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Ex.13.4 Q2 Surface Areas and Volumes Solution - NCERT Maths Class 9

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Question

Find the surface area of a sphere of diameter:

i. \(14 \, \rm cm \)

ii. \(21 \, \rm cm \)

iii. \(3.5 \, \rm m \)

 Video Solution
Surface-Areas-And-Volumes
Ex exercise-13-4 | Question 2

Text Solution

Reasoning:

Surface area of the sphere of radius \(r\) is equal \(4\) times the area of the circle of radius \(r. \)

\(\begin{align}S = 4\pi {r^2} \end{align}\)

What is known?

Diameter of the sphere.

What is unknown?

Surface area of the sphere.

Steps:

(i)  Diameter ( \(2r\) ) \(= 14{\rm\,{ cm}}\)

     Radius (\(r\)) \(= 7\,\rm\,{cm}\)

Surface area 

\[\begin{align}&= 4\pi {r^2}\\ & = 4 \times \frac{{22}}{7} \times 7 \times 7\\&= 616\,\rm{cm^2} \end{align}\]

(ii) Diameter(\(2 r\)) = \(21 \rm\,{cm}\)

     Radius (\(r\)) = \({\frac{21}{2} \, \mathrm {cm}}\)

Surface area

\[\begin{align}&={4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}} \\&={1386 \, \mathrm{cm}^{2}}\end{align}\]

(iii) Diameter (\(2 r\)\(= 3.5\rm\,  m \)

      Radius (\(r\)) \(= \frac{3.5}{2}\rm\, m\) 

Surface area

\[\begin{align} &={4 \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2}} \\ &={38.5 \mathrm{m}^{2}}\end{align}\]

Answer:

Surface areas

\(\begin{align}&\text{(i) 616} \rm\,{cm^2}\\&\text{(ii) 1386}\rm\,{cm^2}\\&\text{(iii) 38.5}\,\rm\,{m^2} \end{align}\)