noteswale
Noteclerical

Surface areas and volumes

Lesson 97 of 113

1. Ye topic hai kya?

Surface area = bahar kitna rang / tin / canvas (cm²). Volume = andar kitna paani / metal (cm³). Paper: band dabba, khula bucket, tent, ball pighla ke rod, ice-cream (cone+scoop), pipe ki deewar, frustum bucket. π = 22/7 jab r 7 ka multiple; warna 3.14.

Yahan lock

CSA / LSA = curved ya 4 deewar. TSA = CSA + jo bases khuli hain. Combo: volume add; glued face do baar mat. Recast: volume same, surface generally naya.

Yahan kya nahi

2D sector/arc/segment, ring leftover. Tangent PA=PB. Compass. Tower 30°. Dice opposite / n³ painted cubes. Mean–median Statistics agle pe.

Pehle: band ya khula? Kitni faces? Diameter diya to r = d/2. Cone mein l slant, h axis. Hemisphere curved 2πr², solid TSA 3πr² — ye sphere 4πr² nahi.
Linear ×k → surface ×k², volume ×k³. Edge double → paint 4 guna, fill 8 guna. Ye scale hai — “3×3×3 paint karke kaato, 2-face kitni” Dice/cube pe hai.
SolidCurved / lateralTotal surfaceVolume
Cuboid2h(l+b)2(lb+bh+hl)lbh
Cube4a²6a²a³
Cylinder2πrh2πr(h+r)πr²h
Coneπrl, l=√(r²+h²)πr(l+r)(1/3)πr²h
Sphere—4πr²(4/3)πr³
Hemisphere (solid)2πr²3πr²(2/3)πr³
Frustumπ(R+r)lπ(R+r)l + πR² + πr²(1/3)πh(R²+Rr+r²)

2. Cuboid aur cube

Cuboid = l, b, h. 6 rectangles, opposite equal. Cube = har edge a, 6 squares. Space diagonal cuboid √(l²+b²+h²), cube a√3. Face diagonal cube a√2 — space se mix mat.

l h b cuboid

Upar = top l×b. Saamne = front l×h. Side = b×h. Cube mein teeno square.

Cuboid: V = lbh, LSA 2h(l+b) (4 deewar), TSA 2(lb+bh+hl) = LSA + top + bottom.
Cube: V = a³, LSA 4a², TSA 6a².
Example

Cuboid 8 cm × 6 cm × 4 cm. LSA, TSA, volume?

Solution

  1. LSA

    2h(l+b) = 2×4×(8+6) = 8×14 = 112 cm². Kamre ki 4 deewar, floor/ceiling nahi.

  2. TSA

    2(48+24+32) = 2×104 = 208 cm². Gift wrap: 6 faces.

  3. V

    8×6×4 = 192 cm³. 208 paint nahi, 192 fill nahi — units alag.

Answer112 cm², 208 cm², 192 cm³
Example

Cube edge 6 cm. TSA aur volume?

Solution

  1. TSA

    6×36 = 216 cm². LSA = 4×36 = 144 cm². Space diagonal 6√3.

  2. V

    6³ = 216 cm³. Number 216 same — cm² ≠ cm³. Paper “equal” mat bolna.

Answer216 cm², 216 cm³ (units alag)
Example

Kamra 8 m × 6 m × 4 m. 4 deewar + ceiling whitewash (floor nahi). Area?

Solution

  1. 4 deewar

    LSA = 2h(l+b) = 2×4×14 = 112 m². Sirf ye Mensuration-2D wala “4 walls” hai.

  2. + ceiling

    lb = 48. Kul 2h(l+b)+lb = 160 m². TSA 208 nahi (floor bhi), LSA 112 nahi (ceiling miss).

Answer160 m²
Door/window area ghatana. Open box (bina dhakkan): TSA se lb hatao. Ye n³ painted-cube counting nahi.

3. Cylinder aur cone

Cylinder ka curved kholo to rectangle: ek side 2πr, doosri h. Cone ka curved = bade circle ka sector — lekin yahan formula πrl hai; θ/360 wala 2D chapter mat copy.

r h cylinder h l r cone

Cone: l slant (hypotenuse), h seedha, r base. l = √(r²+h²) — Pythagoras, tower-elevation nahi.

Cylinder: CSA 2πrh, TSA 2πr(h+r), V πr²h.
Cone: l = √(r²+h²), CSA πrl, TSA πr(l+r), V (1/3)πr²h — same r,h ke cylinder ka 1/3. Height se CSA mat nikaalo.
Example

Cylinder r = 7 cm, h = 15 cm, π = 22/7. CSA, TSA, volume?

Solution

  1. CSA

    2×(22/7)×7×15 = 660 cm². Label around, lids nahi.

  2. TSA

    2πr(h+r) = 2×(22/7)×7×22 = 968 cm². Do circles 2×154 = 308; 660+308 = 968.

  3. V

    (22/7)×49×15 = 2310 cm³. Trap: 2πr² = 308 (do lids). 2D πr² = 154 sirf ek base — Areas chapter.

Answer660 cm², 968 cm², 2310 cm³
Example

Cone r = 7 cm, h = 24 cm, π = 22/7. Slant, CSA, TSA, volume?

Solution

  1. l

    √(49+576) = √625 = 25 cm. 7-24-25. h=24 slant nahi.

  2. CSA

    (22/7)×7×25 = 550 cm².

  3. TSA

    πr(l+r) = (22/7)×7×32 = 704 cm² (550+154).

  4. V

    (1/3)×(22/7)×49×24 = 1232 cm³. Bina 1/3 ke 3696 — same r,h cylinder.

Answerl=25 cm; 550 cm²; 704 cm²; 1232 cm³
Example

Open bucket (bina dhakkan), r=7 cm, h=15 cm, π=22/7. Bahar paint?

Solution

  1. CSA + neeche

    2πrh + πr² = 660 + 154 = 814 cm². Closed TSA 968 dono lids. Sirf CSA 660 andar-bahar nahi — stem “open bucket, bahar” = curved + ek base.

Answer814 cm²

4. Sphere aur hemisphere

Sphere = poori ball, flat face zero. Hemisphere = aadha; curved + ek circular face. Solid ball kaatne pe TSA 3πr²; katori (thin bowl) alag wording.

r sphere 4πr² hemisphere 3πr²

Curved bowl = 2πr². Ellipse = flat face πr². Dono milake solid TSA 3πr².

Sphere: 4πr², (4/3)πr³. Solid hemisphere: CSA 2πr², TSA 3πr², V (2/3)πr³. Paper “solid hemisphere” → 3πr², 4πr² mat.
Example

Sphere r = 21 cm, π = 22/7. Surface aur volume?

Solution

  1. SA

    4×(22/7)×441 = 4×22×63 = 5544 cm². Trap: πr² = 1386 (2D disk, Areas). 2πr² = 2772 (hemi curved). r=7 pe 4πr²=616 — Areas ke r=14 wale πr² jaisa number, yahan 3D skin.

  2. V

    (4/3)×(22/7)×9261. 9261/7=1323; (4/3)×22×1323 = 88×441 = 38808 cm³.

Answer5544 cm², 38808 cm³
Example

Solid hemisphere r = 21 cm. CSA, TSA, volume?

Solution

  1. CSA

    2πr² = 2772 cm².

  2. TSA

    3πr² = 4158 cm². 4πr²=5544 poori sphere.

  3. V

    Aadha sphere = 19404 cm³.

Answer2772 cm², 4158 cm², 19404 cm³
d=42 → r=21. 4πr² mein 42 mat daalna. Volume (4/3)πr³ hai, (4/3)πr² nahi.

5. Combination, recast, pipe, frustum

Do solids chipkao: volume add. Jo face chipki hai hawa mein nahi — surface se hatao. Recast (pighla ke naya shape): metal ka volume same; naya surface naya formula. Pipe: andar kaata cylinder, volume πh(R²−r²) — ye 2D ring leftover nahi, us ring-area × height hai.

hemisphere cone

Toy / ice-cream: flat glued. Volume = cone + hemisphere. Paint = πrl + 2πr² (πr² chipki, skip).

Capsule: cylinder + 2 hemispheres = cylinder + sphere. TSA 2πrh + 4πr² = 2πr(h+2r).
Tent canvas (floor nahi): 2πrh + πrl.
Frustum (cone ka upper hissa kaat do): l = √[h²+(R−r)²] — cone wala √(R²+h²) nahi. CSA π(R+r)l, V (1/3)πh(R²+Rr+r²).
Example

Wooden toy: cone r=7, h=24 glued on hemisphere r=7. π=22/7. Volume aur painted TSA?

Solution

  1. Volumes

    Cone 1232. Hemisphere (2/3)×(22/7)×343 = 2156/3. Total 1232 + 2156/3 = 5852/3 cm³.

  2. Paint

    Cone CSA 550 + hemi CSA 308 = 858 cm². Base 154 chipki — mat jodna. Poori sphere 616 galat (doosra half nahi).

Answer5852/3 cm³, 858 cm²
Example

Tent: cylinder r=7, h=9, upar cone r=7, h=24. Canvas bina floor? Volume? π=22/7.

Solution

  1. Cone l

    Pehle se 25 cm (7-24-25).

  2. Canvas

    2πrh + πrl = 2×(22/7)×7×9 + 550 = 396 + 550 = 946 cm². Floor πr² mat. Poora TSA of both solids alag.

  3. V

    πr²×9 + 1232 = 1386 + 1232 = 2618 cm³.

Answer946 cm², 2618 cm³
Example

Capsule: cylinder r=21, h=14, dono end hemisphere. π=22/7. TSA aur volume?

Solution

  1. TSA

    2πrh + 4πr² = 2×(22/7)×21×14 + 5544 = 1848 + 5544 = 7392 cm². Glued circles andar, bahar nahi.

  2. V

    πr²h + (4/3)πr³ = 19404 + 38808 = 58212 cm³.

Answer7392 cm², 58212 cm³
Example

Sphere r=21 pighla, cylinder r=7. Height? Same sphere se cone r=21, height?

Solution

  1. Volume same

    (4/3)π(21)³ = π(7)² h. π cancel. (4/3)×9261 = 49h → 12348 = 49h → h = 252 cm.

  2. Cone same r

    (4/3)πr³ = (1/3)πr² H → 4r = H → H = 84 cm. Surface naya hoga — recast pe TSA conserve nahi.

Answerh=252 cm; cone H=84 cm
r R h l

Frustum: do radii R > r, height h, slant l. l hypotenuse of h aur (R−r), poore cone ke r se nahi.

Example

Frustum R=14, r=7, h=24, π=22/7. l, CSA, TSA, volume?

Solution

  1. l

    √[24²+(14−7)²] = √(576+49) = √625 = 25. Trap: √(14²+24²) cone-from-vertex hai, frustum nahi.

  2. CSA

    π(R+r)l = (22/7)×21×25 = 1650 cm².

  3. TSA

    1650 + π(196+49) = 1650 + (22/7)×245 = 1650+770 = 2420 cm². 770 = do bases π(R²+r²). Areas ka ring 770 tha R=21,r=14 pe (π(R²−r²), 245 same coincidence). Yahan 14 aur 7 ka ring hota 462, TSA nahi.

  4. V

    (1/3)πh(R²+Rr+r²) = (1/3)×(22/7)×24×(196+98+49) = 8×(22/7)×343 = 8×22×49 = 8624 cm³.

Answerl=25; CSA 1650 cm²; TSA 2420 cm²; V 8624 cm³
Example

Pipe: outer R=4, inner r=3, h=14, π=22/7. Metal volume aur TSA (andar+bahar+2 rings)?

Solution

  1. Metal V

    πh(R²−r²) = (22/7)×14×(16−9) = 44×7 = 308 cm³. 2D annulus area × height — square-inscribed leftover 42 nahi.

  2. TSA

    2πh(R+r) + 2π(R²−r²) = 2π[14×7 + 7] = 2π×105 = 2×(22/7)×105 = 660 cm².

Answer308 cm³, 660 cm²
8 cubes edge 3 cm pighla ke 1 cube: volume 8×27=216 → a=6. Ye recast hai. 6-face paint karke 27 chhote kaatna Dice pe hai.

6. Sawal — basic se pro

Q1 · Basic

Cube edge 6 cm. TSA?

  1. 216 cm²
  2. 144 cm²
  3. 36 cm²
  4. 216 cm³

Solution

  1. 6a²

    216 cm². B LSA. D volume (number same, unit galat).

Answer216 cm²
Q2 · Basic

Cuboid 8×6×4. Volume?

  1. 192 cm³
  2. 208 cm²
  3. 112 cm²
  4. 48 cm³

Solution

  1. lbh

    192. B TSA. C LSA.

Answer192 cm³
Q3 · Basic

Cylinder r=7, h=15, π=22/7. CSA?

  1. 660 cm²
  2. 968 cm²
  3. 2310 cm³
  4. 154 cm²

Solution

  1. 2πrh

    660. B TSA. D ek base (2D).

Answer660 cm²
Q4 · Basic

Cone r=7, h=24. Slant l?

  1. 25 cm
  2. 24 cm
  3. 31 cm
  4. 7 cm

Solution

  1. √(r²+h²)

    25. B height. C r+h.

Answer25 cm
Q5 · Medium

Sphere r=21, π=22/7. Surface?

  1. 5544 cm²
  2. 1386 cm²
  3. 2772 cm²
  4. 38808 cm³

Solution

  1. 4πr²

    5544. B disk. C hemi curved. D volume.

Answer5544 cm²
Q6 · Medium

Solid hemisphere r=21. TSA?

  1. 4158 cm²
  2. 2772 cm²
  3. 5544 cm²
  4. 1386 cm²

Solution

  1. 3πr²

    4158. B CSA. C sphere.

Answer4158 cm²
Q7 · Medium

Toy: cone r=7 h=24 + hemisphere r=7. Painted TSA?

  1. 858 cm²
  2. 1012 cm²
  3. 704 cm²
  4. 616 cm²

Solution

  1. 550+308

    858. B 550+308+154 glued. C cone TSA. D sphere.

Answer858 cm²
Q8 · Medium

Sphere r=21 recast, cylinder r=7. h?

  1. 252 cm
  2. 84 cm
  3. 21 cm
  4. 63 cm

Solution

  1. (4/3)r³ = r_cyl² h

    12348/49=252. B cone same-r height.

Answer252 cm
Q9 · Pro

Frustum R=14, r=7, h=24. Volume (π=22/7)?

  1. 8624 cm³
  2. 1650 cm²
  3. 770 cm²
  4. 1232 cm³

Solution

  1. (1/3)πh(R²+Rr+r²)

    8624. B CSA. C π(R²+r²) bases / ring-lookalike. D cone r=7 h=24.

Answer8624 cm³
Q10 · Pro

3×3×3 cube 6 taraf paint, phir 27 pieces — exactly 2 faces painted. Is chapter ka sawal?

  1. Nahi — Dice / cube reasoning
  2. Haan, TSA = 6×9
  3. Volume 27 a³
  4. Frustum

Solution

  1. Paint-count

    A. Yahan 3×3×3 ka TSA 6×9=54 cm² ho sakta hai agar edge 3, lekin “kitni chhoti cubes 2-face” Dice pe.

AnswerA

Practice

Pehle khud, phir neeche kholo.

12345 678910

P1

Cube a=5. Volume?

  1. 125 cm³
  2. 150 cm²
  3. 100 cm²
  4. 25 cm³

Solution

  1. a³

    125. B TSA. C LSA.

Answer125 cm³
P2

Cuboid 8×6×4. TSA?

  1. 208 cm²
  2. 192 cm³
  3. 112 cm²
  4. 48 cm²

Solution

  1. 2(48+24+32)

    208.

Answer208 cm²
P3

Cylinder r=7 h=15. Volume (π=22/7)?

  1. 2310 cm³
  2. 660 cm²
  3. 968 cm²
  4. 1540 cm³

Solution

  1. πr²h

    2310. D h=10 wala.

Answer2310 cm³
P4

Cone r=7 h=24. Volume?

  1. 1232 cm³
  2. 3696 cm³
  3. 550 cm²
  4. 704 cm²

Solution

  1. (1/3)πr²h

    1232. B cylinder same r,h.

Answer1232 cm³
P5

Sphere r=21. Volume?

  1. 38808 cm³
  2. 5544 cm²
  3. 19404 cm³
  4. 1386 cm²

Solution

  1. (4/3)πr³

    38808. C hemisphere.

Answer38808 cm³
P6

Hemisphere r=21. CSA?

  1. 2772 cm²
  2. 4158 cm²
  3. 5544 cm²
  4. 1386 cm²

Solution

  1. 2πr²

    2772. B TSA. D disk.

Answer2772 cm²
P7

Tent r=7, cyl h=9, cone h=24. Canvas (no floor)?

  1. 946 cm²
  2. 1100 cm²
  3. 550 cm²
  4. 2618 cm³

Solution

  1. 396+550

    946. B +floor 154. D volume.

Answer946 cm²
P8

Pipe R=4, r=3, h=14. Metal volume?

  1. 308 cm³
  2. 660 cm²
  3. 154 cm²
  4. 42 cm²

Solution

  1. πh(R²−r²)

    308. D 2D square leftover. C disk r=7.

Answer308 cm³
P9

Frustum R=14 r=7 h=24. CSA?

  1. 1650 cm²
  2. 2420 cm²
  3. 8624 cm³
  4. 550 cm²

Solution

  1. π(R+r)l

    l=25 → 1650. B TSA. D cone CSA r=7.

Answer1650 cm²
P10

60° sector r=21 ka area is chapter mein?

  1. Nahi — Areas related to circles
  2. Haan, cone CSA
  3. Sphere 4πr²
  4. Frustum l

Solution

  1. 2D pizza

    A. Cone CSA πrl 3D curved hai, θ/360 nahi.

AnswerA

← Index · Areas related to circles

Agle topic: Statistics.