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Surface Areas and Volumes
Introduction
Surface Area of a Combination of Solids
Practice: Surface Area of Combined Solids
Volume of a Combination of Solids
Practice: Volume of Combined Solids
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Surface Areas and VolumesSurface Area of a Combination of Solids

Reading time: ~30 min

Back at the dairy, Riya's uncle has bought rust-proof paint for the tank. Paint is sold by the square metre, so he needs the tank's surface area — and no formula from Class IX fits a tank with rounded ends. Whenever a new problem looks hard, try breaking it into smaller problems you have already solved. Taken apart, the tank is one cylinder and two hemispheres.

Fig. 12.4 is live. Before the pieces are glued back together, find what will disappear: tap every flat face that will end up hidden inside a join.

Once the pieces are joined, the four flat faces are pressed against each other inside the tank, where no paint can reach. All that is left on the outside is the curved surface of each hemisphere and the curved surface of the cylinder. So the total surface area (TSA) of the tank is the of three curved surface areas (CSA):

TSA of the tank = CSA of one hemisphere + CSA of the cylinder + CSA of the other hemisphere

Tip

Never add the total surface areas of the pieces. When two solids are glued, the faces that touch vanish inside the join. Add only the surfaces that stay on the outside — here, the three curved ones.

A Toy That Spins Smoothly

At the Diwali mela, Aman's grandfather carves wooden toys. For a round-bottomed top, he takes a hemisphere and a cone and joins them flat face to flat face. But not just any cone will do.

Fig. 12.5 is live. The hemisphere has radius 3.5 cm. Drag the radius of the cone until the toy is smooth all the way round.

Radius of the cone: ${rc} cm

So the base radius of the cone must be the radius of the hemisphere. Only then do the two flat faces cover each other exactly and disappear into the join.

Now the toy has to be painted. Paint covers only what shows, so we need the toy's surface area, made of the curved surface of the hemisphere and the curved surface of the cone:

surface area of the toy = CSA of hemisphere + CSA of cone

Note

Every example ahead is built from these Class IX results. For a part of radius r:

  • Cylinder of height h: curved surface 2\pi r h
  • Hemisphere: curved surface 2\pi r^2, half of the sphere's 4\pi r^2
  • Cone: curved surface \pi r l, where the slant height is l = \sqrt{r^2 + h^2}

Rasheed's Lattu

Example 1 : Rasheed got a lattu (a playing top) for his birthday, but it came without any colour, so he decided to colour it with his crayons. The top is a cone surmounted by a hemisphere (Fig. 12.6). The whole top is 5 cm tall, and its diameter is 3.5 cm. How much area does he have to colour? (Take \pi = \frac{22}{7})

Fig. 12.6 — Rasheed's top: a hemisphere on a cone of height h and slant height l.

Solution : The top has exactly the shape of the toy in Fig. 12.5, so we can use the result found there:

TSA of the top = CSA of hemisphere + CSA of cone

Both parts have radius r = \frac{3.5}{2} cm = 1.75 cm. The curved surface of the hemisphere is

2\pi r^2 = \left(2 \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2}\right) \text{ cm}^2

The height h of the cone is the height of the whole top minus the height of the hemisphere, which is its radius:

h = \left(5 - \frac{3.5}{2}\right) \text{ cm} = \text{ cm}

So the slant height l of the cone is

l = \sqrt{r^2 + h^2}

= \sqrt{(1.75)^2 + (3.25)^2} \text{ cm} \approx \text{ cm}

and the curved surface of the cone is

\pi r l = \left(\frac{22}{7} \times \frac{3.5}{2} \times 3.7\right) \text{ cm}^2

Add the two curved surfaces, taking out the common factor \frac{22}{7} \times \frac{3.5}{2} = \frac{11}{2}:

\text{TSA} = \frac{11}{2}\,(3.5 + 3.7) \text{ cm}^2

= \frac{11}{2} \times 7.2 \text{ cm}^2 \approx \text{ cm}^2

Tip

The area to colour is not the sum of the total surface areas of the cone and the hemisphere: their two flat faces are glued together and never get coloured.

The Decorative Block

Example 2 : The decorative block in Fig. 12.7 is made of two solids: a cube of edge 5 cm, with a hemisphere of diameter 4.2 cm fixed on its top. Find the total surface area of the block. (Take \pi = \frac{22}{7})

Fig. 12.7 — the decorative block. The circle under the hemisphere is hidden.

Predict first. Compared with the plain cube, the block has:

less surface area
the same surface area
more surface area

Right. The hemisphere hides a circle of area \pi r^2 on the top face, but adds its own curved surface of 2\pi r^2, twice as much. Let us measure the difference.

Solution : The total surface area of the cube is 6 \times (\text{edge})^2 = 6 \times 5 \times 5 \text{ cm}^2 = 150 \text{ cm}^2.

The circle under the hemisphere is covered, so it is no longer part of the surface. So

surface area of the block = TSA of cube − of hemisphere + CSA of hemisphere

= 150 - \pi r^2 + 2\pi r^2

= (150 + \pi r^2) \text{ cm}^2

= 150 \text{ cm}^2 + \left(\frac{22}{7} \times \frac{4.2}{2} \times \frac{4.2}{2}\right) \text{ cm}^2

= (150 + 13.86) \text{ cm}^2 = \text{ cm}^2

Painting the Rocket

Example 3 : A wooden toy rocket is a cone mounted on a cylinder (Fig. 12.8). The whole rocket is 26 cm tall, and the cone is 6 cm tall. The base of the cone has diameter 5 cm, while the base of the cylinder has diameter 3 cm. The conical part is to be painted orange and the cylindrical part blue. Find the area painted with each colour. (Take \pi = 3.14)

Fig. 12.8 is live. Before you calculate, paint the rocket: pick a paint, then tap each surface that gets painted. Two of the surfaces can only be seen from below.

Solution : Call the cone's radius r, its slant height l and its height h, and the cylinder's radius r' and height h'. Then r = 2.5 cm, h = 6 cm, r' = 1.5 cm, h' = 26 - 6 = 20 cm, and

l = \sqrt{r^2 + h^2} = \sqrt{2.5^2 + 6^2} \text{ cm} = \text{ cm}

The cone's base rests on the top of the cylinder, but the cone's base is than the cylinder's. So a ring of the cone's base is left uncovered — the ring you painted from below.

So the area painted orange = CSA of the cone + base area of the cone − base area of the cylinder

= \pi r l + \pi r^2 - \pi (r')^2

= \pi\,[(2.5 \times 6.5) + (2.5)^2 - (1.5)^2] \text{ cm}^2

= 3.14 \times 20.25 \text{ cm}^2 = \text{ cm}^2

The area painted blue = CSA of the cylinder + the area of base of the cylinder (its top is hidden under the cone)

= 2\pi r' h' + \pi (r')^2 = \pi r'(2h' + r')

= (3.14 \times 1.5)(2 \times 20 + 1.5) \text{ cm}^2

= 4.71 \times 41.5 \text{ cm}^2 = \text{ cm}^2

Mayank's Bird-Bath

Example 4 : Mayank made a bird-bath for his garden: a cylinder with a hemispherical hollow at its top end (Fig. 12.9). The cylinder is 1.45 m tall and its radius is 30 cm. Find the total surface area of the bird-bath. (Take \pi = \frac{22}{7})

Solution : Let h be the height of the cylinder and r the common radius of the cylinder and the hollow. The bird-bath stands on its base, so its surface is the outside of the cylinder together with the inside of the hemispherical hollow — the flat top has been scooped away.

Fig. 12.9 — the bird-bath.

TSA of the bird-bath = CSA of cylinder + CSA of hemisphere

= 2\pi r h + 2\pi r^2 = 2\pi r(h + r)

= 2 \times \frac{22}{7} \times 30 \times (145 + 30) \text{ cm}^2

= 33000 \text{ cm}^2 = \text{ m}^2

Speed Round: Which Surfaces Show?

For each solid, picture which faces are hidden in the joins and which stay outside, then pick its total surface area.

Get five right to move on.

Sina