This MCQ module is based on: Surface Areas of Combinations of Solids
TOPIC 29 OF 35
Surface Areas of Combinations of Solids
🎓 Class 10
Mathematics
CBSE
Theory
Ch 12 — Surface Areas and Volumes
⏱ ~30 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Surface Areas of Combinations of Solids
Targeting Class 10 level in Mensuration, with Intermediate difficulty.
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12.1 Introduction
In Class IX you met five basic solids — the cuboid, the cube, the cylinder, the cone and the sphere — and you computed their surface areas and volumes. Real life is rarely made of one shape alone. Look around: a test-tube is a cylinder capped by a hemisphere; a circus tent is a cylinder topped with a cone; a gulab-jamun has two hemispheres joined by a cylinder; an oil-can is a capsule — two hemispheres on each end of a cylinder. This chapter studies combinations of solidsi.
Fig. 12.1 — The five basic solids you have studied earlier
Basic Formulas Recap
- Cuboid: TSA = 2(lb + bh + hl); Volume = l · b · h
- Cube: TSA = 6a²; Volume = a³
- Cylinder: CSA = 2πrh; TSA = 2πr(r + h); Volume = πr²h
- Cone: CSA = πrℓ (ℓ = slant); TSA = πr(r + ℓ); Volume = ⅓πr²h
- Sphere: SA = 4πr²; Volume = (4/3)πr³
- Hemisphere: CSA = 2πr²; TSA = 3πr²; Volume = (2/3)πr³
12.2 Surface Area of a Combination of Solids
When we place one solid on top of another, or embed one in another, the total surface area of the combined solid is not simply the sum of the separate total surface areas: the two surfaces that join each other get hidden and must be excluded.
Principle
TSA of combination = sum of the exposed surfaces only. For each joined pair, one flat surface on each side is hidden at the joint — do not count either.Example 1 — Toy (cone on hemisphere)
Rasheed got a toy in the shape of a cone mounted on a hemisphere with the same base radius. The radius of the hemisphere is 3.5 cm and the total height of the toy is 15.5 cm. Find the total surface area of the toy. (Use π = 22/7.)
Fig. 12.6 — Toy: hemisphere at base with a cone on top
Solution. Hemisphere radius r = 3.5 cm. Height of cone h = 15.5 − 3.5 = 12 cm. Slant ℓ = √(r² + h²) = √(12.25 + 144) = √156.25 = 12.5 cm.
TSA of toy = CSA of cone + CSA of hemisphere = πrℓ + 2πr² = πr(ℓ + 2r) = (22/7)(3.5)(12.5 + 7) = (22/7)(3.5)(19.5).
\[\text{TSA}=\tfrac{22\cdot 3.5\cdot 19.5}{7}=\tfrac{1501.5}{7}=214.5\ \text{cm}^2.\]The toy's total exposed surface is 214.5 cm².
Example 2 — Cubical block with hemisphere scoop
The decorative block shown is made of two solids — a cube and a hemisphere. The cube has edge 5 cm. The largest possible hemisphere is mounted on one face of the cube. Find the total surface area of the block. (Use π = 22/7.)
Solution. Maximum hemisphere radius = half the edge = 2.5 cm. The hemisphere sits on one face, so its flat circular base is hidden, but it covers a circular region of area πr² of that face. Outside that region, the face area still counts.
TSA = (TSA of cube) − (πr² hidden under hemisphere) + (CSA of hemisphere) = 6(5²) − π(2.5)² + 2π(2.5)² = 150 + π(2.5)² = 150 + (22/7)(6.25) = 150 + 19.64 ≈ 169.64 cm².
Example 3 — Wooden article (cylinder minus two hemispheres)
A wooden article is made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 10 cm and its base radius 3.5 cm, find the total surface area of the article.
Fig. 12.8 — Solid cylinder with a hemisphere scooped out from each end
Solution. TSA = CSA of cylinder + CSA of two hemispheres (now internal surfaces) = 2πrh + 2(2πr²) = 2πr(h + 2r) = 2(22/7)(3.5)(10 + 7) = 2(22/7)(3.5)(17) = 374 cm²? Let's verify: 2(22/7)(3.5) = 22; 22 × 17 = 374. So TSA = 374 cm².
Example 4 — Water-filled rocket
A rocket is in the shape of a right circular cylinder closed at the lower end, with a cone attached to the top. The cylinder has diameter 6 cm and height 12 cm. The cone has slant height 5 cm. Find the total surface area and the curved-only area visible from outside. (π = 3.14.)
Solution. r = 3, h_cyl = 12, ℓ = 5. CSA of cylinder = 2π(3)(12) = 72π = 226.08 cm². CSA of cone = π(3)(5) = 15π = 47.1 cm². Base of cylinder (closed end) = πr² = 9π = 28.26 cm². Total exterior = 72π + 15π + 9π = 96π ≈ 301.44 cm².
Example 5 — Bird bath
Mayank makes a bird-bath shaped as a cylinder with a hemispherical depression at one end. The cylinder has height 1.45 m and radius 30 cm. Find the total surface area of the bird-bath. (Use π = 22/7.)
Solution. r = 0.30 m, h = 1.45 m. Total SA = CSA of cylinder + CSA of hemisphere + base area (closed bottom).
For NCERT: TSA = 2πrh + 2πr² = 2πr(h + r) = 2(22/7)(0.3)(1.45 + 0.3) = 2(22/7)(0.3)(1.75). Computing: 2 × 22 × 0.3 × 1.75 / 7 = 23.1/7 = 3.3 m². So the total exposed surface is 3.3 m².
Activity — Build your own combination
Take a cylindrical plastic cup and a rubber ball that fits snugly on top. Measure the cup's height h, cup radius r, and verify the ball's radius also equals r. Predict the total exposed surface area of the "cup + half-ball dome" combination, then measure it approximately using a piece of paper wrap. Check: TSA = CSA of cup + CSA of hemisphere + base of cup = 2πrh + 2πr² + πr² = πr(2h + 3r).
For r = 4 cm, h = 10 cm: TSA = π(4)(20 + 12) = 128π ≈ 402 cm². The hidden joint (flat disc at the rim) does NOT count from either solid — only the curved half-ball and the open top ring contribute.
Exercise 12.1
(Use π = 22/7 unless stated otherwise.)
Q1. Two cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Edge a: a³ = 64 ⇒ a = 4. Cuboid: l = 8, b = 4, h = 4. TSA = 2(lb + bh + hl) = 2(32 + 16 + 32) = 2(80) = 160 cm².
Q2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and total height of the vessel is 13 cm. Find the inner surface area.
r = 7, h_cyl = 13 − 7 = 6 cm. Inner SA = CSA of cylinder + CSA of hemisphere = 2πrh + 2πr² = 2π(7)(6) + 2π(49) = 84π + 98π = 182π = (182)(22)/7 = 572 cm².
Q3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. Total height of the toy is 15.5 cm. Find the TSA.
As in Example 1: 214.5 cm².
Q4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the SA of the solid.
Max diameter = 7 cm ⇒ r = 3.5. TSA = 6a² − πr² + 2πr² = 6(49) + π(12.25) = 294 + (22/7)(12.25) = 294 + 38.5 = 332.5 cm².
Q5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter ℓ of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid in terms of ℓ.
r = ℓ/2. SA = 6ℓ² − πr² (remove disc) + 2πr² (add hemisphere inner surface) = 6ℓ² + πr² = 6ℓ² + πℓ²/4 = (ℓ²/4)(24 + π) cm².
Q6. 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 14 mm and diameter 5 mm. Find its surface area.
r = 2.5 mm. Cylinder length = 14 − 2(2.5) = 9 mm. SA = 2πrh + 2(2πr²) = 2πr(h + 2r) = 2(22/7)(2.5)(9 + 5) = 2(22/7)(2.5)(14) = 220 mm².
Q7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m, and slant height of top is 2.8 m, find the area of canvas used. Find the cost at ₹500 per m².
r = 2, h = 2.1, ℓ = 2.8. Canvas = CSA cylinder + CSA cone = 2πrh + πrℓ = πr(2h + ℓ) = (22/7)(2)(4.2 + 2.8) = (22/7)(2)(7) = 44 m². Cost = 44 × 500 = ₹22 000.
Q8. From a solid cylinder of height 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and base is hollowed out. Find the TSA of the remaining solid to the nearest cm².
r = 0.7, h = 2.4. Slant ℓ = √(0.49 + 5.76) = √6.25 = 2.5. TSA = CSA cylinder + base + CSA cone (interior) = 2πrh + πr² + πrℓ = πr(2h + r + ℓ) = (22/7)(0.7)(4.8 + 0.7 + 2.5) = (22/7)(0.7)(8) = 17.6 ≈ 18 cm².
Q9. A wooden article as in Example 3 (cylinder with two hemispherical scoops). Cylinder height 10 cm, radius 3.5 cm. Find TSA.
374 cm² (as derived earlier).
Competency-Based Questions
Q1. Apply the combination principle: a cake is a cylinder of diameter 20 cm, height 8 cm, topped by a hemispherical dome of the same radius. Find the area to be iced (no icing on the base).
L3 Applyr = 10, h = 8. Iced area = CSA of cylinder + CSA of hemisphere = 2πrh + 2πr² = 2π(10)(8) + 2π(100) = 160π + 200π = 360π ≈ 1131 cm².
Q2. Analyse why two hemispheres joined at their flat faces (forming a sphere) have less total surface area than the sum of their individual TSAs.
L4 AnalyseIndividual TSAs = 3πr² each, total = 6πr². But once joined, the two flat discs (each πr²) become hidden — no longer part of the surface. Remaining exposed = 6πr² − 2πr² = 4πr² = the sphere's SA. Always consistent.
Q3. Evaluate: a candle is made of a cylinder (height 15 cm, radius 2 cm) with a cone-shape tip (height 3 cm, same radius). Compare its TSA to that of a plain cylinder of same total height.
L5 EvaluateCandle: CSA_cyl + CSA_cone + base = 2π(2)(15) + π(2)(√(4+9)) + 4π = 60π + 2π√13 + 4π ≈ 60π + 7.21π + 4π ≈ 71.21π ≈ 223.7 cm². Plain cylinder (height 18 cm, r 2): TSA = 2π(2)(18) + 2π(4) = 72π + 8π = 80π ≈ 251.3 cm². The coned candle has less exposed area.
Q4. Design a problem about a dumbbell: two spheres of radius 3 cm connected by a cylindrical rod of length 10 cm and radius 1 cm. Find the total surface area.
L6 CreateTSA = 2(sphere SA) − 2(joint disc of rod radius) + CSA rod = 2(4π·9) − 2π(1²) + 2π(1)(10) = 72π − 2π + 20π = 90π ≈ 282.74 cm². Each sphere has a tiny disc hidden where the rod enters, which we subtract; the cylindrical rod contributes its CSA (the two end discs are inside the spheres).
Assertion & Reason
A: When a cone of same base radius is placed on a hemisphere, the total surface area of the combination equals πr(ℓ + 2r).
R: The base circle of the cone coincides with the rim of the hemisphere — both discs are internal and are excluded.
R: The base circle of the cone coincides with the rim of the hemisphere — both discs are internal and are excluded.
Answer: A. CSA(cone) + CSA(hemisphere) = πrℓ + 2πr² = πr(ℓ + 2r).
A: Scooping a hemispherical cavity from a cube increases its surface area.
R: Losing a disc of area πr² but gaining a curved area of 2πr² yields net increase of πr².
R: Losing a disc of area πr² but gaining a curved area of 2πr² yields net increase of πr².
Answer: A. 2πr² − πr² = +πr² per cavity.
A: If a sphere is inscribed in a cube (touching all six faces), the cube's surface is fully visible.
R: The sphere touches each face at only one point, so it does not occlude any flat portion of the face.
R: The sphere touches each face at only one point, so it does not occlude any flat portion of the face.
Answer: A. Point-contact has zero area.
Frequently Asked Questions
What is the curved surface area of a cylinder?
Curved surface area of a cylinder = 2 pi r h, where r is the radius and h is the height. Total surface area adds 2 pi r squared for the two circular ends.
What is the total surface area of a cone?
Total surface area of a cone = pi r l + pi r squared, where l is the slant height. Curved surface area alone is pi r l.
What is the surface area of a hemisphere?
Curved surface area of a hemisphere = 2 pi r squared. Total surface area (including flat base) = 3 pi r squared.
How do you avoid double-counting in combined solids?
Only include the areas actually exposed to the outside - the region where two solids join is not part of the outer surface, so exclude it from both solids.
What is slant height of a cone?
Slant height l is the distance from the apex to any point on the base circle. It satisfies l squared = r squared + h squared (Pythagoras theorem).
Do we need to memorize all these formulas?
Yes. Surface area and volume formulas for cylinder, cone, sphere, hemisphere and cuboid/cube are mandatory for CBSE Class 10 boards.
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