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Question Number 15623 by tawa tawa last updated on 12/Jun/17

Answered by ajfour last updated on 12/Jun/17

Commented by ajfour last updated on 12/Jun/17

BD^2 =BC^( 2) −CD^2 AC=BC= r =7 h^2 = 49−16=33 h=(√(33)) . Area of ΔBCD=(1/2)×4×(√(33)) Area of sector ABC =(1/2)r^2 θ cos θ=(4/7) ⇒ θ=cos^(−1) ((4/7)) Shaded area = Area of sector ABC −Area of ΔBCD = (1/2)(7)^2 cos^(−1) ((4/7))−2(√(33)) Shaded area =((49)/2)cos^(−1) ((4/7))−2(√(33)) ≈ 12.093 .

$$\:\:{BD}^{\mathrm{2}} ={BC}^{\:\mathrm{2}} −{CD}^{\mathrm{2}} \\ $$$$\:\:\:{AC}={BC}=\:{r}\:=\mathrm{7} \\ $$$$\:\:\:\:\:\:\:{h}^{\mathrm{2}} =\:\mathrm{49}−\mathrm{16}=\mathrm{33} \\ $$$$\:\:\:\:\:\:\:{h}=\sqrt{\mathrm{33}}\:. \\ $$$$\:{Area}\:{of}\:\Delta{BCD}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}×\sqrt{\mathrm{33}} \\ $$$$\:{Area}\:{of}\:{sector}\:{ABC}\:=\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \theta \\ $$$$\:\:\:\:\:\:\:\mathrm{cos}\:\theta=\frac{\mathrm{4}}{\mathrm{7}}\:\Rightarrow\:\:\theta=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{7}}\right) \\ $$$$\:{Shaded}\:{area}\:=\:{Area}\:{of}\:{sector}\:{ABC} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{Area}\:{of}\:\Delta{BCD} \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{7}\right)^{\mathrm{2}} \mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{7}}\right)−\mathrm{2}\sqrt{\mathrm{33}} \\ $$$${Shaded}\:{area}\:=\frac{\mathrm{49}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{7}}\right)−\mathrm{2}\sqrt{\mathrm{33}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\approx\:\mathrm{12}.\mathrm{093}\:. \\ $$

Commented by tawa tawa last updated on 12/Jun/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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