Question

A toy is in the form of a right circular with a hemispere on one end and a cone on the other . The height and the radius of base of the cylindrcal part are 13 cm and 5cm respectively. The radius of hemisphere and base of the conical part are some as that of the cylinder . Calculate the surface area of the toy , if the height of the cone is 12 cm.


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Answer 1

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$\sf \scriptsize{( i ) \: Radius \: of \: hemisperical \: part \: = 5 cm. \: } \\ \sf \scriptsize{∴Curved \: surface \: area \: of \: hemispherical \: portion \: = 2 \: \pi \: r^{2} } \\ \sf \scriptsize{ = 2 \times \frac{22}{7} \times 5 \times 5 = \: \red{ \frac{1100}{7} \: cm^{2} }} \\ \\ \sf \scriptsize{(ii) Height \: \: of \: cylindrical \: part = \: 13 cm }\\ \sf \scriptsize{ And \: radius \: = 5 cm } \\ \sf \scriptsize{∴Curved \: surface \: area \: of \: cylindrical \: portion = 2 \: \pi \: r \: h} \\ \sf \scriptsize{ = 2 \times \frac{22}{7} \times 5 \times 13 = \red{ \frac{2860}{7} cm^{2} }} \\ \\ \sf \scriptsize{ (iii)Height \: of \: conical \: part \: = 12 \: cm.} \\ \sf \scriptsize{And \: radius \: = 5 \: cm.} \sf \scriptsize{ slant \: height \: = \sqrt{r^{2} + h^{2} } = \sqrt{5^{2} + 12^{2} } = \sqrt{25 + 144} } \\ \sf \scriptsize{ = \sqrt{169} = \red{ 13 \: cm }} \\ \\ \sf \scriptsize{ \therefore \: curved \: survace \: area \: = \pi \: r \: l \: = \frac{22}{7} \times 5 \times 13 = \red{\frac{1430}{7} = \: cm^{2}} } \\ \\ \sf \scriptsize{ \therefore \: Total \: surface \: area \: of \: the \: toy \: = ( \frac{1100}{7} + \frac{2860}{7} + \frac{1430}{7}) \: cm^{2} } \\ \sf \scriptsize{ = \frac{5390}{7} cm^{2} \red{= 770 \: cm^{2}} }$

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