Question

the diameter of a sphere is decreased by 25% by what percent does it curved surface area decrease .who can tell me the answer first​

Answer 1

Solution :

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For a sphere having a radius r ;

Curved surface area > 4 π r² .

Now , the value of r decreases by 25%

New value to r -

> 75% of r

> 75/100 r

> ¾ r.  

New curved surface area >

> 4 π ( ¾ r )² .

Change of surface area :

> 4 π r² - 4 π( 9/16 r² )

> 4 π r² [ 1 - 9/16 ]

> 4 π r² × 7/16

> π r² × 7/4

> 7/4 π r².  

Change in percentage :

> (7/4) π r² × 1/ ( 4 π r² ) × 100%

> 700/16%

> 43.75 %

This is the required answer.  

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Answer 2

Answer:

Given :-

• The diameter of a sphere is decreased by 25%.

Find Out :-

• Percentage decrease in curved surface area.

Solution :-

Let, the diameter of the sphere be d.

Then, the radius of the sphere is $\dfrac{d}{2}$.

As we know that,

✠ Curved surface area = 4πr² ✠

Then,

⇨ C.S.A = $4{\pi}(\dfrac{d}{2})^{2}$

⇨ C.S.A = $4{\pi} \times \dfrac{{d}^{2}}{4}$

➟ C.S.A = πd²

It is given that diameter is decreased by 25%.

So, new diameter,

↣ $d - \dfrac{d}{4}$

↣ $\dfrac{4d - d}{4}$

➡ $\dfrac{3d}{4}$

Now, new curved surface area will be,

➣ ${\pi}(\dfrac{3d}{4})^{2}$

➣ ${\pi}\dfrac{9{d}^{2}}{16}$

➲ $\dfrac{9}{16}{\pi}{d}^{2}$

Then, change in area,

➻ ${\pi}{d}^{2} - (\dfrac{9}{16}){\pi}{d}^{2}$

➽ $\dfrac{7}{16}{\pi}{d}^{2}$

Now, find the percentage decrease in curved surface area,

➪ $\dfrac{(\dfrac{7}{16}){\pi}{d}^{2}}{{\pi}{d}^{2}} \times 100$

➪ $\dfrac{7}{16} \times 100$

➪ $\dfrac{700}{16}$

➥ 43.75%

Hence, the percentage decrease in curved surface area is 43.75%.