The-figure-below-represents-a-design-on-the-windows-of-a-building-The-curved-part-XY-is-an-arc-of-a-circle-The-rise-of-the-segmental-arc-is-10cm-its-span-is-100cm-and-XZ-ZY-120cm-calculate-i-th

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Question Number 204873 by necx122 last updated on 29/Feb/24

$$Thefigurebelowrepresentsadesign$$$$onthewindowsofabuilding.The$$$$curvedpartXYisanarcofacircle.$$$$Theriseofthesegmentalarcis10cm,$$$$itsspanis100cmandXZ=ZY=120cm.$$$$calculate:$$$$\left(i\right)theradiusofthecircle$$$$\left(ii\right)theareaofthesegmentalcap,$$$$correctto2significantfigures.$$$$\left(iii\right)thetotalareaofthedesign,correct$$$$to3significantfigures.$$

Commented by necx122 last updated on 29/Feb/24

Commented by necx122 last updated on 29/Feb/24

I will really need our help on this. Thanks in advance.

Answered by A5T last updated on 29/Feb/24

$$ConstructtheperpendicularbisectorofXY,$$$$thenZandO,centreofcircle,lieonit.$$$$R^2={50}^2+{(R-10)}^2\Rightarrow R^2={50}^2+R^2-20R+100$$$$\Rightarrow 20R=2600\Rightarrow R=130cm$$$${100}^2=2\times {130}^2-2\times {130}^{2}cosXOY$$$$\Rightarrow cosXOY=\frac{119}{169}\Rightarrow sinXOY=\frac{120}{169}$$$$\Rightarrow Areaofsector=\frac{{sin}^{-1}\left(\frac{120}{169}\right)}{360}\times {130}^2\pi \approx 6671.9699$$$$AreaofXOY=65\times 130\times \frac{120}{169}=6000$$$$\Rightarrow Areaofsegmentalarc\approx 671.9699{cm}^2$$$$\Rightarrow Totalareaofdesign\approx 60\times 120\times \frac{5\sqrt{119}}{72}-671.9699$$$$500\sqrt{119}-671.9699\approx 4782.389$$$$[{100}^2=2\times {120}^2-2\times {120}^{2}cosXZY\Rightarrow cosXZY=\frac{47}{72}$$$$\Rightarrow sinXZY=\frac{5\sqrt{119}}{72}]$$

Commented by necx122 last updated on 29/Feb/24

$$Thisisclear,understandableand$$$$precise.Thankyousir.$$

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