Question

the volume of two spheres are in the ratio 64:27 the difference of their surface area is the sum of their raddi 7 unit is

Answer 1

Here is your solution

Ratio of volumes=64:27

Therefore ratio of radii=4:3

sum of radii =>7=4+3

Therefore  radii of 2 spheres = 4 and 3

difference in surface areas=?
$= 4\pi \: r1 {}^{2} - 4\pi \: r2 {}^{2} \\ = 4\pi(r1 {}^{2} -r2 {}^{2}) \\ = 4 \times \frac{22}{7} (4 {}^{2} - 3 {}^{2} ) \\ \\ = \frac{88}{7} (16 - 9) \\ \\ = \frac{88}{\cancel7} \times \cancel7 \\ \\ = 88unit$
Difference in surface areas is 88 unit

hope it helps you

Answer 2

let x and y be radii of sheres respectively.

vol of 1st shere =4/3×pi×x^3=64

vol of 2nd shere=4/3×pi×y^3=27

so we have ratio of their vol=64:27

x^3/y^3=64:27
x^3/y^3=(4/3)^3
x/y=4/3
x=4y/3------------(1)


Now we have 4pi (x^2-y^2)=x+y
4pi(x+y)(x-y)=x+y

x-y=1/4pi
x-y=7/88
. ... x=7/88+y-------------(2)


solving 1 and 2
we have x=29/88cm and y=21/88cm