Question-78216

Question Number 78216 by TawaTawa last updated on 15/Jan/20

Commented by TawaTawa last updated on 15/Jan/20

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle}. \\ $$

Commented by mr W last updated on 15/Jan/20

radius =R DC=x area of rectangle=xR (x/8)=(8/(2R)) ⇒xR=(8^2 /2)=area of rectangle

$${radius}\:={R} \\ $$$${DC}={x} \\ $$$${area}\:{of}\:{rectangle}={xR} \\ $$$$\frac{{x}}{\mathrm{8}}=\frac{\mathrm{8}}{\mathrm{2}{R}} \\ $$$$\Rightarrow{xR}=\frac{\mathrm{8}^{\mathrm{2}} }{\mathrm{2}}={area}\:{of}\:{rectangle} \\ $$

Commented by Tony Lin last updated on 15/Jan/20

Commented by Tony Lin last updated on 15/Jan/20

(r−8cosθ)^2 +(8sinθ)^2 =r^2 rcosθ=4 Area=r∙8cosθ=32

$$\left({r}−\mathrm{8}{cos}\theta\right)^{\mathrm{2}} +\left(\mathrm{8}{sin}\theta\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${rcos}\theta=\mathrm{4} \\ $$$${Area}={r}\centerdot\mathrm{8}{cos}\theta=\mathrm{32} \\ $$

Commented by TawaTawa last updated on 15/Jan/20