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1011_F5MATH_NORMAL_P1(ALL)

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ABCD and CEFG are squares of side lengths 8 cm and 6 cmhttps://www.flickr.com/photos/130230543@N06/16189046311/in/photostream/(a) Find the length of AG(b) Show that AC//EG. Hence find the area of triangleAEGFigure 5 shows a right-angled triangleABC with AB=3, BC=4 and CA=5. Starting from B, perpendiculars... 顯示更多 ABCD and CEFG are squares of side lengths 8 cm and 6 cm https://www.flickr.com/photos/130230543@N06/16189046311/in/photostream/ (a) Find the length of AG (b) Show that AC//EG. Hence find the area of triangleAEG Figure 5 shows a right-angled triangleABC with AB=3, BC=4 and CA=5. Starting from B, perpendiculars are drawn to AC and BC alternatively. The feet of perpendiculars on AC are labelled X1, X2, X3, X4, ... while those on BC are labelled Y1, Y2, Y3, Y4, ... The triangles with a side lying on BC are then shaded. https://www.flickr.com/photos/130230543@N06/16165099676/in/photostream/ (a) Show that triangleAX1B and triangleBY1X1 are both similar to triangleABC (b) Find the area of triangleBX1Y1 (c) Find the sum of the areas of all the shaded triangles, assuming that the process of drawing perpendiculars goes on infinitely https://www.flickr.com/photos/130230543@N06/16004075678/ Let the coordinates of A be (k,0) (i) Write down the coordinates of B and D (ii) Find the area of ABCD in terms of k (iii) If the area of ABCD is 32, find all possible values of k Leave the answers in surd form if necessary 更新: https://www.flickr.com/photos/130230543@N06/16014819750/

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(1a) AG2＝AB2＋BG2==> AG2＝82＋(8＋6)2==> AG2＝64＋196＝260==> AG＝2√65 cm (1b) ∠ACB＝∠EGC＝45° ?? (prop. of square)∴ AC//EG ????????? (corr. ∠s are equal)Hence, area of △AEG＝△CEG＝6*6/2＝18 cm2 (2a) Let ∠C＝θ, so ∠A＝90°－θ, and ∠ABX1＝θ, therefore,∠BX1Y1＝∠ABX1＝θ ??? (alt. ∠s, AB//X1Y1)and ∠X1BY1＝90°－θIn △s AX1B, BY1X1, ABC∠AX1B＝∠BY1X1＝∠ABC＝90° ??? (given)∠X1BA＝∠Y1X1B＝∠BCA＝θ ???? (proved)∠BAX1＝∠X1BY1＝∠CAB＝90°－θ ?? (proved)∴ △AX1B～△BY1X1～△ABC (2b) In △ABC, we get, sin θ＝AB/AC＝3/5 and cos θ＝BC/AC＝4/5 ∴ Area of △BX1Y1 ＝(1/2)(X1Y1)(BY1) ＝(1/2)(BX1 cos θ)(BX1 sin θ) ＝(1/2)(AB cos θ cos θ)(AB cos θ sin θ) ＝(1/2)(3)(4/5)(4/5)(3)(4/5)(3/5) ＝864/625 (sq. units) (2c) As △BX1Y1～△Y1X2Y2～△Y2X3Y3～△Y3X4Y4～?? and Y1X2＝Y1X1 cos θ＝BX1 cos θ cos θ＝BX1 cos2θ Y2X3＝Y2X2 cos θ＝Y1X2 cos2θ＝(BX1 cos2θ) cos2θ Y3X4＝Y3X3 cos θ＝Y2X3 cos2θ＝(BX1 cos2θ) cos?θ ∴ Area of △BX1Y1：Area of △Y1X2Y2：Area of △Y2X3Y3：?? ＝(BX1)2：(BX1 cos2θ)2：(BX1 cos?θ)2：?? ＝1：(16/25)2：(16/25)?：?? ∴ the sum of the areas of all shaded triangles is : Area of △BX1Y1＋Area of △Y1X2Y2＋Area of △Y2X3Y3＋?? ＝864/625＋(864/625)(16/25)2＋(864/625)(16/25)?＋?? ＝(864/625) / [1－(16/25)2] ＝864/(625－256) ＝96/41 (sq. units) (3i) As the x-coordinate of B is k, so the y-coordinate of B is (2k2－12k). So the equation of line BD is y＝2k2－12k. Solve 2x2－12x＝2k2－12k, we get, x2－6x－k(k－6)＝0 ==> (x－k)(x＋k－6)＝0 ==> x＝k or x＝6－k ∴ the coordinates of B is (k, 2k2－12k), D is (6－k, 2k2－12k). (3ii) The area of ABCD is : ＝(AB)(BD) ＝(6－k－k)[0－(2k2－12k)] ＝4k(3－k)(6－k) ＝4k3－36k2＋72k (sq. units) (3iii) 4k3－36k2＋72k＝32 ==> k3－9k2＋18k－8＝0 ==> (k－2)(k2－7k＋4)＝0 ==> k＝2 or k＝(7－√33)/2 or k＝(7＋√33)/2 (reject, as 0
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