Question

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

Answer 1

$\large{\underbrace{\sf{\red{Answer:}}}}$

$\huge{\boxed{\sf{\orange{ 43 \dfrac{3}{4} \% \:or\: 43.75 \%}}}}$

$\large{\underbrace{\sf{\purple{Explanation:}}}}$

Let the radius of the sphere be $\displaystyle{\sf{ \dfrac{ \pi}{2}cm.}}$

Then its diameter = $\implies{\sf{ \big( \dfrac{ \pi}{2} \big) cm.}}$

Curved surface area of the original sphere

⠀⠀:$\implies{\sf{ 4 \pi \dfrac{r}{2}^{2} = \pi r^{2}cm^{2}}}$

New diameter(decreased) of the sphere

⠀:$\implies{\sf{ r-r \times \dfrac{25}{100}}}$

⠀⠀:$\implies{\sf{ r- \dfrac{r}{4}= \dfrac{3r}{4}}}$

${ \therefore}$ Radius of the new sphere

:$\implies{\sf{ \dfrac{1}{2} \big( \dfrac{3r}{4} \big) = \frac{3r}{8}cm }}$

${ \therefore}$ New curved surface area of the sphere

:$\implies{\sf{ 4\pi \big( \frac{3r}{8}cm \big) = \dfrac{9\pi r^{2}}{16}cm^{2}}}$

${ \therefore}$ Decrease in the original curved surface area

:$\implies{\sf{ \pi r^{2} - \dfrac{9 \pi r^{2}}{16}}}$

:$\implies{\sf{ \pi r^{2} - \dfrac{7 \pi r^{2}}{16}}}$

${ \therefore}$ Percentage of decrease in the original curved surface area

:$\implies{\sf{ \dfrac{\dfrac{7 \pi r^{2}}{16}}{ \pi r^{2}} \times 100 \%}}$

$\large{:}\implies{\boxed{\sf{\pink{ 43 \dfrac{3}{4} \%\:or\:43.75 \%}}}}$

Hence, the original curved surface area decreases by 43.75%.

Answer 2

$\huge\underline{\mathbb\red{A}\green{N}\mathbb\blue{S}\purple{W}\mathbb\orange{E}\pink{R}}\:$

Let D be the diameter of the sphere, and the radius of the sphere r is D2.

The curved surface area of the sphere is A=4πr2=4πD2=πD2.

Since the diameter is decreased by 25 , the new diameter is 75100×d=3d4.

New surface area =4πr2=4π(3d2⋅4)2=916πd2

Decrease in surface area =πd2−916πd2=716πd2

So the % decrease in Surface Area= DecreaseinsurfaceareaSurfacearea×100%

=7πd216πd2×100=716×100=43.75%

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