Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 69494 by TawaTawa last updated on 24/Sep/19

Commented by TawaTawa last updated on 24/Sep/19

$$\mathrm{Please}\mathrm{help}\mathrm{me}\mathrm{find}\mathrm{the}\mathrm{Area}\mathrm{and}\mathrm{Perimeter}\mathrm{of}\mathrm{the}\mathrm{shaded}\mathrm{part}.$$

Commented by mind is power last updated on 24/Sep/19

$$PB=QD=15-R=15-\frac{15\sqrt{3}}{2}$$$$P\stackrel{⌢}{Q}=R.\frac{\pi }{6}=\frac{15\pi }{6}$$$$perimeter=2PB+2BC+P\stackrel{⌢}{Q}=2(15-\frac{15\sqrt{3}}{2})+2.15+\frac{15\pi }{6}=60-15\sqrt{3}+\frac{15\pi }{6}$$

Commented by TawaTawa last updated on 24/Sep/19

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

Answered by mind is power last updated on 24/Sep/19

$${BD}^2={AB}^2+{AD}^2-2ADABcos\left(BAD\right)$$$$={15}^2+{15}^2-2.15.15.\frac{1}{2}={15}^2\Rightarrow {BD}^2={15}^z\Rightarrow BD=15cm$$$$AT=R=ADcos\left(BAD/2\right)=\frac{15\sqrt{3}}{2}$$$$letS=areaofABCD$$$$S=2.AreaABD=2.(BD.AT)/2=15.\frac{15}{2}$$$$\frac{\pi }{6}=\frac{1}{12}.2\pi$$$$S^{\prime }=AreaofArcofcircl=\pi .r^2.\left(\frac{\pi }{6.2\pi }\right)=\frac{\pi r^2}{12}=\frac{\pi }{12}.\frac{{15}^2.3}{4}=\frac{225\pi }{16}$$$$Areaofred=S-S^{\prime }=\frac{225}{2}-\frac{225\pi }{16}=\frac{1800-225\pi }{16}$$

Commented by TawaTawa last updated on 24/Sep/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir},\mathrm{what}\mathrm{of}\mathrm{perimeter}?$$

Commented by mind is power last updated on 24/Sep/19

$$SoorytherismistackR=15cos\left(30\right)=\frac{15\sqrt{3}}{2}$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com