Chapter 11: Mensuration

(a) Given: Length of cuboidal box  (l) = 60 cm

Breadth of cuboidal box  (b)  = 40 cm

Height of cuboidal box  (h)  = 50 cm

Total surface area of cuboidal box =  2×(lb+bh+hl)

= 2×(60×40+40×50+50×60)

= 2×(2400+2000+3000)

= 14800 cm²

(b)  Length of cubical box (l) = 50 cm

Breadth of cubicalbox (b) = 50 cm

Height of cubicalbox (h) = 50 cm

Total surface area of cubical box = 6(side)²

= 6(50×50)

= 6×2500

= 15000

Surface area of the cubical box is 15000 cm²

From the result of (a) and (b), cuboidal box requires the lesser amount of material to make.

Length of suitcase box, l = 80 cm,

Breadth of suitcase box, b= 48 cm

And Height of cuboidal box , h = 24 cm

Total surface area of suitcase box = 2(lb+bh+hl)

= 2(80×48+48×24+24×80)

= 2 (3840+1152+1920)

= 2×6912

= 13824

Total surface area of suitcase box is 13824 cm²

Area of Tarpaulin cloth = Surface area of suitcase

l×b = 13824

l ×96 = 13824

l = 144

Required tarpaulin for 100 suitcases = 144×100 = 14400 cm = 144 m

Hence tarpaulin cloth required to cover 100 suitcases is 144 m.

Surface area of cube = 600 cm² (Given)

Formula for surface area of a cube = 6(side)²

Substituting the values, we get

6(side)² = 600

(side)² = 100

Or side = ±10

Since side cannot be negative, the measure of each side of a cube is 10 cm

Length of cabinet, l = 2 m, Breadth of cabinet, b = 1 m and Height of cabinet, h  = 1.5 m

Area painted = Total surface area of the cabinet – Area of bottom

Total surface area of the cabinet =  2(lb+bh+hl )

= 2(2×1+1×1.5+1.5×2)

= 2(2+1.5+3.0)

= 13 m²

Area of bottom = Length × Breadth

= 2 × 1

= 2 m²

Area painted = 13 – 2 = 11 m²

The required surface area of the cabinet is 11m².

Length of wall, l  = 15 m, Breadth of wall, b = 10 m and Height of wall, h  = 7 m

Total Surface area of classroom = lb+2(bh+hl )

= 15×10+2(10×7+7×15)

= 150+2(70+105)

= 150+350

= 500

Now, Required number of cans =  Area of hall/Area of one can

= 500/100 = 5

Therefore, 5 cans are required to paint the room.

Similarity

Both figures have the same length and the same height

Difference

The first figure has circular bottom and top

The second figure has square bottom and top

The first figure is a cylinder and the second figure is a cube

Diameter of cylinder = 7 cm (Given)

Radius of cylinder, r  =  7/2  cm

Height of cylinder, h = 7 cm

Lateral surface area of cylinder = 2πrh

=  2×(22/7)×(7/2)×7 = 154

So, Lateral surface area of cylinder is 154 cm²

Now, lateral surface area of cube =  4 (side)²=4×7² = 4×49 = 196

Lateral surface area of cube is 196 cm²

Hence, the cube has a larger lateral surface area.

Radius of cylindrical tank, r  = 7 m

Height of cylindrical tank , h = 3 m

Total surface area of cylindrical tank = 2πr(h+r)

=  2×(22/7)×7(3+7)

= 44×10 = 440

Therefore, 440  m² metal sheet is required.

Lateral surface area of hollow cylinder = 4224  cm²

Width of rectangular sheet = 33 cm and say l be the length of the rectangular sheet

Lateral surface area of cylinder = Area of the rectangular sheet

4224 =  b × l

4224 = 33 × l

l = 4224/33 = 128 cm

So the length of the rectangular sheet is 128 cm.

Also, Perimeter of rectangular sheet =  2(l+b)

= 2(128+33)

= 322 cm

The perimeter of the rectangular sheet is 322 cm.

Diameter of road roller, d = 84 cm

Radius of road roller, r = d/2 = 84/2 = 42 cm

Length of road roller, h = 1 m = 100 cm

Formula for Curved surface area of road roller = 2πrh

= 2×(22/7)×42×100 = 26400

Curved surface area of the road roller is 26400 cm²

Again, Area covered by the road roller in 750 revolutions = 26400×750cm²

= 1,98,00,000cm²

= 1980 m²  [∵ 1 m²= 10,000 cm²]

Hence the area of the road is 1980 m².

Diameter of the cylindrical container , d = 14 cm

Radius of cylindrical container, r = d/2 = 14/2  = 7 cm

Height of cylindrical container = 20 cm

Height of the label, say h = 20–2–2 (from the figure)

= 16 cm

Curved surface area of label = 2πrh

=  2×(22/7)×7×16

= 704

Hence, the area of the label is 704 cm².