L M G F To prove: + = 80 onstruction: Label angles and roof: = opposite angles = opposite angles = 60 quadrilateral ( + ) = 60 + = Given: arallelogram as shown. =. M is the midpoint of [] 6 7 M To prove: M = 90 onstruction: Label angles,,,,, 6, 7 roof: = opposite sides = = = M = = = alternate = imilarly M = = 6 = alternate = = 80 ( + ) = 80 + = 90 ctive Maths (trands ): h 9 olutions 7

8 + = 7 = 80 angles in a Δ = 80 7 = (a) Given: arallelogram as shown. X To prove: Y = X Y 6 onstruction: Label angles,,,,, 6 roof: = 6 alternate = opposite sides = given Δ Y Δ X () (b) Y = X corresponding sides. 8. Given: arallelogram. M is the midpoint of []. X M Y To prove: MX = MY onstruction: Label angles,,, roof: = alternate M = M given = vertically opposite Δ YM Δ XM () MX = MY corresponding sides. 9. (i) Given: arallelogram. T is the midpoint of []. T U To prove: Δ T Δ TU onstruction: Label angles,,, roof: = alternate T = T given = vertically opposite Δ T Δ TU () (ii) U = corresponding sides (iii) U = + U = opposite sides U = part (ii) U =. 0. (i) Given: quare FG.,,, are the midpoints of the sides G F 8 ctive Maths (trands ): h 9 olutions

9 To prove: Δ Δ G onstruction: Label angles,,,,, 6 roof: = both right angles = G diagonal bisect G = Δ Δ G (H) (ii) Given: ame as part (i) To prove: Δ G Δ onstruction: Label angles 7, 8 roof: 7 = 8 diagonals meet at night angles G = square G = diagonals bisected Δ G Δ (H) (iii) It can be shown that Δ F Δ F Δ Δ etc. Using a similar proof on part (i) Total area = 8 area Δ F = 8 8 = 6 cm.. Given: arallelogram as shown, [X] [] and [Y] [] Y X To prove: Δ Δ onstruction: Label angles,,,,, 6 6 roof: = alternate s and X and Y X Y = alternate = 6 equal angles implies the rd angles are equal = opposite side Δ Δ () = corresponding sides xercise 9... onsider Δs F and. F = alternate F =. alternate F =. vertically opposite ΔF is similar to Δ. = F. F =.... T = T. corresponding = T. corresponding T = common angle. Δ is similar to ΔT = T T ctive Maths (trands ): h 9 olutions 9

10 .. (i) F G = G FG = G FG. FG = G. (cross-multiply) Label angles + = 90 as Δ is right-angled. + = 90 straight angle + = + = + = 90 as Δ is right-angled + = 90. straight angle + = + = lso = both 90 Δ is similar to Δ... (i) onsider Δ and ΔF = F. alternate = F. alternate = F vertically opposite. 6. Δ is similar to ΔF F = (ii) From (i) Δ is similar to ΔF F =. =. F ut = opposite sides of a parallelogram. =. F T (ii) edrawing Δs 0 ctive Maths (trands ): h 9 olutions We have = similar triangles theorem... We note that. FG = G. could also be written as = = G FG. Δ T is similar to Δ as = T common angle and = T both 90 = T lso = T both 90 Δ is similar to Δ T. = T y x y = T T x = y T = y x

11 . 7. Δs F and = F F Δs F and = F F = =.. (i) In Δ T and Δ W T = W.. alternate angles T = W opposite angles in a parallelogram. (ii) ΔT is similar to ΔW. W. 8. (i) T = T.. alternate T = T both 90 Δ T is similar to Δ T. (ii) Δ T is similar to Δ T T = T T T = T.. 9. In Δ and Δ = both 90 = common angle. Δ is similar to Δ. = also = both 90 Δ is similar to Δ and also Δ is similar to Δ = =.. 0. (a) In Δ and Δ =... angles at the circle subtended by the same are =... as bisects Δ is similar to Δ (b) s Δ is similar to Δ =. =. x ctive Maths (trands ): h 9 olutions x T x Let T = x, T = x = x _ From (i) W = T _ W = x x = = W ut = = W : W : xercise 9... Given: quare and F. y y x F To prove: rea F = (area ) onstruction: Mark sides of as x and F as y. y x x y x

12 roof: Using ythagoras Theorem x + x = y x = y (area of ) = area of F.. In Δ, = + In Δ, = + ut = = + = + =.. Given: Triangle as shown. roof: + = + = ythagoras ythagoras = + ( + ) + ( + ) = + + = + ythagoras.. Given: ircle with centre, tangents [] and [] as shown. To prove: + = + roof: = + ythagoras Theorem = = + ythagoras Theorem = = + = +.. Given: Triangles and as shown. To prove: = onstruction: raw the line. Label angles and roof: = right angles is a common side = radii Δ Δ (H) = corresponding sides. 6. In Δ = + In semicircle rea = pr To prove: + = + onstruction: raw and. ctive Maths (trands ): h 9 olutions = ( p ) = p

13 In semicircle we similarly have rea = p and in semicircle, rea = p o is p = p + p? Is p = p + p? Is = y ythagoras Theorem = + ut = so = = True. 8. onstruction:, the perpendicular bisector of In Δ, = ut Δ is equilateral so = and also = = ( ) = = = =. 9. M T N onstruction: [MN] through T such that [MN] [] and [MN] [] Join T to,, and y ythagoras Theorem: TM = T M = T M T M = T M T T = M M imilarly TN = T N = T N o T T = N N ut M M = N [N] as M = N and M = N T T = T T T + T = T + T ctive Maths (trands ): h 9 olutions

14 . 0. y ythagoras Theorem = F + F Now F = F = nd by ythagoras Theorem F = F + = + F + xercise Given: Triangle F inside circle as shown. onstruction: Label angle. raw extension from as shown and label angle roof: + = 80 opposite angles in cyclic quadrilateral = alternate angle + = 80 straight angle + = 80 + = + =.. Given: iameter []. entre. 6 F To prove: Δ Δ F onstruction: Label angles,,,,, 6 roof: = 6 = 90 given + straight angle = F radii is a common side Δ Δ F.. Given: iagram as shown. () To prove: + = 90 onstruction: Mark equal radii. Label angle roof: + = 90 angle in a semi circle = isosceles Δ + = 90.. Given: hords [] and [] of a circle as shown. X 6 To prove: = To prove: Δ X and Δ X are similar ctive Maths (trands ): h 9 olutions

15 onstruction: Mark angles,,,,, 6 roof: = 6 angles standing on the same arc = angles standing on the same arc = vertically opposite Δ X and Δ X are similar.. Given: yclic quadrilateral as shown. To prove: onstruction: onstruct the quadrilateral and label angles,,, roof: = 90 angle in a semi circle = 90 angle in a semi circle also = = 90 angle in a semi circle is a rectangle = alternate angles (ii) This holds and is a rectangle. 7. Given: iagram as shown To prove: = onstruction: Label angles,, roof: + = 80 opposite + = 80 straight angle + = + =. 6. (i) Given: iameter [] and [] of circle as shown. To rove: = onstruction: raw [] and [] roof: = radii = given [] is a common side Δ Δ () = corresponding angle ctive Maths (trands ): h 9 olutions

16 . 8. (i) Given: iagram as shown To rove: = q onstruction: Label angles, roof: = q isosceles Δ = + q = q (ii) To prove: = q onstruction: Label angles, roof: = + but = = q isosceles Δ (radii) = q. 9. Given: is the centre of a circle. M is the midpoint of [], as shown. To prove: M = onstruction: Label angles,, roof: M = M given = radii M is a common side M Δ M Δ M () = corresponding angles + = angle at the centre = =. 0. (i) Given: iameter [] of circle with centre. = as shown. To prove: = 90 onstruction: Join to. Label angles,. roof: = 90 angles in a semi circle = = 90 straight angle (ii) To prove: = roof: = right angles = given is a common side Δ Δ = corresponding sides is the midpoint of [] 6 ctive Maths (trands ): h 9 olutions

17 .. Given: uadrilateral as shown. To prove: is a cyclic quadrilateral onstruction: Label angles,,, roof: = 60 quadrilateral = = 90 given = 60 + = 80 o opposite sides sum to 80 is a cyclic quadrilateral.. (i) Given: iagram as shown. To prove: onstruction: Label angles, roof: = = 90 angles in a semicircle (ii) To prove: = roof: Label angles,, = above = corresponding is common to Δ and Δ Δ and Δ are similar = but = radii = = ctive Maths (trands ): h 9 olutions 7