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CBSE · NCERT

Class 10 Maths – Chapter 13: Surface Areas & Volumes

Calculate surface areas (CSA, TSA) and volumes of 3D solids: cube, cuboid, cylinder, cone, sphere, hemisphere, frustum, and combinations of these solids. Use π = 22/7 unless stated otherwise.

Exercises: 13.1–13.5·Total Questions: 38

Exercise 13.1 9 Questions – Cubes & Cuboids

ðŸ" Formulas: Cube: SA=6a², Vol=a³ · Cuboid: SA=2(lb+bh+hl), Vol=l×b×h. Area is in cm²/m², volume in cm³/m³.
Q 1Combined Solid

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

Volume 64 â‡' edge = 4 cm. Cuboid: l=8, b=4, h=4. SA = 2(32+16+32) = 2(80) = 160 cm²

Q 2–9Applications

Q2: A vessel is in form of hollow hemisphere mounted by hollow cylinder. Inner diameter=14cm, total height=13cm. Find inner SA. â†' r=7. Cylinder h=13âˆ'7=6. CSA(cyl)=2Ï€(7)(6)=264, CSA(hemi)=2Ï€(49)=308. Total=572 cm²

Q5: A hemispherical depression is cut out from one face of a cubical wooden block of edge 21 cm. Find SA of remaining solid after cutting hemisphere of largest possible diameter. â†' r=10.5. SA=6(21²)âˆ'Ï€(10.5)²+2Ï€(10.5)² = 2646+Ï€(110.25) ≈ 2992.37 cm²

Q8: A hemispherical dome of a building needs painting. Circumference of base=17.6 m. Cost=₹5/100cm². â†' r=17.6/2Ï€=2.8m. CSA=2Ï€(2.8)²=49.28m². Cost=₹24,640

Exercise 13.2 8 Questions – Cylinder + Cone Combinations

ðŸ" Formulas: Cylinder: CSA=2Ï€rh, TSA=2Ï€r(r+h), Vol=Ï€r²h. Cone: CSA=Ï€rl, TSA=Ï€r(l+r), Vol=â…"Ï€r²h, l=√(r²+h²).
Q 1Pencil

A solid is in shape of cone standing on hemisphere with both radii = 1 cm. Height of solid = 5 cm. Find volume (Ï€=3.14).

Cone height = 5âˆ'1 = 4 cm. Vol(cone) = â…"Ï€(1)²(4) = 4Ï€/3. Vol(hemi) = (2/3)Ï€(1)³ = 2Ï€/3. Total = 2Ï€ = 6.28 cm³

Q 2–8Combinations

Q3: A gulab jamun contains sugar syrup up to 30% of volume. Cylinder with two hemispherical ends, l=5cm, d=2.8cm. Find syrup in 45 gulab jamuns. â†' r=1.4, cyl h=5âˆ'2.8=2.2. Vol(total)=Ï€r²h+(4/3)Ï€r³ = Ï€(1.96×2.2+3.66) = 7.97π≈25.05cm³. Syrup=7.515cm³/unit. ×45=338cm³

Q6: A solid iron pole consists of cylinder of height 220cm and base dia 24cm surmounted by another cylinder of height 60cm and radius 8cm. Find mass. â†' Vol=Ï€[12²×220+8²×60]=π×35520. Mass=35520π×8≈892.26kg

Exercise 13.3 9 Questions – Cone, Sphere & Hemisphere

Q 1Sphere Melted

A metallic sphere of radius 4.2 cm is melted and recast into shape of cylinder of radius 6 cm. Find height.

Vol(sphere) = (4/3)Ï€(4.2)³ = (4/3)Ï€(74.088). Vol(cyl) = Ï€(6)²h = 36Ï€h. Equate: 36Ï€h = (4/3)Ï€(74.088) â‡' h = 4×74.088/(3×36) = 2.744 cm

Q 2–9Applications

Q3: A 20 m deep well with diameter 7 m. Earth dug out is spread to form 22 m × 14 m platform. Find height. â†' Vol(earth)=Ï€(3.5)²×20=245Ï€=770m³. Platform h=770/(22×14)=2.5 m

Q6: A cylindrical bucket 32 cm high, 18 cm radius, filled with sand. Emptied on ground forming conical heap. Sand heap height=24 cm. Find radius and slant height. â†' Vol=Ï€(18)²(32)=10368Ï€. Cone=(1/3)Ï€r²(24)=8Ï€r². 8Ï€r²=10368Ï€ â‡' r=36cm. l=√(36²+24²)=√1872=12√13 cm

Q8: Water in canal 6 m wide, 1.5 m deep flowing at 10 km/h. How much area will it irrigate in 30 min if 8 cm standing water needed? â†' Vol in 30min=6×1.5×5000=45000m³. Area=45000/0.08=562500 m²

Exercise 13.4 5 Questions – Spheres & Hemispheres

ðŸ" Formulas: Sphere: SA=4Ï€r², Vol=(4/3)Ï€r³. Hemisphere: CSA=2Ï€r², TSA=3Ï€r², Vol=(2/3)Ï€r³.
Q 1Sphere Surface

A drinking glass is in shape of frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find capacity.

r₁=2, râ‚‚=1, h=14. Vol(frustum)=â…"Ï€h(r₁²+r₂²+r₁râ‚‚)=â…"Ï€(14)(4+1+2)=(14/3)Ï€(7)=102.67 cm³

Q 2–5Applications

Q3: Shanti Sweets Stall was placing order for cardboard boxes. Two sizes: 25×20×5cm and 15×12×5cm. 5% extra for overlap. 1000 boxes total. Cardboard cost ₹4/1000cm². â†' Total SA for both types with overlap = 1.05×[(1450)+(540)]×1000 = ₹8,358

Q5: A container shaped like right circular cylinder of diameter 12 cm, height 15 cm, full of ice cream. Ice cream to be filled in cones of height 12 cm and diameter 6 cm with hemispherical top. Find number of cones. â†' Vol(cyl)=Ï€(6)²(15)=540Ï€. Vol(cone+hemi)=â…"Ï€(3)²(12)+(2/3)Ï€(3)³=36Ï€+18Ï€=54Ï€. Number=10

Exercise 13.5 (Optional) 7 Questions – Frustum

ðŸ" Frustum Formulas: Vol=â…"Ï€h(r₁²+r₂²+r₁râ‚‚) · CSA=Ï€l(r₁+râ‚‚) where l=√[h²+(r₁âˆ'râ‚‚)²] · TSA=CSA+Ï€r₁²+Ï€r₂²
Q 1–7Frustum Problems

Q1: A copper rod of diameter 1 cm and length 8 cm is drawn into wire of length 18 m of uniform thickness. Find thickness. â†' Vol=Ï€(0.5)²(8)=2Ï€. Wire: Ï€r²×1800=2Ï€ â‡' r²=2/1800 â‡' r=1/30. Diameter=2/30=0.067 cm

Q5: A bucket is in form of frustum of cone of height 30 cm with radii 10 cm and 20 cm. Find capacity and cost of milk at ₹40/litre. â†' Vol=â…"Ï€(30)(100+400+200)=10Ï€(700)=21991cm³=21.99L. Cost=₹879.60

Q7: Derivation of frustum volume formula using similar triangles.

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