Question

The diameter of a sphere is decreased by 25 percent. By what per cent does its curved surface area decrease?​

Answer 1

Answer:

The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease? cm. Hence, the original curved surface area decreases by 43.75%.

plz refer the attachment to get it better

Answer 2

$\large{\underline{\sf{\red{Answer:-}}}}$

$\large\boxed{\underline{{\rm 43.75 \% }}}$

$\large{\underbrace{\sf{\pink{Solution:}}}}$

Let the,

• Diameter of sphere be d.

• Radius of sphere be r.

• Original CSA = $\rm{4\pi \: (\dfrac{d}{2})^{2}  =\pi d^{2} }$

Now,

Diameter of the square is decreased by 25%

New diameter = $\rm{d-\dfrac{25}{100} d=\dfrac{3d}{4}  }$

                           $\rm{d=(1-\dfrac{25}{100})d  }$

                         = $\rm{(\dfrac{100-25}{100}  )d }$

                         = $\rm{(\dfrac{75}{100}  )d }$

                         = $\rm{(\dfrac{3}{4}  )d }$

New CSA (A) = $\rm{A=4 \times \pi \times (\dfrac{3}{8} d)^{2} }$

                     = $\rm{A=4 \times \pi \times \dfrac{9}{64} d^{2} }$

                     = $\rm{A=\dfrac{9}{64} \pi  \:d^{2} }$

Percentage of area decreased:-

$\large\boxed{\underline{\purple{\rm \dfrac{Decrease\:in\:area}{Original\:area} \times 100 }}}$

$\red{\implies\:\:}$ $\rm{\dfrac{A-A}{A}\times 100 \% }$

$\red{\implies\:\:}$ $\rm{\dfrac{\pi \:d^{2}-\dfrac{9}{16} \pi \:d^{2}  }{\pi \:d^{2} }\times 100 \% }$

$\red{\implies\:\:}$ $\rm{\dfrac{\pi \:d^{2}(1-\dfrac{9}{16} )  }{\pi \:d^{2} }\times 100 \% }$

$\red{\implies\:\:}$ $\rm{(1-\dfrac{9}{16})\times 100 }$

$\red{\implies\:\:}$ $\rm{\dfrac{16-9}{16}\times 100 \% }$

$\red{\implies\:\:}$ $\rm{\dfrac{7}{16}\times 100 \% }$

$\red{\implies\:\:}$ $\rm{\dfrac{7}{4}\times 25 \% }$

$\red{\implies\:\:}$ $\rm{\dfrac{175}{4} \% }$

$\red{\implies\:\:}$ $\rm{43.75 \% }$

Brainliest plz ♥

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