Question

The radius of circular ends of a solid frustum of a cone are 33 cm and 27 cm. Its slant height is 10 cm. Find the volume and the total surface area of frustum.

Answer 1

Solution:-
The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm.
So, let the radius of the upper end be 'R' and radius of the lower end be 'r'.
R = 33 cm and r = 27 cm and slant height (l) = 10 cm
Now, we know that slant height of frustum, l = √h²+(R - r)²  where h is the height of the frustum.
10 = √h²+(33 - 27)²
10² = h² + 36
h² = 100 - 36
h² = 64
h = 8 cm
volume of the frustum = πh/3(R²+r²+Rr)
= (22/7)*(8/3){33²+27²+(33*27)}
= (22/7)*(8/3)*2709
= 476784/21
Volume = 22704 cm³
Total surface area of the frustum = π{R² + r² +l(R + r)}
= 22/7{33² + 27² + 10(33 + 27)}
= 22/7*(1089 + 729 + 600)
= 22/7*2418
= 53196/7
Total surface area = 7599.43 sq cm
Answer.
 

Answer 2

The radius of circular ends of a solid frustum of a cone are 33 cm and 27 cm.
So, suppose R = 33 cm and r = 27 cm and slant height; l = 10 cm
Now we know that slant height of frustum, l = h2+(R−r)2‾‾‾‾‾‾‾‾‾‾‾‾√ where h is the height of frustum
10 = h2+(33−27)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√⇒102 = h2+36⇒h2 = 100−36 = 64⇒h = ±64‾‾‾√ = ±8
As the height can't be negative so neglect h = -8.
So, h = 8 cm
So, volume of frustum = πh3(R2+r2+Rr) = 227×83(332+272+33×27) = 227×83×2709 = 22704 cm3
And total surface area of frustum = π[R2+r2+l(R+r)] = 227[332+272+10(33+27)] =