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Mensuration: Surface Area and Volume - Compute surface area and volume of spheres and hemispheres in applications

Grade 9CBSE

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

🔑Concepts

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A sphere is a perfectly round geometrical object in three-dimensional space that is the surface of a completely round ball. The surface area of a sphere depends solely on its radius rr. It is exactly four times the area of a circle with the same radius: S=4πr2S = 4 \pi r^2.

Diagram of a sphere showing radius r and a central cross-section
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A hemisphere is exactly half of a sphere. When a sphere is cut through its center, two hemispheres are formed. Unlike a sphere, a hemisphere has two types of surface areas: the Curved Surface Area (CSA) which is 2πr22 \pi r^2 and the Total Surface Area (TSA), which includes the flat circular base (3πr23 \pi r^2).

Diagram of a hemisphere showing the curved top and the circular base
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The volume of a sphere represents the amount of space occupied by it. It is calculated as V=43πr3V = \frac{4}{3} \pi r^3. For a hemisphere, the volume is exactly half of the sphere's volume, given by V=23πr3V = \frac{2}{3} \pi r^3.

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In practical applications, if you are asked to find the cost of painting a hemispherical dome, you calculate the CSA (2πr22 \pi r^2). If you are asked to find the cost of painting a solid hemispherical bowl (including the base), you use TSA (3πr23 \pi r^2).

📐Formulae

SurfaceAreaofaSphere=4πr2Surface Area of a Sphere = 4 \pi r^2

CurvedSurfaceArea(CSA)ofaHemisphere=2πr2Curved Surface Area (CSA) of a Hemisphere = 2 \pi r^2

TotalSurfaceArea(TSA)ofaHemisphere=3πr2Total Surface Area (TSA) of a Hemisphere = 3 \pi r^2

AreaofthecircularbaseofaHemisphere=πr2Area of the circular base of a Hemisphere = \pi r^2

Radius(r)=Diameter(d)2Radius (r) = \frac{Diameter (d)}{2}

💡Examples

Problem 1:

Find the surface area of a sphere of radius 7cm7 cm. (Use π=227\pi = \frac{22}{7})

Solution:

  1. Given: Radius (rr) = 7cm7 cm.
  2. Formula for Surface Area of a Sphere = 4πr24 \pi r^2.
  3. Substitute the values: SA=4×227×(7)2SA = 4 \times \frac{22}{7} \times (7)^2.
  4. SA=4×227×7×7SA = 4 \times \frac{22}{7} \times 7 \times 7.
  5. SA=4×22×7=616cm2SA = 4 \times 22 \times 7 = 616 cm^2.

Explanation:

To find the surface area of a sphere, identify the radius and substitute it into the formula 4πr24 \pi r^2. Since the radius is a multiple of 7, using π=227\pi = \frac{22}{7} makes the calculation simpler by allowing for cancellation.

Problem 2:

Find the Total Surface Area (TSA) of a hemisphere of radius 10cm10 cm. (Use π=3.14\pi = 3.14)

Solution:

  1. Given: Radius (rr) = 10cm10 cm.
  2. Formula for TSA of a Hemisphere = 3πr23 \pi r^2.
  3. Substitute the values: TSA=3×3.14×(10)2TSA = 3 \times 3.14 \times (10)^2.
  4. TSA=3×3.14×100TSA = 3 \times 3.14 \times 100.
  5. TSA=3×314=942cm2TSA = 3 \times 314 = 942 cm^2.

Explanation:

For a solid hemisphere, we must account for both the curved surface (2πr22 \pi r^2) and the flat circular base (πr2\pi r^2), which totals 3πr23 \pi r^2. Using 3.143.14 for π\pi is convenient here because the radius squared (100100) easily clears the decimals.

Problem 3:

A hemispherical bowl made of brass has an inner diameter of 10.5cm10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100cm2100 cm^2.

Hemispherical bowl with internal diameter 10.5 cm

Solution:

Inner radius r=10.52=5.25cmr = \frac{10.5}{2} = 5.25 cm. Inner surface area of the bowl (CSA) =2πr2= 2 \pi r^2 =2×227×5.25×5.25cm2= 2 \times \frac{22}{7} \times 5.25 \times 5.25 cm^2 =2×227×214×214cm2= 2 \times \frac{22}{7} \times \frac{21}{4} \times \frac{21}{4} cm^2 =11×3×214=173.25cm2= \frac{11 \times 3 \times 21}{4} = 173.25 cm^2 Cost of tin-plating 100cm2=Rs16100 cm^2 = Rs 16 Cost of tin-plating 1cm2=Rs161001 cm^2 = Rs \frac{16}{100} Total cost =173.25×16100=Rs27.72= 173.25 \times \frac{16}{100} = Rs 27.72

Explanation:

Since the tin-plating is done on the inside surface of a bowl, we calculate the Curved Surface Area (CSA). The diameter is converted to radius, then the area is found and multiplied by the rate per unit area.

Problem 4:

The radius of a spherical balloon increases from 7cm7 cm to 14cm14 cm as air is being pumped into it. Find the ratio of the surface areas of the balloon in the two cases.

Two spheres representing a balloon expanding from radius 7 to radius 14

Solution:

Let r1=7cmr_1 = 7 cm and r2=14cmr_2 = 14 cm. Surface area S1=4πr12S_1 = 4 \pi r_1^2 Surface area S2=4πr22S_2 = 4 \pi r_2^2 Ratio =4πr124πr22=r12r22= \frac{4 \pi r_1^2}{4 \pi r_2^2} = \frac{r_1^2}{r_2^2} =7×714×14=12×2=14= \frac{7 \times 7}{14 \times 14} = \frac{1}{2 \times 2} = \frac{1}{4} Therefore, the ratio is 1:41:4.

Explanation:

The surface area of a sphere is proportional to the square of its radius. When the radius doubles (77 to 1414), the surface area increases by a factor of 22=42^2 = 4.