Question

A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹4989.60. If the cost of white-washing is ₹20 per square meter, find the (i) inside surface area of the dome, (ii) volume of the air inside the dome​

Answer 1

Answer:

Hope it will help you

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Step-by-step explanation:

Total cost = 498.96 Rs. => Area whitewashed = 249.48 Rs. And according to the question, the area whitewashed is Curved surface area of hemisphere. => 523.908 m³ (approx.)

Answer 2

$\star \; {\underline{\boxed{\pmb{\red{\frak{ \; Given \; :- }}}}}}$

• Cost of white - washing = Rs.4989.60
• Rate of white - washing = Rs.20

$\star \; {\underline{\boxed{\pmb{\blue{\frak{ \; To \; Find \; :- }}}}}}$

• Inside surface area of the dome = ?
• Volume of the air inside the dome = ?

$\star \; {\underline{\boxed{\pmb{\orange{\frak{ \; SolutioN \; :- }}}}}}$

$\maltese$ Formula Used :

• ${\underline{\boxed{\pmb{\sf{ CSA{\small_{(Hemisphere)}} = 2 \pi {r}^{2} }}}}}$

• ${\underline{\boxed{\pmb{\sf{ Volume{\small_{(Hemisphere)}} = \dfrac{2}{3} \pi {r}^{3} }}}}}$

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$\maltese$ How to Calculate :

$\longrightarrow$ Here, we can Calculate the Surface Area by dividing the Cost with the Rate . And for calculating the Air sinsid the dome we need to calculate the Volume . As we know for calculating it we should get the Radius first .So, Let's Solve :

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$\maltese$ Calculating the Surface Area :

${\longmapsto{\qquad{\sf{ Surface \; Area = \dfrac{ Cost }{Rate} }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ Surface \; Area = \dfrac{ 4989.60 }{20} }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ Surface \; Area = \dfrac{ 498960 }{20 \times 100} }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ Surface \; Area = \cancel\dfrac{ 498960 }{2000} }}}} \\ \\ \\ \ {\qquad \; \; {\therefore \; {\underline{\boxed{\pmb{\pink{\sf{ Surface \; Area = 249.48 \; {m}^{2} }}}}}}}}$

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$\maltese$ Calculating the Radius :

${\dashrightarrow{\qquad{\sf{ CSA = 2 \pi {r}^{2} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ 249.48 = 2 \times \dfrac{22}{7} \times {r}^{2} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ 249.48 \times 7 = 2 \times 22 \times {r}^{2} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ 1746.36 = 44 \times {r}^{2} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ \dfrac{1746.36}{44} = {r}^{2} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ \cancel\dfrac{1746.36}{44} = {r}^{2} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ 39.69 = {r}^{2} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ \sqrt{39.69} = r }}}} \\ \\ \\ \ {\qquad \; \; {\therefore \; {\underline{\boxed{\pmb{\purple{\sf{ Radius = 6.3 \; cm }}}}}}}}$

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$\maltese$ Calculating the Volume of air :

${\implies{\qquad{\sf{ Volume = \dfrac{2}{3} \pi {r}^{3} }}}} \\ \\ \\ \ {\implies{\qquad{\sf{ Volume = \dfrac{2}{3} \times \dfrac{22}{7} \times {6.3}^{3} }}}} \\ \\ \\ \ {\implies{\qquad{\sf{ Volume = \dfrac{2}{3} \times \dfrac{22}{7} \times \bigg( {\dfrac{63}{10}} \bigg)^{3} }}}} \\ \\ \\ \ {\implies{\qquad{\sf{ Volume = \dfrac{2}{3} \times \dfrac{22}{\cancel7} \times \dfrac{\cancel{63}}{10} \times \dfrac{63}{10} \times \dfrac{63}{10} }}}} \\ \\ \\ \ {\implies{\qquad{\sf{ Volume = \dfrac{2}{3} \times 22 \times \dfrac{9}{10} \times \dfrac{63}{10} \times \dfrac{63}{10} }}}} \\ \\ \\ \ {\implies{\qquad{\sf{ Volume = \dfrac{1571724}{3000} }}}} \\ \\ \\ \ {\implies{\qquad{\sf{ Volume = \cancel\dfrac{1571724}{3000} }}}} \\ \\ \\ \ {\qquad \; \; {\therefore \; {\underline{\boxed{\pmb{\red{\sf{ Volume = 523.9 \; {m}^{3} \; \bigg\lgroup Approx. \bigg\rgroup }}}}}}}}$

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$\maltese$ Therefore :

❛❛ Surface Area of the Dome is 249.48 cm² and the Volume of air inside the dome is 523.9 cm³ . ❜❜

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