the volumes of two spheres are in the ratio 64:27 and the sum of their radii is 7 cm. what is the difference in their total surface areas?
Let the radii of the spheres be r and R, where r < R.
We are given that the sum of their radii is 7 cm, so we have r + R = 7. (1)
We are also given that the volumes of the spheres are in the ratio 64:27, so we have (4/3)πr^3 : (4/3)πR^3 = 64 : 27.
Canceling out the common factors, we get r^3 : R^3 = 64 : 27.
Taking the cube root of both sides, we get r : R = 4 : 3. (2)
From equation (1), we have r = 7 - R.
Substituting this into equation (2), we get (7-R) : R = 4 : 3.
Cross multiplying, we get 3(7-R) = 4R.
Expanding, we get 21 - 3R = 4R.
Combining like terms, we get 21 = 7R.
Dividing by 7, we get R = 3.
Substituting this into equation (1), we get r + 3 = 7.
Subtracting 3 from both sides, we get r = 4.
The radii of the spheres are 4 cm and 3 cm.
The total surface area of a sphere is given by 4πr^2, so the total surface area of the sphere with radius r = 4 cm is 4π(4^2) = 64π.
The total surface area of the sphere with radius R = 3 cm is 4π(3^2) = 36π.
The difference in their total surface areas is 64π - 36π = 28π.
Therefore, the difference in their total surface areas is 28π.