Question

↓ Please help me out with this ↓

The volume of the two spheres are in the ratio 64:27. Find the difference of their surface areas, if the sum of their radii is 7.​

Answer 1

Answer:

hey dear .....

Let the radii of the two spheres be r1 cm and r2 cm respectively. Then,

From (1) and (2)

straight r subscript 1 space equals space 4 space cm

straight r subscript 2 space equals space 3 space cm

therefore Difference of their surface areas

equals 4 πr subscript 1 squared minus 4 πr subscript 2 squared

equals space 4 πr left parenthesis straight r subscript 1 squared minus straight r subscript 2 squared right parenthesis

squared minus straight r subscript 2 squared right parenthesisequals space 4 straight pi space left parenthesis 4 squared minus 3 squared right parenthesis

equals space 4 straight pi left parenthesis 16 minus 9 right parenthesis

equals 4.22 over 7.7

equals space 88 space cm squared...

hope it's helpful for you....

Answer 2

$\huge\mathfrak{Bonjour!!}$

$\huge\bold\pink{Solution:-}$

☘ Answer:-

☞ The difference of their surface areas = 88 cm²

☘ Step-by-step explanation:-

Let the radii of two spheres be r1 cm and r2 cm respectively.

and let the volumes of those two spheres be V1 and V2 respectively. Then,

V1/V2 = 64/27

$= ) \frac{ \frac{4}{3}\pi \: r1^{3} }{ \frac{4}{3}\pi \: r2^{3} } = \frac{64}{27}$

$= ) \frac{r1^{3} }{r2^{3} } = \frac{4^{3} }{3^{3} }$

$= )( \frac{r1}{r2} )^{3} = ( \frac{4}{3} )^{3}$

$= ) \frac{r1}{r2} = \frac{4}{3}$

$= )r1 = \frac{4}{3} r2 \: .....(i)$

But,

r1 + r2 = 7 [Given]

=> 4/3 r2 + r2 = 7

=> 7/3 r2 = 7

=> 7 × 3/7 = 3 cm.

Therefore,

r1 = 4/3 × 3 = 4 cm.

Now,

Let S1 and S2 be the surface areas of two spheres. Then,

S1 = 4πr1² = 4π × 4 × 4 = 64π cm².

and

S2 = 4πr2² = 4π × 3 × 3 = 36π cm².

Therefore,

☞ S1 - S2 = 64π - 36π = 28π cm² = 28 × 22/7 cm² = 88 cm²

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