求人幫忙圖形問題~

1.
The original pyramid and the small pyramid are similar.

(Volume of the original right pyramid) : (Volume of the small pyramid)
= 20³ : 12³
= 125 : 27

(Volume of the frustum pyramid) : (Volume of the original pyramid)
= (125 - 27) : 125
= 98 : 125


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2.
AC = AC (common side)
AB = AD (given)
BC = DC (given)
Hence, ΔABC ≡ ΔADC (SSS)

∠D = ∠B = 70° (corr. ∠s, ΔABC≡ ΔADC)

ΔADC is a isos. Δ with AD = DC (given)
∠DAC = ∠DCA = x (base ∠sof isos. Δ)

In ΔADC :
∠D + ∠DAC + ∠DCA = 180° (∠ sum of Δ)
70° + x + x = 180°
x = 55°


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3.
AC = BC (given)
∠A = ∠B = 90° (given)
CD = CD (common sides)
ΔACD ≡ ΔBCD (RHS)


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4.
∠b = ∠c (alt. ∠s, PR // QS)
∠a = ∠d (alt. ∠s, PQ // RS)
QR = QR (common side)
ΔPQR ≡ ΔSQR (ASA)

PR = RS (corr. sides, ΔPQR ≡ ΔSQR)


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5.
Area of ΔABC = Area of ΔABC (same Δ)
(1/2) x AB x CP = (1/2) x AC x BQ (formula of area of Δ)
But AB = AC (given)
Hence, CP = BQ

BQ = CP (proved)
∠AQB = ∠APC = 90° (given)
AB = AC (given)
ΔABQ ≡ ΔACP (RHS)


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6.
(a)
∠BCD + 80° = 180° (adj. ∠son a st. line)
∠EAB+ 80° = 180° (adj. ∠son a st. line)
Hence, ∠BCD = ∠EAB

∠EAB = ∠BCD (proved)
EA = BC (given)
AB = CD (given)
Hence, ΔEAB = ΔBCD (SAS)

(b)
∠CBD = y (corr. ∠s, ΔEAB≡ ΔBCD)
z + y = 80° (ext. ∠ of ΔEAB)

z + θ + ∠CBD + 60° = 180° (adj. ∠son a st. line)
z + θ + y + 60° = 180°
θ + (z + y) + 60° = 180°
θ + 80° + 60° = 180°
θ = 40°

BE= BD (corr. sides, ΔEAB ≡ ΔBCD)
∠BDE = ∠BED (base ∠s of isos. Δ BDE)

θ + ∠BDE + ∠BED = 180° (adj. ∠son a st. line)
40° + ∠BDE + ∠BDE = 180°
∠BDE = ∠BED = 70°

75° + ∠BDE + ∠BDC = 180° (adj. ∠son a st. line)
75° + 70° + ∠BDC = 180°
∠BDC = 35°
But z = ∠BDC
Hence, z = 35°

It has been shown that :
z + y = 80°
Hence, y = 45°

∠BED + y + x = 180° (adj. ∠son a st. line)
70° + 45° + x = 180°
x = 65°

但總是神龍見首不見尾

好一個小學級 2 級的 YAHOO!知識+ 管理員
有大志，但 ......

2014-04-02 01:46:11 補充：
並非「神龍見首不見尾」
只是「老鼠天平自稱之」