F.2 數學題

1.
ABCD is a square.
∠DAB = ∠ABC = ∠BCD = ∠CDA = 90°

ADE is an equilateral triangle.
∠DAE = ∠EDA = ∠AED = 60°

∠BAE = ∠BAD + ∠DAE = 90° + 60° = 150°

The diagonal BD bisects ∠ABC.
∠ABM = 90° / 2 = 45°

CD = DE
Hence, ΔBCE is an isosceles triangle.
The two base angles ∠DCE = ∠DEC

∠CDE = ∠CDA + ∠EDA = 90° + 60° = 150°

∠DCE + ∠DEC + ∠CDE = 180° (∠ sum of Δ)
∠DEC + ∠DEC + 150° = 180°
∠DEC = 15°

∠MEA = ∠AED - ∠DEC = 60° - 15° = 45°

∠BAE + ∠ABM + ∠BME + ∠MEA = 360° (int. ∠sum of quad.)
150° + 45° + ∠BME + 45° = 360°
∠BME = 120°

The four interior angles of quadrilateral ABME :
∠BAE = 150°
∠ABM = 45°
∠MEA = 45°
∠BME = 120°


====
2.
Sum of interior angles of pentagon ABCDE :
∠EAB + ∠ABC + (reflex ∠BCD)+ ∠CDE + ∠DEA = (2 × 5 - 4) × 90°
(180° - a) + (180° - b) + (360° - c) + (180° - d) + (180° - e) = 540°
1080° - (a + b + c + d + e) = 540°
a + b + c + d + e = 540°


====
3.
n = m + 5 ..... [1]
(2n - 4) × 90° = 2 × (2m - 4) × 90° ...... [2]

Put [1] into [2] :
[2(m + 5) - 4] × 90° = 2 × (2m - 4) × 90°
2m + 6 = 4m - 8
2m = 14
m = 7

Put m = 7 into [1] :
n = 7 + 5
n = 12