Question Number 204873 by necx122 last updated on 29/Feb/24
$${The}\:{figure}\:{below}\:{represents}\:{a}\:{design} \\ $$$${on}\:{the}\:{windows}\:{of}\:{a}\:{building}.\:{The} \\ $$$${curved}\:{part}\:{XY}\:{is}\:{an}\:{arc}\:{of}\:{a}\:{circle}. \\ $$$${The}\:{rise}\:{of}\:{the}\:{segmental}\:{arc}\:{is}\:\mathrm{10}{cm}, \\ $$$${its}\:{span}\:{is}\:\mathrm{100}{cm}\:{and}\:{XZ}={ZY}=\mathrm{120}{cm}. \\ $$$${calculate}: \\ $$$$\left({i}\right)\:{the}\:{radius}\:{of}\:{the}\:{circle} \\ $$$$\left({ii}\right)\:{the}\:{area}\:{of}\:{the}\:{segmental}\:{cap}, \\ $$$${correct}\:{to}\:\mathrm{2}\:{significant}\:{figures}. \\ $$$$\left({iii}\right)\:{the}\:{total}\:{area}\:{of}\:{the}\:{design},\:{correct} \\ $$$${to}\:\mathrm{3}\:{significant}\:{figures}. \\ $$
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
![Construct the perpendicular bisector of XY, then Z and O,centre of circle, lie on it. R^2 =50^2 +(R−10)^2 ⇒R^2 =50^2 +R^2 −20R+100 ⇒20R=2600⇒R=130cm 100^2 =2×130^2 −2×130^2 cosXOY ⇒cosXOY=((119)/(169))⇒sinXOY=((120)/(169)) ⇒Area of sector=((sin^(−1) (((120)/(169))))/(360))×130^2 π≈6671.9699 Area of XOY=65×130×((120)/(169))=6000 ⇒Area of segmental arc≈671.9699cm^2 ⇒Total area of design≈60×120×((5(√(119)))/(72))−671.9699 500(√(119))−671.9699≈4782.389 [100^2 =2×120^2 −2×120^2 cosXZY⇒cosXZY=((47)/(72)) ⇒sinXZY=((5(√(119)))/(72))]](https://www.tinkutara.com/question/Q204877.png)
$${Construct}\:{the}\:{perpendicular}\:{bisector}\:{of}\:{XY}, \\ $$$${then}\:{Z}\:{and}\:{O},{centre}\:{of}\:{circle},\:{lie}\:{on}\:{it}. \\ $$$${R}^{\mathrm{2}} =\mathrm{50}^{\mathrm{2}} +\left({R}−\mathrm{10}\right)^{\mathrm{2}} \Rightarrow{R}^{\mathrm{2}} =\mathrm{50}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{20}{R}+\mathrm{100} \\ $$$$\Rightarrow\mathrm{20}{R}=\mathrm{2600}\Rightarrow{R}=\mathrm{130}{cm} \\ $$$$\mathrm{100}^{\mathrm{2}} =\mathrm{2}×\mathrm{130}^{\mathrm{2}} −\mathrm{2}×\mathrm{130}^{\mathrm{2}} {cosXOY} \\ $$$$\Rightarrow{cosXOY}=\frac{\mathrm{119}}{\mathrm{169}}\Rightarrow{sinXOY}=\frac{\mathrm{120}}{\mathrm{169}} \\ $$$$\Rightarrow{Area}\:{of}\:{sector}=\frac{{sin}^{−\mathrm{1}} \left(\frac{\mathrm{120}}{\mathrm{169}}\right)}{\mathrm{360}}×\mathrm{130}^{\mathrm{2}} \pi\approx\mathrm{6671}.\mathrm{9699} \\ $$$${Area}\:{of}\:{XOY}=\mathrm{65}×\mathrm{130}×\frac{\mathrm{120}}{\mathrm{169}}=\mathrm{6000} \\ $$$$\Rightarrow{Area}\:{of}\:{segmental}\:{arc}\approx\mathrm{671}.\mathrm{9699}{cm}^{\mathrm{2}} \\ $$$$\Rightarrow{Total}\:{area}\:{of}\:{design}\approx\mathrm{60}×\mathrm{120}×\frac{\mathrm{5}\sqrt{\mathrm{119}}}{\mathrm{72}}−\mathrm{671}.\mathrm{9699} \\ $$$$\mathrm{500}\sqrt{\mathrm{119}}−\mathrm{671}.\mathrm{9699}\approx\mathrm{4782}.\mathrm{389} \\ $$$$\left[\mathrm{100}^{\mathrm{2}} =\mathrm{2}×\mathrm{120}^{\mathrm{2}} −\mathrm{2}×\mathrm{120}^{\mathrm{2}} {cosXZY}\Rightarrow{cosXZY}=\frac{\mathrm{47}}{\mathrm{72}}\right. \\ $$$$\left.\Rightarrow{sinXZY}=\frac{\mathrm{5}\sqrt{\mathrm{119}}}{\mathrm{72}}\right] \\ $$
Commented by necx122 last updated on 29/Feb/24
$${This}\:{is}\:{clear},\:{understandable}\:{and} \\ $$$${precise}.\:{Thank}\:{you}\:{sir}. \\ $$