F.4 TRIGONOMETRY M2

15 Area of △ABC = (1/2)(2a)(2a)sin(π/3) = √3a^2

Area of ADEB

= Area AOD + Area BOE + Area of Sector ODE

= 2(1/2)(a)(a)sin(π/3) + (1/2)(a^2)(π/3)

= √3a^2/2 + πa^2/6

So, the area of shaded region

= Area of △ABC - Area of ADEB

= √3a^2 - √3a^2/2 - πa^2/6

= √3a^2/2 - πa^2/6

= a^2(√3/2 - π/6)

16(a) x + x + xθ = 10

θ = (10 - 2x)/x

(b) A

= (1/2)(x^2)θ

= (1/2)(x^2)[(10 - 2x)/x]

= 5x - x^2

(c) dA/dx = 5 - 2x, d^2A/dx^2 = -2

So, when x = 2.5, A has the max. value 6.25 cm^2


x = √(2A/θ)