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Surface Areas and Volumes - Surface Area of a Right Circular Cylinder

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Right Circular Cylinder is a solid generated by the revolution of a rectangle about one of its sides. Visually, it consists of two congruent and parallel circular bases joined by a curved surface. The line joining the centers of the two circular bases is perpendicular to the bases, which is why it is called a 'Right' circular cylinder.

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The radius rr of the cylinder is the radius of its circular base. The height hh (or length) is the perpendicular distance between the two circular bases. In visual representations, hh is the vertical span and rr is the horizontal distance from the center of the base to the boundary.

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The Curved Surface Area (CSA) is the area of the side of the cylinder, excluding the top and bottom circular faces. If you imagine cutting the curved surface vertically and flattening it out, it forms a rectangle. The length of this rectangle is equal to the circumference of the circular base 2πr2\pi r, and its breadth is equal to the height hh.

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The area of each circular base is πr2\pi r^2. Since a closed cylinder has two bases (top and bottom), the total area of the bases combined is 2πr22\pi r^2.

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The Total Surface Area (TSA) of a cylinder is the sum of the Curved Surface Area and the areas of the two circular bases. Visually, this covers every part of the outer skin of a solid cylinder, like a sealed tin can.

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If a cylinder is 'open' at the top (like a drinking glass), the surface area consists of the Curved Surface Area plus only one circular base area: 2πrh+πr22\pi rh + \pi r^2.

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Units of Measurement: Since surface area measures the extent of a 2D surface on a 3D object, it is always expressed in square units, such as cm2cm^2 or m2m^2. Always ensure rr and hh are in the same units before calculating.

📐Formulae

Circumferenceofthebase=2πrCircumference of the base = 2\pi r

Areaofthebase=πr2Area of the base = \pi r^2

CurvedSurfaceArea(CSA)=2πrhCurved Surface Area (CSA) = 2\pi rh

TotalSurfaceArea(TSA)=2πrh+2πr2=2πr(h+r)Total Surface Area (TSA) = 2\pi rh + 2\pi r^2 = 2\pi r(h + r)

Where π≈227 or 3.14\text{Where } \pi \approx \frac{22}{7} \text{ or } 3.14

💡Examples

Problem 1:

A cylindrical pillar is 5050 cm in diameter and 3.53.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs.12.50Rs. 12.50 per m2m^2.

Solution:

  1. First, convert all dimensions to the same unit (meters). Radius r=diameter2=502=25r = \frac{diameter}{2} = \frac{50}{2} = 25 cm =0.25= 0.25 m. Height h=3.5h = 3.5 m.
  2. Calculate the Curved Surface Area (CSA): CSA=2πrh=2×227×0.25×3.5CSA = 2\pi rh = 2 \times \frac{22}{7} \times 0.25 \times 3.5 CSA=2×22×0.25×0.5=44×0.125=5.5CSA = 2 \times 22 \times 0.25 \times 0.5 = 44 \times 0.125 = 5.5 m2m^2.
  3. Calculate the total cost: Cost=Area×Rate=5.5×12.50=68.75Cost = Area \times Rate = 5.5 \times 12.50 = 68.75. Final Answer: Rs.68.75Rs. 68.75.

Explanation:

To solve this, we first identified that only the curved surface needs painting. We converted the diameter to radius and ensured all units were in meters to match the rate given per m2m^2. We then applied the CSA formula and multiplied the result by the unit cost.

Problem 2:

The total surface area of a right circular cylinder is 462462 cm2cm^2. Its curved surface area is one-third of its total surface area. Find the radius and height of the cylinder.

Solution:

  1. Given TSA=462TSA = 462 cm2cm^2 and CSA=13×TSACSA = \frac{1}{3} \times TSA.
  2. Calculate CSA: CSA=13×462=154CSA = \frac{1}{3} \times 462 = 154 cm2cm^2.
  3. Find the area of the two bases: 2×Base Area=TSA−CSA=462−154=3082 \times \text{Base Area} = TSA - CSA = 462 - 154 = 308 cm2cm^2.
  4. Find the radius rr: 2πr2=308  ⟹  2×227×r2=308  ⟹  r2=308×744=7×7=492\pi r^2 = 308 \implies 2 \times \frac{22}{7} \times r^2 = 308 \implies r^2 = \frac{308 \times 7}{44} = 7 \times 7 = 49. So, r=7r = 7 cm.
  5. Find the height hh using CSA: 2πrh=154  ⟹  2×227×7×h=154  ⟹  44h=154  ⟹  h=15444=3.52\pi rh = 154 \implies 2 \times \frac{22}{7} \times 7 \times h = 154 \implies 44h = 154 \implies h = \frac{154}{44} = 3.5 cm. Final Answer: r=7r = 7 cm, h=3.5h = 3.5 cm.

Explanation:

The problem provides a relationship between TSA and CSA. We used this to find the numerical value of CSA and the area of the bases. From the base area, we derived the radius, and then used the radius in the CSA formula to find the height.