Question

Curved Surface area of a right circular cylinder is 6.6cm2. If the radius of the
base of cylinder is 0.7cm, find the height.

​

Answer 1

• The height of the cylinder = 1.5 cm.

Given :

• Curved surface area (C.S.A.) of a right circular cylinder = 6.6 cm².
• The radius of the base of cylinder = 0.7 cm.

To Find :

• The height of the cylinder.

Diagram :

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(1,6)\qbezier(5,1)(5,1)(5,6)\qbezier(1,1)(2.9,0.3)(5,1)\qbezier(1,6)(2.9,5.2)(5,6)\qbezier(1,6)(2.9,6.5)(5,6)\put(6,2.5){\vector(0,1){3.4}}\put(6.4,3){\large\sf h}\put(6,2.5){\vector(0,-1){1.6}}\put(2.9,6){\vector(1,0){2}}\put(2.9,6.5){\large\sf r = 0.7 m}\end{picture}$

Solution :

Let,

The height of the cylinder be h.

We know that,

Curved surface area (C.S.A.) of the cylinder = 2πrh

Given that,

Curved surface area (C.S.A.) of the cylinder = 6.6 cm²

$\implies 2\pi rh = 6.6 \: {cm}^{2}$

We have,

• Radius of the base of cylinder = 0.7 cm.

$\implies 2 \times \dfrac{22}{7} \times 0.7 \: cm \times h = 6.6 \: {cm}^{2}$

$\implies 2 \times \dfrac{22}{ \cancel7} \times \dfrac{ \cancel7}{10} \: cm \times h = \dfrac{66}{10} \: {cm}^{2}$

$\implies 2 \times \dfrac{22}{1} \times \dfrac{1}{ \cancel{10}} \: cm \times h \times \cancel{10}= 66 \: {cm}^{2}$

$\implies 2 \times 22 \times 1 \: cm\times h = 66 \: {cm}^{2}$

$\implies h = \dfrac{66 \: \cancel{ {cm}}^{2} }{2 \times 22 \: \cancel{cm}}$

$\implies h = \cancel \dfrac{66}{44} \: cm$

$\implies h = \cancel\dfrac{33}{22} \: cm$

$\implies h = \dfrac{3}{2} \: cm$

$\implies h = 1.5 \: cm$

Hence,

The height of the cylinder is 1.5 cm.