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# Question

# Answer

a) AC = ED = 40 cm

Since ED and EC are both radius to a circle, ED = EC = 40 cm

Looking at △ ACE, since AC = EC = 40cm, and ∠ACE = 90°, △ ACE is a right isosceles △,

∴ ∠AEC + ∠EAC +90° = 180°

∠AEC + ∠EAC = 90°

Since ∠AEC = ∠EAC, ∠AEC = 45°

45° = (45 x π/180) rad

= π/4 rad

≈ **0.785 rad (Proven)**

b) ⌒ABC = 1/2 x 2π(20)

= 20π cm

⌒DC / 2π(40) = 45/360

⌒DC = 45/360 x 2π(40)

= 1/8 x 80π

= 10π cm

(AE)^{2} = (AC)^{2} + (EC)^{2}

(AE)^{2} = 40^{2} + 40^{2
}AE = √3200

AD = AE – ED

AD = √3200 – 40

≈ 16.57 cm

Perimeter of shaded region = ⌒ABC + ⌒DC + AD

= 20π + 10π + 16.57

= 30π + 16.57

≈ **110.82 cm**

c) AC = EC = 40 cm (S)

∠BEC = ∠BAC = 45° (A)

BC is a common side of both △s (S)

∴ △ABC and △EBC are congruent by **SAS** congruence test

d) Area of △ ABC = 1/2 x 20 x 20

= 200 cm^{2}

Area of semi-circle ABC = 1/2 x π x 20 x 20

= 200π cm^{2
}Areas of ⌒AB and ⌒BC = (200π – 200) cm^{2}

45/360 = Area of DEC / π(40)^{2}

1/8 = Area of DEC / 1600π

Area of DEC = 1/8 x 1600π

= 200π cm^{2
}Area of DBC = Area of DEC – Area of △EBC

Area of DBC = (200π – 200) cm^{2}

Area of shaded region = Areas of ⌒AB and ⌒BC + Area of DBC

= (200π – 200) + (200π – 200)

= (400π – 400) cm^{2}

≈ **856.64 cm ^{2}**

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