Exercise 10.2 — Surface Area of Combined Solids

Surface areas of combination of solids.

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Exercise 10.2 — Surface Areas of Combinations of Solids

Exercise 10.2 is from Chapter 10, Mensuration, of Class 10 Mathematics — part of the CBSE, Telangana, and Andhra Pradesh board syllabi. This exercise focuses on finding the total surface area (TSA) of solids formed by combining two or more basic 3D shapes such as cones, cylinders, hemispheres, and cubes.

In real life, many objects — toys, capsules, storage tanks, wooden articles — are not simple cubes or spheres. They are combinations of two or more shapes joined together. The key idea is: when two solids are joined, the circular faces at the joint are no longer exposed, so we only add the curved surface areas (CSA) of the visible outer parts.

Cone + Hemisphere Cylinder + Cone + Hemisphere Capsule Shape Two Cubes → Cuboid Storage Tank Volume Ratio: Sphere : Cylinder : Cone

💡 Core Rule: Total Surface Area of a combination = Sum of the Curved Surface Areas (CSA) of each individual solid that is visible from outside. The flat circular joining faces are NOT counted — they are hidden inside.

Essential Formulas You Must Know

Before solving the problems, revise these six formulas. Every question in Exercise 10.2 uses one or more of them.

Cone (CSA)

πrl
where l = √(r² + h²)

Hemisphere (CSA)

2πr²

Cylinder (CSA)

2πrh

Cuboid (TSA)

2(lh + bh + lb)

Sphere (Volume)

(4/3)πr³

Cone (Volume)

(1/3)πr²h

📌 Slant Height Reminder: For a cone, always calculate slant height l first using the Pythagorean theorem: l = √(r² + h²). Do not confuse height h with slant height l — this is the most common mistake.

Understanding Combinations of Solids

When two solids are placed together (one mounted on top of the other, or one embedded into the other), the outer visible surface is what we measure. The diagram below shows the two most common combinations tested in board exams:

Left: Toy (cone on hemisphere) — Right: Capsule / Tank (cylinder with two hemispheres)

For the toy (cone + hemisphere): Total surface area = CSA of Cone + CSA of Hemisphere. The flat circular base of the cone sits exactly on the flat face of the hemisphere — both are hidden, so neither is counted.

For the capsule/tank (cylinder + 2 hemispheres): Total surface area = CSA of Cylinder + 2 × CSA of Hemisphere. The two circular ends of the cylinder are each covered by a hemisphere, so again the flat circles vanish from the calculation.

Exercise 10.2 — All 8 Problems Solved Step by Step

Problem 1

Toy: Cone mounted on a hemisphere (diameter = 6 cm, height of cone = 4 cm). Find the surface area of the toy.

Given Information

Diameter of cone base = 6 cm → radius r = 3 cm
Height of cone h = 4 cm
Hemisphere has the same radius r = 3 cm (base touches base)

Step 1 — Find the Slant Height of the Cone

l = √(r² + h²) = √(3² + 4²) = √(9 + 16) = √25 ∴ l = 5 cm

Step 2 — CSA of Cone

CSA of Cone = πrl = 3.14 × 3 × 5 = 47.1 cm²

Step 3 — CSA of Hemisphere

CSA of Hemisphere = 2πr² = 2 × 3.14 × 3 × 3 = 56.52 cm²

Step 4 — Total Surface Area of the Toy

TSA = CSA of Cone + CSA of Hemisphere = 47.1 + 56.52 ∴ Surface Area of Toy = 103.62 cm²

✅ Answer: 103.62 cm²

Problem 2

Solid = Right circular cylinder + hemisphere at one end + cone at the other end. Radius = 8 cm, cylinder height = 10 cm, cone height = 6 cm. Find total surface area.

Given Information

Common radius r = 8 cm
Height of cone h = 6 cm
Height of cylinder h = 10 cm

Cone l

√(8²+6²)

= √100 = 10 cm

CSA Cone

251.2

cm²

CSA Cylinder

502.4

cm²

CSA Hemisphere

401.92

cm²

Slant of Cone: l = √(8² + 6²) = √100 = 10 cm CSA of Cone = πrl = 3.14 × 8 × 10 = 251.2 cm² CSA of Cylinder = 2πrh = 2 × 3.14 × 8 × 10 = 502.4 cm² CSA of Hemisphere = 2πr² = 2 × 3.14 × 8 × 8 = 401.92 cm² ───────────────────────────────────────────── TSA = 251.2 + 502.4 + 401.92 ∴ Total Surface Area = 1155.52 cm²

✅ Answer: 1155.52 cm²

Problem 3

Medicine Capsule = Cylinder + 2 Hemispheres at each end. Total length = 14 mm, width = 5 mm. Find surface area.

Key Insight: Finding the Cylinder Height

The total length includes both hemispheres (each of radius 2.5 mm) plus the cylinder height.

Width = 5 mm → radius r = 5/2 = 2.5 mm Height of cylinder h = total length − 2 × r h = 14 − 2 × 2.5 = 14 − 5 = 9 mm

Surface Area Calculation

SA = 2πrh + 2 × 2πr² = 2πr(h + 2r)

SA = 2 × 3.14 × 2.5 × (9 + 5) = 5 × 3.14 × 14 ∴ Surface Area of Capsule = 219.8 mm²

✅ Answer: 219.8 mm²

💡 Elegant simplification: 2πrh + 2×2πr² = 2πr(h + 2r). Always try to factor out 2πr to simplify your calculation. This saves time in board exams.

Problem 4

Two cubes, each of volume 64 cm³, are joined end to end. Find the surface area of the resulting cuboid.

Step 1 — Find the Side of Each Cube

Volume of cube = a³ = 64 cm³ a = ∛64 = 4 cm

Step 2 — Dimensions of the Cuboid Formed

Dimension	Value	Reason
Length (l)	8 cm	4 + 4 (two cubes joined)
Breadth (b)	4 cm	Same as cube side
Height (h)	4 cm	Same as cube side

Step 3 — Surface Area of the Cuboid

TSA = 2(lh + bh + lb) = 2(8×4 + 4×4 + 8×4) = 2(32+16+32) = 2×80 = 160 cm²

✅ Answer: 160 cm²

⚠️ Common mistake: Students sometimes calculate 2 × TSA of single cube (2 × 96 = 192 cm²). That is wrong — the two touching square faces (each 4×4 = 16 cm²) are no longer exposed, so we subtract 2 × 16 = 32 cm² → 192 − 32 = 160 cm². The cuboid formula directly gives the correct answer.

Problem 5

Storage tank = cylinder + hemisphere at each end. External diameter = 1.4 m, cylinder length = 8 m. Find cost of painting at Rs.20 per m².

Given Information

Diameter = 1.4 m → radius r = 0.7 m
Length of cylinder h = 8 m
Rate of painting = Rs. 20 per m²

Surface Area Using the Elegant Formula

SA = 2πrh + 2 × 2πr² = 2πr(h + 2r)

SA = 2 × (22/7) × 0.7 × (8 + 1.4) = 2 × 22 × 0.1 × 9.4 = 2 × 22 × 0.94 Surface Area = 41.36 m²

Cost of Painting

Cost = Surface Area × Rate = 41.36 × 20 ∴ Cost of Painting = Rs. 827.20

✅ Answer: Rs. 827.20

📌 Why use 22/7 here? Because radius = 0.7 m = 7/10, and 22/7 × 7/10 = 22/10 = 2.2. This makes arithmetic cleaner. Use 22/7 when radius has 7 as a factor; use 3.14 otherwise.

Problem 6

A sphere, a cylinder, and a cone have the same radius and same height (height = diameter of sphere). Find the ratio of their volumes.

Setting Up the Volumes

Since height = diameter of sphere = 2r for both the cylinder and cone:

Solid	Formula	Substitution (h = 2r)	Volume
Sphere	(4/3)πr³	—	(4/3)πr³
Cylinder	πr²h	πr²(2r)	2πr³
Cone	(1/3)πr²h	(1/3)πr²(2r)	(2/3)πr³

Finding the Ratio

Sphere : Cylinder : Cone = (4/3)πr³ : 2πr³ : (2/3)πr³ = (4/3) : 2 : (2/3) ÷ πr³ throughout = 4 : 6 : 2 × 3 to clear fractions ∴ Sphere : Cylinder : Cone = 2 : 3 : 1

✅ Answer: 2 : 3 : 1

✅ Classic result — must memorise for boards: When sphere, cylinder, and cone share the same radius and height (height = diameter), their volumes are always in the ratio 2 : 3 : 1. This is a frequently asked 1-mark question in Telangana, AP, and CBSE exams.

Problem 7

A hemisphere is scooped out from one face of a cube such that the diameter of the hemisphere equals the edge of the cube (edge = a). Find the surface area of the remaining solid.

Understanding the Shape

After scooping the hemisphere, the top face of the cube loses a circle of area π(a/2)², but gains the curved inner bowl surface of the hemisphere.

Radius of hemisphere r = a/2 (since diameter = a) TSA of cube = 6a² Area of circular hole removed = π(a/2)² = πa²/4 CSA of hemisphere added = 2πr² = 2π(a/2)² = πa²/2 ────────────────────────────────────────────── SA of remaining solid = 6a² − πa²/4 + πa²/2 = 6a² + πa²/4 ∴ SA = a²(6 + π/4)

Surface Area = a²(6 + π/4)

✅ Answer: a²(6 + π/4)

💡 Why +πa²/4 and not −? Removing the hemisphere adds curved surface (we can now see the inner bowl) but removes the flat circle. Net change = +πa²/2 − πa²/4 = +πa²/4. The remaining solid is larger in surface area than the original cube, which is a subtle but important insight.

Problem 8

A wooden article is made by scooping a hemisphere from each end of a solid cylinder. Cylinder length = 10 cm, radius = 3.5 cm. Find the total surface area.

Key Insight: What Surfaces are Visible?

When a hemisphere is scooped from each end, the flat circular ends of the cylinder are removed and replaced by the curved inner bowls of the hemispheres. The lateral (curved) surface of the cylinder is still fully visible.

r = 3.5 cm, h = 10 cm CSA of cylinder = 2πrh = 2 × (22/7) × 3.5 × 10 = 2 × 22 × 0.5 × 10 = 220 cm² CSA of one hemisphere = 2πr² = 2 × (22/7) × 3.5 × 3.5 = 2 × 22 × 0.5 × 3.5 = 77 cm² TSA of article = CSA of cylinder + 2 × CSA of hemisphere = 220 + 2 × 77 = 220 + 154 ∴ TSA of article = 374 cm²

✅ Answer: 374 cm²

Quick Summary — All 8 Answers at a Glance

Q.	Shape Combination	Key Dimensions	Answer
1	Cone + Hemisphere (Toy)	r=3, h=4, l=5 cm	103.62 cm²
2	Cone + Cylinder + Hemisphere	r=8, h_cone=6, h_cyl=10 cm	1155.52 cm²
3	Cylinder + 2 Hemispheres (Capsule)	r=2.5, h=9 mm	219.8 mm²
4	Two Cubes → Cuboid	a=4 cm, l=8 cm	160 cm²
5	Cylinder + 2 Hemispheres (Tank)	r=0.7, h=8 m	Rs. 827.20
6	Volume Ratio Sphere:Cylinder:Cone	same r, h=2r	2 : 3 : 1
7	Cube with Hemisphere scooped out	edge = a, r = a/2	a²(6 + π/4)
8	Cylinder with 2 Hemispheres scooped	r=3.5, h=10 cm	374 cm²

Common Mistakes to Avoid in Board Exams

Forgetting to find slant height l first — The curved surface area of a cone uses l (slant height), not h (vertical height). Always compute l = √(r² + h²) as your first step.
Using TSA instead of CSA — When two solids are joined, use only the curved (lateral) surface areas. The flat circular joining faces are interior and invisible.
Diameter vs Radius confusion — Every formula uses radius r. If the problem gives diameter d, divide by 2 immediately and note it.
Wrong π value — Use π = 22/7 when radius has 7 as a factor (e.g., 3.5, 0.7, 7). Use π = 3.14 otherwise. Do not mix them in the same problem.
Capsule height error — For a capsule, the cylinder height = total length − 2r (not total length). Many students forget to subtract the two hemispherical ends.
Cube joined problem — Do not simply double the TSA of one cube. The two touching faces are hidden; the cuboid formula automatically handles this.

Board Exam Tips & What This Lesson Prepares You For

Exercise 10.2 is a high-weightage topic in Telangana SSC, AP SSC, and CBSE Class 10 Board Exams. Questions from this exercise appear almost every year — typically as 4-mark or 5-mark problems. The cone-hemisphere toy (Problem 1) and the sphere-cylinder-cone volume ratio (Problem 6) are especially popular.

Always draw a rough sketch of the combined solid — it helps you identify which surfaces are exposed and which are hidden.
Write out your formula before substituting numbers. Examiners award partial marks for correct formula even if arithmetic is wrong.
The answer for Problem 6 (ratio 2:3:1) should be memorised as a direct fact — it can be asked as a 1-mark fill-in-the-blank.
For painting/cost problems, always compute the surface area first, then multiply by rate. Never skip the surface area step.

🎯 After mastering Exercise 10.2, practise Exercise 10.3 on Volumes of Combinations of Solids, which uses the same shapes but asks for volume instead of surface area. You can also revisit the basic formulas in Introduction to Mensuration or strengthen your geometry foundation with Coordinate Geometry.