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Surface Areas and Volumes - Surface Area of a Combination of Solids

Grade 10CBSE

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

🔑Concepts

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Combination of Solids: A combined solid is formed by joining two or more basic three-dimensional shapes like cylinders, cones, spheres, or cubes. Visually, this involves attaching these shapes at their common faces, such as a hemisphere placed on the flat face of a cylinder to create a capsule shape.

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Total Surface Area (TSA) Calculation: The TSA of a combined solid is the sum of the curved surface areas (CSA) of the individual parts that remain visible after they are joined. Crucially, the surfaces that are hidden or overlapped at the point of contact are not included in the total surface area calculation.

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Cone and Hemisphere Combination: In objects like a spinning top or an ice-cream cone, a cone is mounted on a hemisphere. The total surface area consists of the CSA of the cone (the sloping outer side) and the CSA of the hemisphere (the rounded bottom). The flat circular base where they meet is inside the solid and thus excluded.

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Cylinder and Hemisphere Combination: A common example is a storage tank or a capsule where hemispheres are attached to the ends of a cylinder. The surface area is calculated as the sum of the CSA of the cylinder and the CSA of the two hemispheres. Visually, the flat circular ends of the cylinder disappear into the join.

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Hollowing Out (Cavities): When a shape like a cone or hemisphere is 'scooped out' from a larger solid like a cube or cylinder, the surface area actually increases. This is because the inner walls of the cavity (the inner CSA of the removed shape) are now exposed to the outside. Visually, this looks like a solid block with a hole or depression inside it.

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Slant Height of Cones: For combinations involving cones, the slant height ll is required for the surface area formula πrl\pi rl. Visually, ll is the hypotenuse of a right-angled triangle formed by the vertical height hh and the radius rr of the cone's base, calculated using the Pythagorean theorem.

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Common Radius: In most Grade 10 problems, when two solids are joined, they share a common radius rr. This ensures that the edges of the two shapes align perfectly at the junction, such as a cylinder and cone having bases of identical size.

📐Formulae

CSA of Cylinder = 2πrh2\pi rh

CSA of Cone = πrl\pi rl, where l=r2+h2l = \sqrt{r^2 + h^2}

CSA of Hemisphere = 2πr22\pi r^2

Surface Area of Sphere = 4πr24\pi r^2

TSA of Cube = 6a26a^2 (where aa is the edge length)

TSA of Cuboid = 2(lb+bh+hl)2(lb + bh + hl)

Area of Circle (Base) = πr2\pi r^2

💡Examples

Problem 1:

A toy is in the form of a cone of radius 3.53.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.515.5 cm. Find the total surface area of the toy.

Solution:

  1. Radius of cone and hemisphere (rr) = 3.53.5 cm.
  2. Total height of the toy = 15.515.5 cm.
  3. Height of the conical part (hh) = Total height - Radius of hemisphere = 15.5−3.5=1215.5 - 3.5 = 12 cm.
  4. Slant height of the cone (ll): l=r2+h2=(3.5)2+122=12.25+144=156.25=12.5 cml = \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\text{ cm}
  5. TSA of toy = CSA of cone + CSA of hemisphere
  6. TSA = πrl+2πr2=πr(l+2r)\pi rl + 2\pi r^2 = \pi r(l + 2r)
  7. TSA = 227×3.5×(12.5+2×3.5)\frac{22}{7} \times 3.5 \times (12.5 + 2 \times 3.5)
  8. TSA = 11×(12.5+7)=11×19.5=214.5 cm211 \times (12.5 + 7) = 11 \times 19.5 = 214.5\text{ cm}^2.

Explanation:

To solve this, we first identify that the flat circular faces of the cone and hemisphere are joined and hidden. Thus, we only add the curved surface areas. We must subtract the hemisphere's radius from the total height to find the cone's vertical height, then use Pythagoras to find the slant height.

Problem 2:

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 1414 mm and the diameter of the capsule is 55 mm. Find its surface area.

Solution:

  1. Diameter = 55 mm, so Radius (rr) = 2.52.5 mm.
  2. Total length = 1414 mm.
  3. Length of cylindrical part (hh) = Total length - (Radius of left hemisphere + Radius of right hemisphere) = 14−(2.5+2.5)=914 - (2.5 + 2.5) = 9 mm.
  4. Surface area of capsule = CSA of cylinder + 2×2 \times CSA of hemisphere
  5. SA = 2πrh+2(2πr2)=2πrh+4πr2=2πr(h+2r)2\pi rh + 2(2\pi r^2) = 2\pi rh + 4\pi r^2 = 2\pi r(h + 2r)
  6. SA = 2×227×2.5×(9+2×2.5)2 \times \frac{22}{7} \times 2.5 \times (9 + 2 \times 2.5)
  7. SA = 2×227×2.5×142 \times \frac{22}{7} \times 2.5 \times 14
  8. SA = 22×2.5×4=220 mm222 \times 2.5 \times 4 = 220\text{ mm}^2.

Explanation:

The capsule is a combination of one cylinder and two hemispheres. Since the hemispheres are at the ends, the total surface area is the sum of the curved surface area of the cylinder and the curved surface areas of both hemispheres. We calculate the cylinder's height by subtracting the radii of the two ends from the total length.