Surface Area of a Combination of Solids, Class 10

The for­mu­las you need#

Recap table#

In the table, $a$ is the edge of a cube; $l$, $b$, $h$ are the length, breadth and height of a cuboid; $r$ is a radius, $h$ a height and $l$ the slant height of a cone.

For a cone the slant height comes from Pythago­ras' the­o­rem, because the height, the radius and the slant height form a right-angled tri­an­gle:

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

Unless a ques­tion says oth­er­wise, take $\displaystyle \pi = \frac{22}{7}$. When the radius is a mul­ti­ple of 7 (or of 3.5), this value makes the arith­metic come out neatly.

Curved sur­face or total sur­face?#

The total sur­face area of a hemi­sphere, $3\pi r^2$, is its curved dome $2\pi r^2$ plus its flat cir­cu­lar face $\pi r^2$. When solids are joined, the flat faces are usu­ally the ones that dis­ap­pear. That is why com­bined-solid prob­lems almost always use the curved sur­face areas, plus only those flat faces that are still exposed.

The key idea: hid­den faces#

State­ment#

Sur­face area of a com­bined solid = sum of the areas of all the faces that are still on the out­side after join­ing.

It is not the sum of the total sur­face areas of the sep­a­rate pieces. When two solids are stuck together along a flat face, that face is now inside the object: nobody can paint it or touch it.

Why this is true#

Sur­face area mea­sures the outer skin of an object. Pic­ture wrap­ping the fin­ished object in paper. The paper touches only the exposed faces. The cir­cle where a cone meets a hemi­sphere is sealed on both sides, so no paper reaches it. So we add the exposed parts and leave out every face that lies along a join. If a joined face is only partly cov­ered, as when a small hemi­sphere sits on a large cube face, the uncov­ered part of that face is still exposed and must be counted.

A method in steps#

1. Draw the object and name each basic solid in it.
2. Write down every dimen­sion you need. Find any miss­ing one, such as a slant height $l = \sqrt{h^2 + r^2}$ or a cylin­der length (total length minus the ends).
3. For each piece, decide which of its faces are still exposed: the curved part, a flat base, or only part of a flat face.
4. Write the sur­face area as a sum (and dif­fer­ence) of those exposed parts. Fac­torise when you can, for exam­ple $2\pi r h + 2\pi r^2 = 2\pi r(h + r)$.
5. Sub­sti­tute and sim­plify, then give the unit ($\text{cm}^2$ or $\text{m}^2$).
6. If the ques­tion asks for a cost, mul­ti­ply the area by the rate per unit area.

Worked exam­ples#

Exam­ple 1: A toy made of a cone on a hemi­sphere#

A wooden toy is a cone of radius 3.5 cm mounted on a hemi­sphere of the same radius. The total height of the toy is 15.5 cm. Find its total sur­face area. If paint­ing costs ₹2 per $\text{cm}^2$, find the cost of paint­ing it.

Dimen­sions. The hemi­sphere's height equals its radius, so the height of the cone is $15.5 - 3.5 = 12$ cm. Then

$$l = \sqrt{12^2 + 3.5^2} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5 \text{ cm}.$$

Exposed faces. The curved sur­face of the cone and the curved sur­face of the hemi­sphere. The shared cir­cle is hid­den.

$$\displaystyle \begin{aligned} \text{CSA of hemisphere} &= 2\pi r^2 = 2 \times \frac{22}{7} \times 3.5 \times 3.5 = 77 \text{ cm}^2 \\ \text{CSA of cone} &= \pi r l = \frac{22}{7} \times 3.5 \times 12.5 = 137.5 \text{ cm}^2 \\ \text{Total surface area} &= 77 + 137.5 = 214.5 \text{ cm}^2 \end{aligned}$$

Cost of paint­ing $= 214.5 \times 2 =$ ₹429.

Answer: 214.5 $\text{cm}^2$; ₹429.

Exam­ple 2: A cap­sule#

A cap­sule is a cylin­der with a hemi­sphere at each end. Its total length is 20 mm and its diam­e­ter is 7 mm. Find the area of its outer sur­face.

Dimen­sions. $r = 3.5$ mm. Each hemi­sphere adds 3.5 mm to the length, so the cylin­der is $20 - 2 \times 3.5 = 13$ mm long.

Exposed faces. The curved sur­face of the cylin­der and the two hemi­spher­i­cal domes. Both flat ends of the cylin­der are hid­den.

$$\displaystyle \begin{aligned} S &= 2\pi r h + 2\pi r^2 + 2\pi r^2 = 2\pi r(h + 2r) \\ &= 2 \times \frac{22}{7} \times 3.5 \times (13 + 7) \\ &= 22 \times 20 = 440 \text{ mm}^2 \end{aligned}$$

Notice that $h + 2r$ is just the total length, 20 mm. So the sur­face of a cap­sule equals $2\pi r \times (\text{total length})$, the same as an open cylin­der of that full length.

Answer: 440 $\text{mm}^2$.

Exam­ple 3: A cube with a hemi­sphere on top#

A dec­o­ra­tive block is a cube of edge 10 cm. A hemi­sphere of diam­e­ter 7 cm is fixed on the cen­tre of its top face. Find the total sur­face area of the block.

Exposed faces. All six faces of the cube except the cir­cle of radius 3.5 cm under the hemi­sphere, plus the dome of the hemi­sphere.

$$\displaystyle \begin{aligned} S &= 6a^2 - \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2 \\ &= 6 \times 10^2 + \frac{22}{7} \times 3.5 \times 3.5 \\ &= 600 + 38.5 = 638.5 \text{ cm}^2 \end{aligned}$$

Why this works: the top face is only partly cov­ered, so we remove just the cir­cle, not the whole face.

Answer: 638.5 $\text{cm}^2$.

Exam­ple 4: Can­vas for a tent, with cost#

A tent has a cylin­dri­cal lower part of radius 7 m and height 4 m, with a con­i­cal top of the same radius and slant height 10 m. Find the area of can­vas needed (there is no can­vas floor). Find the cost of the can­vas at ₹120 per $\text{m}^2$.

Exposed faces. The curved wall of the cylin­der and the curved sur­face of the cone. The cir­cle where they meet is hid­den, and the floor has no can­vas.

$$\displaystyle \begin{aligned} \text{Canvas} &= 2\pi r h + \pi r l \\ &= 2 \times \frac{22}{7} \times 7 \times 4 + \frac{22}{7} \times 7 \times 10 \\ &= 176 + 220 = 396 \text{ m}^2 \end{aligned}$$

Cost $= 396 \times 120 =$ ₹47,520.

Answer: 396 $\text{m}^2$ of can­vas, cost­ing ₹47,520.

Exam­ple 5: A cylin­der with a con­i­cal hol­low#

A solid cylin­der of radius 3 cm and height 10 cm has a cone of the same radius and height 4 cm hol­lowed out from its top face. Find the total sur­face area of the remain­ing solid, first in terms of $\pi$ and then using $\displaystyle \pi = \frac{22}{7}$.

Dimen­sions. Slant height of the hol­low: $l = \sqrt{4^2 + 3^2} = 5$ cm.

Exposed faces. The curved sur­face of the cylin­der and its bot­tom cir­cle are unchanged. The top cir­cle has been removed com­pletely (the hol­low has the same radius). In its place is the inner curved sur­face of the cone, which is now on the out­side.

$$\displaystyle \begin{aligned} S &= 2\pi r h + \pi r^2 + \pi r l \\ &= \pi(2 \times 3 \times 10 + 3^2 + 3 \times 5) \\ &= \pi(60 + 9 + 15) = 84\pi \text{ cm}^2 \\ &= 84 \times \frac{22}{7} = 264 \text{ cm}^2 \end{aligned}$$

Why this works: hol­low­ing out removes a flat face but cre­ates a new curved face, so the sur­face area can go up even though mate­r­ial is removed.

Answer: $84\pi$ $\text{cm}^2$, that is 264 $\text{cm}^2$.

Com­mon mis­takes#

• Adding total sur­face areas of the pieces instead of only the exposed faces, so the hid­den cir­cles get counted twice.
• Using the height of a cone where the slant height is needed in $\pi r l$.
• For­get­ting that a hemi­sphere's height is its radius when work­ing out the height of the cone or the length of the cylin­der.
• Using the diam­e­ter as the radius. Always halve the diam­e­ter first.
• Remov­ing the whole top face of a cube when a hemi­sphere cov­ers only a cir­cle of it.
• Leav­ing out the new inner sur­face cre­ated when a solid is hol­lowed out.
• Mix­ing units, such as a radius in metres and a height in cen­time­tres, or giv­ing an area in $\text{cm}^3$.

Try these#

1. A solid is a cylin­der of radius 7 cm and height 10 cm with a hemi­sphere of the same radius on top. Its flat base is exposed. Find its total sur­face area. Answer: 902 $\text{cm}^2$.
2. Two cubes of edge 5 cm are joined face to face. Find the sur­face area of the cuboid formed. Answer: 250 $\text{cm}^2$.
3. A cone of radius 6 cm and height 8 cm stands on top of a cylin­der of radius 6 cm and height 10 cm. Find the total sur­face area in terms of $\pi$. Answer: $216\pi$ $\text{cm}^2$.
4. A hemi­sphere of diam­e­ter 14 cm is fixed on one face of a cube of edge 14 cm. Find the total sur­face area. Answer: 1330 $\text{cm}^2$.
5. For the toy in Exam­ple 1, what would paint­ing cost at ₹2 per $\text{cm}^2$ if only the cone were painted? Answer: ₹275.

Key terms#

Com­bined solid

A solid made by join­ing two or more basic solids such as cubes, cylin­ders, cones and hemi­spheres.

Curved sur­face area (CSA)

The area of the curved part of a solid only, with­out its flat faces.

Total sur­face area (TSA)

The area of every face of a sin­gle solid, curved and flat together.

Hid­den face

A face, or part of a face, that lies along a join and is no longer on the out­side.

Slant height

The dis­tance from the ver­tex of a cone to any point on the edge of its base, $l = \sqrt{h^2 + r^2}$.

Hemi­sphere

Half of a sphere, cut through its cen­tre; its curved sur­face is $2\pi r^2$.

Lat­eral sur­face area

The area of the side faces of a solid, exclud­ing the top and bot­tom.

Com­mon ques­tions#

Is the sur­face area of a com­bined solid the sum of the sep­a­rate sur­face areas?#

No. Faces that are joined together are hid­den, so you add only the parts that remain on the out­side. Sum­ming total sur­face areas counts the joined faces twice.

When do I use $2\pi r^2$ and when $3\pi r^2$ for a hemi­sphere?#

Use $2\pi r^2$ when the flat face is joined to another solid or is open. Use $3\pi r^2$ only for a solid hemi­sphere stand­ing alone with its flat face exposed.

Can hol­low­ing out a solid increase its sur­face area?#

Yes. A flat face is removed, but a new inner curved sur­face appears. In Exam­ple 5 the solid cylin­der had $78\pi$ $\text{cm}^2$ of sur­face before and $84\pi$ $\text{cm}^2$ after.

Should I use $\displaystyle \pi = \frac{22}{7}$ or $\pi = 3.14$?#

Use what­ever the ques­tion states. If it says noth­ing, $\displaystyle \frac{22}{7}$ is usual when the radius is a mul­ti­ple of 7 or 3.5; oth­er­wise an answer in terms of $\pi$ is also accept­able unless a num­ber is asked for.

How do I han­dle cost ques­tions?#

Find the area to be cov­ered first, check that its unit matches the rate (for exam­ple, per $\text{m}^2$), then mul­ti­ply the area by the rate.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
3. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.