Question

The diameter of a sphere is increased by 15% .By what % does its surface
area and volume increases.​

Answer 1

Answer:

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Answer 2

Answer:

Surface area = 32.25%  

volume = 52.08%

Given problem:

The diameter of a sphere is increased by 15% .By what % does its surface

area and volume increases.​

Step-by-step explanation:

Let d be the initial diameter of the sphere

then initial radius of the sphere = $\frac{d}{2}$

increased diameter = d + 15% of d

                                 = d + $\frac{15d}{100}$  = d+ 0.15d = 1.15d

hence, increased radius = 1.15d / 2                      

Percentage of increased surface area  

initially surface area of the sphere = 4πr²  

                                                         = $4\pi (\frac{d}{2})(\frac{d}{2} )$ = πd²    

surface area with increased radius =  $4\pi \frac{1.15d}{2}(\frac{ 1.15d}{2} )$  

                                              = π (1.15d)²

                                              = 1.3225 πd²    

increased surface area = 1.3225 πd² - πd²  = 0.3225 πd²

percentage of increased surface area = (increased area/initial area)100

                                  = $\frac{0.3225\pi d^{2} }{\pi d ^{2} } (100)$  

                                  = 0.3225(100)

                                  =  32.25 %

Percentage of increased volume

initially volume of the sphere = $\frac{4}{3} \pi r^{3}$  

                                                 = $\frac{4}{3} \pi (\frac{d}{2}) (\frac{d}{2}) (\frac{d}{2} )$  

                                                 = $\frac{1}{3} \pi \frac{d ^{3} }{2}$ = $\frac{\pi d^{3} }{6}$  

volume of sphere with increased radius = $\frac{4}{3} \pi (\frac{1.15d}{2}) (\frac{1.15d}{2}) (\frac{1.15d}{2} )$

                                                                  = $\frac{1}{3} \pi \frac{(1.15d )^{3} }{2}$  

                                                                  = $\frac{1}{6} \pi (1.5208d^{3} )$    

increased volume = $\frac{1}{6} \pi (1.5208d^{3} )$ - $\frac{\pi d^{3} }{6}$    

                              = $\frac{\pi d ^{3} }{6} [ 1.5208 - 1]$  

                              = $\frac{\pi d ^{3} }{6} [ 0.5208 ]$  

percentage of increased volume = (increased volume/initial volume) (100)

                                                     = $\frac{\frac{\pi d ^{3} }{6} [ 0.5208 ]}{\frac{\pi d^{3} }{6}} (100)$

                                                     = 0.5208(100) = 52.08%