NCERT Mathematics · Class 10 · English · 2026-27
Surface Areas and Volumes — Exercise 12.1
Independently written, step-by-step solutions. Prompts below are concise paraphrases; use the official textbook for the complete question wording.
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Question 1
Two cubes, each of volume 64 cm³, are joined face to face. Find the resulting cuboid’s area.
Step-by-step solution
Use π=22/7 unless the question specifies another value. Count only exposed surfaces for surface area; add or subtract solid volumes for capacity.
Each cube has side ∛64=4 cm. The cuboid measures 8×4×4 cm. Surface area=2(8×4+4×4+8×4)=2×80.
Resource ID: CB-10-M-CH12-EX12.1-Q1
Question 2
An open vessel combines a hemisphere of diameter 14 cm and a cylinder; total height is 13 cm. Find inner area.
Step-by-step solution
Use π=22/7 unless the question specifies another value. Count only exposed surfaces for surface area; add or subtract solid volumes for capacity.
Radius r=7 cm. The hemispherical depth is 7 cm, leaving cylinder height h=13−7=6 cm. The inner area is hemisphere curved area plus cylinder curved area: 2πr²+2πrh=2(22/7)7(7+6). The top is open.
Resource ID: CB-10-M-CH12-EX12.1-Q2
Question 3
A cone on a hemisphere has radius 3.5 cm and total height 15.5 cm. Find toy surface area.
Step-by-step solution
Use π=22/7 unless the question specifies another value. Count only exposed surfaces for surface area; add or subtract solid volumes for capacity.
Cone height=15.5−3.5=12 cm. Its slant height is √(12²+3.5²)=12.5 cm. The common circular base is hidden. Exposed area=πrl+2πr²=(22/7)3.5(12.5+7).
Resource ID: CB-10-M-CH12-EX12.1-Q3
Question 4
A hemisphere sits on a cube of side 7 cm. Find its largest diameter and the solid’s exposed area.
Step-by-step solution
Use π=22/7 unless the question specifies another value. Count only exposed surfaces for surface area; add or subtract solid volumes for capacity.
The greatest fitting diameter is 7 cm, so r=3.5. Begin with cube area 6×7²=294, remove the covered circle πr², and add hemispherical curved area 2πr². Net area=294+πr²=294+38.5.
Resource ID: CB-10-M-CH12-EX12.1-Q4
Question 5
A hemispherical hollow has diameter equal to cube edge l. Find the remaining solid’s surface area.
Step-by-step solution
Use π=22/7 unless the question specifies another value. Count only exposed surfaces for surface area; add or subtract solid volumes for capacity.
Radius is l/2. Start with the six cube faces, remove the circular opening and add the inner curved hemispherical surface: 6l²−π(l/2)²+2π(l/2)².
Resource ID: CB-10-M-CH12-EX12.1-Q5
Question 6
A capsule is 14 mm long and 5 mm in diameter, with two hemispherical ends. Find area.
Step-by-step solution
Use π=22/7 unless the question specifies another value. Count only exposed surfaces for surface area; add or subtract solid volumes for capacity.
Radius=2.5 mm. The two hemispheres together occupy axial length 5 mm, so cylinder length=14−5=9 mm. Area=2πrh+4πr²=2(22/7)(2.5)(9)+4(22/7)(2.5)².
Resource ID: CB-10-M-CH12-EX12.1-Q6
Question 7
A tent has cylinder height 2.1 m, diameter 4 m and cone slant height 2.8 m. Find canvas area and cost at ₹500/m².
Step-by-step solution
Use π=22/7 unless the question specifies another value. Count only exposed surfaces for surface area; add or subtract solid volumes for capacity.
Radius=2 m. No floor is included. Canvas=2πrh+πrl=πr(2h+l)=(22/7)2(4.2+2.8)=44 m². Multiply by ₹500/m².
Resource ID: CB-10-M-CH12-EX12.1-Q7
Question 8
A cylinder of height 2.4 cm and diameter 1.4 cm has a cone of matching height/radius hollowed from one end. Find exposed area to nearest cm².
Step-by-step solution
Use π=22/7 unless the question specifies another value. Count only exposed surfaces for surface area; add or subtract solid volumes for capacity.
r=0.7, h=2.4 and cone slant length l=√(2.4²+0.7²)=2.5. Exposed surfaces are cylinder curve, one intact circular base, and inner cone curve. Total=2πrh+πr²+πrl=πr(2h+r+l)=(22/7)0.7(4.8+0.7+2.5)=17.6.
Resource ID: CB-10-M-CH12-EX12.1-Q8
Question 9
A cylinder of height 10 cm and radius 3.5 cm has a hemisphere scooped from each end. Find area.
Step-by-step solution
Use π=22/7 unless the question specifies another value. Count only exposed surfaces for surface area; add or subtract solid volumes for capacity.
The exposed area is cylinder curved area plus the two curved hollows. There are no intact flat end discs. Thus area=2πrh+4πr²=2(22/7)(3.5)(10)+4(22/7)(3.5)²=220+154.
Resource ID: CB-10-M-CH12-EX12.1-Q9