1011_F5MATH_NORMAL_P1(ALL)

(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) cos⁴θ
∴ Area of △BX1Y1：Area of △Y1X2Y2：Area of △Y2X3Y3：⋯⋯
＝(BX1)²：(BX1 cos²θ)²：(BX1 cos⁴θ)²：⋯⋯
＝1：(16/25)²：(16/25)⁴：⋯⋯
∴ 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)⁴＋⋯⋯
＝(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<k<6)
∴ k＝2 or k＝(7－√33)/2

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