Question

A hollow cylinder is 35 cm in length (height). Its internal and external diameters are 8 cm and 8.8 cm respectively. Find its :
(i) outer curved surface area
(ii) inner curved surface area
(iii) area of cross-section
(iv) total surface area.

Answer 1

The height of the cylinder h = 35 cm
The internal radius r = $\frac{8}{2}$ cm = 4 cm
The external radius R = $\frac{8.8}{2}$ cm = 4.4 cm
(i) Outer curved surface area = 2πRh = 2 × $\frac{22}{7}\times 4.4\times 35c{m}^2$ = 968 cm²
(ii) Inner curved surface area = 2πrh = 2 ×$\frac{22}{7}\times 4\times 35c{m}^2$ = 880 cm²
(iii) The cross-section of a hollow cylinder is like a ring with external radius R = 4.4 cm and internal radius r = 4 cm


∴ Area of cross-section = πR² – πr² = π(R² – r²) = $\frac{22}{7}({4.4}^2-4^2)c{m}^2$
=$\frac{22}{7}(4.4+4)(4.4-4)c{m}^2=\frac{22}{7}\times 8.4\times 0.4c{m}^2$ = 10.56 cm²
(iv) Total surface area
= 2πRh + 2πrh + 2π(R² – r²)
= [968 + 880 + 2(10.56)] cm²
= 1869.12 cm²