三條F.3數學問題 幫幫手

10.
(a)
GD = GE (given)
∠AGD = ∠CGE (vert. opp. ∠s)
∠GAD = ∠GCE (alt. ∠s, AD // BC)
Hence, ΔGAD ≡ ΔGCE (ASA)

(b)
It has been proven that : ΔGAD ≡ ΔGCE
AD = EC (corr. sides of congruent Δs)
BE = EC (given)
Hence, AD = BD

Also, AD // BD
Hence, ABED is a parallelogram. (opp. sides equal and //)


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12.
(a)
ΔCDE is isos. with CD = CE (given)
∠CDE = ∠CED (base ∠s of isos. Δ)

∠CDE + ∠CED + ∠DCE = 180° (∠ sum of Δ)
∠CDE + ∠CDE + 48° = 180°
∠CDE = 66°

ADE // BC (opp. sides of rhombus)
∠DCB = ∠CDE (alt. ∠s, ADE // BC)
∠DCB = 66°

(b)
CD = CB (sides of rhombus)
CD = CE (given)
Hence, ΔCBE is isos. with CB = CE
∠CEF = ∠CBF (base ∠s. of isos. Δ)

∠CEF + ∠CBF + ∠BCE = 180° (∠ sum of Δ)
∠CEF + ∠CEF + (66° + 48°) = 180°
∠CEF = 33°

∠DFE = ∠ECD + ∠CEF (ext. ∠ of Δ)
∠DFE = 48° + 33°
∠DFE = 81°


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16.
(a)
Let h cm be the length of the altitude from A.

h² + (12/2)² =24² (Pythagorean theorem)
h² = 540
h = 6√15

Area of ΔABC :
(1/2) x BC x h = (1/2) x AC x BD
(1/2) x 12 x 6√15 = (1/2) x 24 x BD
BD = 3√15

In ΔBDC :
BD² + DC² =BC²
(3√15)² + DC² =12²
DC² = 9
Hence, DC = 3

(b)(i)
AP // BQ (opp. sides of parallelogram)
∠ADP = ∠CDB (vert. opp. ∠s)
∠DAP = ∠DCB (alt. ∠s, AP // BQ)
∠DPA = ∠DBC (alt. ∠s, AP // BQ)
ΔDBC ~ ΔDPA (AAA)
BC/PA = DC/DA
12/PA = 3/(24 - 3)
PA = 84

BQ = PA (opp. sides of parallegram)
12 + CQ = 84
CQ = 72

(b)(ii)
Area of ΔBPQ
= (1/2) x h x BQ
= (1/2) x (6√15) x 84
= 252√15

Area of ΔBDC
= (1/2) x BD x DC
= (1/2) x 3√15 x 3
= (9√15)/2

Area of quad. CDPQ
= (252√15) - (9√15)/2
= (495√15)/2

there is the answer !
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