2條數學問題.求解答20分!

設 八邊形ABCDEFGH中,
AB = 9, BC = 10, CD = 11, DE = 12, EF = 13, FG = 14, GH = p, HA = q

等角八邊形的每一隻角 = (8 - 2)(180°)/8 = 135°…*

設B = (0 , 0), 即
B 的 X軸座標 = 0
A 的 X軸座標 = -9
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延伸CD使之與X軸相交於M
∠CBM
= 180° - 135° (根據*, 直線上的鄰角)
= 45°
∠BCM
= 180° - 135° (根據*, 直線上的鄰角)
= 45°
則 △BCM是等腰三角形
即 BM = MC
又 BM^2 + MC^2 = BC^2 = 10^2 (畢氐定理)
2BM^2 = 10^2
BM = MC = [(10^2)/2]^(1/2)
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C 的 X軸座標
= 0 + [(10^2)/2]^(1/2) (畢氐定理)
= 50^(1/2)
= 5(2)^(1/2)

D 的 X軸座標 = C 的 X軸座標 = 5(2)^(1/2)

E 的 X軸座標
= 5(2)^(1/2) - [(12^2)/2]^(1/2) (畢氐定理)
= 5(2)^(1/2) - 6(2)^(1/2)
= -(2)^(1/2)

F 的 X軸座標 = -(2)^(1/2) - 13

G 的 X軸座標
= -(2)^(1/2) - 13 - [(14^2)/2]^(1/2) (畢氐定理)
= -(2)^(1/2) - 13 - 7(2)^(1/2)
= -8(2)^(1/2) - 13

H 的 X軸座標 = G 的 X 軸座標 = -8(2)^(1/2) - 13 … (i)

觀察線AH,
H 的 X軸座標 = -9 - [(q^2)/2]^(1/2) … (ii) (畢氐定理)

根據 (i) 和 (ii)
-8(2)^(1/2) - 13 = -9 - [(q^2)/2]^(1/2)
[(q^2)/2]^(1/2) = 8(2)^(1/2) + 4
(q^2)/2 = [8(2)^(1/2) + 4]^2
= 128 + 64(2)^(1/2) + 16
= 144 + 64(2)^(1/2)
q^2 = 288 + 128(2)^(1/2)
q = [288 + 128(2)^(1/2)]^(1/2)
~ [288 + 128(1.41)]^(1/2)
= (468.68)^(1/2)
~ 22 #
[21^2 = 441, 22^2 = 484, (21.5)^2 = 462.5 < 468.68]

第二題諗唔倒…