How much paint does a room need? How much tin makes a can? These ques­tions ask for the sur­face area of a solid: the total area of all the faces that cover it. In this les­son we find sur­face areas of the three solids with flat or straight sides, the cuboid, the cube and the cylin­der, by unfold­ing each one into flat shapes we already know.

Cuboid

A cuboid (a box shape) of length ll, breadth bb and height hh has six rec­tan­gu­lar faces in three equal pairs: top and bot­tom (l×bl \times b), front and back (l×hl \times h), and the two ends (b×hb \times h).

Total surface area=2(lb+bh+hl)\text{Total surface area} = 2(lb + bh + hl)

The four side faces alone (the walls of a room, with­out floor and ceil­ing) make the lat­eral sur­face area:

Lateral surface area=2h(l+b)\text{Lateral surface area} = 2h(l + b)

The longest straight rod that fits inside, cor­ner to oppo­site cor­ner, is the diag­o­nal: l2+b2+h2\sqrt{l^2 + b^2 + h^2}.

Cube

A cube is a cuboid with all edges equal to aa:

Total surface area=6a2,Lateral surface area=4a2,Diagonal=3 a\text{Total surface area} = 6a^2, \qquad \text{Lateral surface area} = 4a^2, \qquad \text{Diagonal} = \sqrt{3}\,a

Exam­ple 1

A room is 88 m long, 66 m wide and 44 m high. Find the cost of white­wash­ing its walls and ceil­ing at ₹25 per m².

Walls=2h(l+b)=2×4×(8+6)=112 m2\text{Walls} = 2h(l + b) = 2 \times 4 \times (8 + 6) = 112 \text{ m}^2

Ceiling=l×b=8×6=48 m2\text{Ceiling} = l \times b = 8 \times 6 = 48 \text{ m}^2

The area to white­wash is 112+48=160112 + 48 = 160 m², and the cost is 160×25=4000160 \times 25 = 4000, that is ₹4000.

Exam­ple 2

The total sur­face area of a cube is 600600 cm². Find its edge and its diag­o­nal.

6a2=600  ⟹  a2=100  ⟹  a=10 cm6a^2 = 600 \implies a^2 = 100 \implies a = 10 \text{ cm}

The diag­o­nal is 103≈17.3210\sqrt{3} \approx 17.32 cm.

Exam­ple 3

Find the length of the longest rod that can be placed in a box 1212 cm by 44 cm by 33 cm.

122+42+32=144+16+9=169=13 cm\sqrt{12^2 + 4^2 + 3^2} = \sqrt{144 + 16 + 9} = \sqrt{169} = 13 \text{ cm}

Right cir­cu­lar cylin­der

A cylin­der of radius rr and height hh has two cir­cu­lar ends and a curved sur­face. Cut the curved sur­face straight down and unroll it: it becomes a rec­tan­gle whose length is the cir­cum­fer­ence 2πr2\pi r and whose width is hh.

Curved surface area=2πrh\text{Curved surface area} = 2\pi r h

Total surface area=2πrh+2πr2=2πr(h+r)\text{Total surface area} = 2\pi r h + 2\pi r^2 = 2\pi r(h + r)

An open pipe has only the curved sur­face; a closed can has the curved sur­face and both ends; a tum­bler has the curved sur­face and one end.

Exam­ple 4

Find the curved and total sur­face areas of a closed cylin­der of radius 77 cm and height 2020 cm.

CSA=2×227×7×20=880 cm2\displaystyle \text{CSA} = 2 \times \frac{22}{7} \times 7 \times 20 = 880 \text{ cm}^2

TSA=880+2×227×7×7=880+308=1188 cm2\displaystyle \text{TSA} = 880 + 2 \times \frac{22}{7} \times 7 \times 7 = 880 + 308 = 1188 \text{ cm}^2

Exam­ple 5

A road roller has diam­e­ter 8484 cm and length 120120 cm. Find the area of road it lev­els in 500500 com­plete turns, in square metres.

In one turn the roller presses down its curved sur­face once.

CSA=2×227×42×120=31680 cm2\displaystyle \text{CSA} = 2 \times \frac{22}{7} \times 42 \times 120 = 31680 \text{ cm}^2

In 500500 turns: 31680×500=1584000031680 \times 500 = 15840000 cm². Since 11 m² =10000= 10000 cm², this is 15841584 m².

Units. Areas con­vert by the square of the length fac­tor: 11 m =100= 100 cm, so 11 m² =100×100=10000= 100 \times 100 = 10000 cm².

Prac­tice

  1. Find the total sur­face area of a cuboid 1010 cm by 88 cm by 55 cm.
  2. Find the total and lat­eral sur­face areas of a cube of edge 77 cm.
  3. A room is 1010 m long, 88 m wide and 55 m high. Find the cost of paint­ing its four walls at ₹15 per m².
  4. Two cubes of edge 55 cm are joined face to face. Find the total sur­face area of the cuboid formed.
  5. Find the length of the longest rod that fits in a box 22 m by 33 m by 66 m.
  6. Find the curved and total sur­face areas of a closed cylin­der of radius 1414 cm and height 1010 cm.
  7. The curved sur­face area of a cylin­der is 13201320 cm² and its radius is 10.510.5 cm. Find its height.
  8. A cylin­dri­cal pil­lar has diam­e­ter 5050 cm and height 3.53.5 m. Find the cost of paint­ing its curved sur­face at ₹12.50 per m².

Answers

Show answers
  1. 2(80+40+50)=3402(80 + 40 + 50) = 340 cm²
  2. 6×49=2946 \times 49 = 294 cm² and 4×49=1964 \times 49 = 196 cm²
  3. Walls 2×5×18=1802 \times 5 \times 18 = 180 m²; cost 180×15=2700180 \times 15 = 2700, that is ₹2700.
  4. The cuboid is 10×5×510 \times 5 \times 5: 2(50+25+50)=2502(50 + 25 + 50) = 250 cm². (Two faces of 2525 cm² each are hid­den in the join: 2×150−50=2502 \times 150 - 50 = 250.)
  5. 4+9+36=49=7\sqrt{4 + 9 + 36} = \sqrt{49} = 7 m
  6. CSA 2×227×14×10=880\displaystyle 2 \times \frac{22}{7} \times 14 \times 10 = 880 cm²; TSA 880+2×616=2112880 + 2 \times 616 = 2112 cm²
  7. 2×227×10.5×h=66h=1320\displaystyle 2 \times \frac{22}{7} \times 10.5 \times h = 66h = 1320, so h=20h = 20 cm.
  8. r=0.25r = 0.25 m: CSA 2×227×0.25×3.5=5.5\displaystyle 2 \times \frac{22}{7} \times 0.25 \times 3.5 = 5.5 m²; cost 5.5×12.50=68.755.5 \times 12.50 = 68.75, that is ₹68.75.