F5 數學一條問題4

第一題：
(a)
弧AB = 弧BC (已知)
AB = BC (等弦對等弧)
故 ΔABC 為一等腰Δ
∠BCA = ∠BAC (等腰Δ兩底角)
所以 ∠BCA = x

∠DBC = ∠DAC (對同弧圓周角)
所以 ∠DBC = y

ΔEBC 中：
∠AEB = ∠BCA + ∠DBC (Δ外角)
所以 ∠AEB = x + y

(b)
∠BAD = ∠BEA = x + y (已證)
∠ABD = ∠EBA (公共角)
ΔBAD ~ ΔBEA (AAA)

AB/BD = BE/AB (相似Δ對應邊)
交叉相乘得：AB² = BE x BD

(c)
把 AB = 9 cm 及 BD = 12 cm 代入上式：
(9 cm)² = BE x (12 cm)
BE = 6.75 cm

DE
= BD - BE
= 12 - 6.75 cm
= 5.25 cm


第二題：
原來水杯的售價 = $120/N

[(120/N) - 2] x (N + 2) = 120 - 8
(120 - 2N)(N + 2) / N = 112
(120 - 2N)(N + 2) = 112N
120N + 240 - 2N² - 4N = 112N
2N² - 4N - 240 = 0
N² - 2N - 120 = 0
(N - 12)(N + 10) = 0
N = 12 或 N = -10(不合)