NCERT Solution: Surface Areas and Volumes
Answer
1. Length of plastic box = 1.5 m
Width of plastic box = 1.25 m
Depth of plastic box = 1.25 m
(i) The area of sheet required to make the box is equal to the surface area of the box excluding the top.
Surface area of the box = Lateral surface area + Area of the base
= 2(l+b)×h + (l×b)
= 2[(1.5 + 1.25)×1.25] + (1.5 × 1.25) m2
= (3.575 + 1.875) m2
= 5.45 m2
The sheet required required to make the box is 5.45 m2
(ii) Cost of 1 m2 of sheet = Rs 20
∴ Cost of 5.45 m2 of sheet = Rs (20 × 5.45) = Rs 109
2. The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of rs 7.50 per m2.
Answer
length of the room = 5m
breadth of the room = 4m
height of the room = 3m
Area of four walls including the ceiling = 2(l+b)×h + (l×b)
= 2(5+4)×3 + (5×4) m2
= (54 + 20) m2
= 74 m2 Cost of white washing = ���7.50 per m2
Total cost = rs (74×7.50) = rs 555
Answer
Perimeter of rectangular hall = 2(l + b) = 250 m
Total cost of painting = ���15000
Rate per m2 = rs10
Area of four walls = 2(l + b) h m2 = (250×h) m2
A/q,
(250×h)×10 = rs15000
⇒ 2500×h = rs15000
⇒ h = 15000/2500 m
⇒ h = 6 m
Thus the height of the hall is 6 m.
Answer
Volume of paint = 9.375 m2 = 93750 cm2
Dimension of brick = 22.5 cm×10 cm×7.5 cm
Total surface area of a brick = 2(lb + bh + lh) cm2
= 2(22.5×10 + 10×7.5 + 22.5×7.5) cm2
= 2(225 + 75 + 168.75) cm2
= 2×468.75 cm2 = 937.5 cm2
Number of bricks can be painted = 93750/937.5 = 100
(i) Lateral surface area of cubical box of edge 10cm = 4×102 cm2 = 400 cm2
Lateral surface area of cuboid box = 2(l+b)×h
= 2×(12.5+10)×8 cm2
= 2×22.5×8 cm2 = 360 cm2
Thus, lateral surface area of the cubical box is greater by (400 – 360) cm2 = 40 cm2
(ii) Total surface area of cubical box of edge 10 cm =6×102cm2=600cm2
Total surface area of cuboidal box = 2(lb + bh + lh)
= 2(12.5×10 + 10×8 + 8×12.5)cm2
= 2(125+80+100) cm2
= (2×305) cm2 = 610 cm2
Thus, total surface area of cubical box is smaller by 10 cm2
(i) Dimensions of greenhouse:
l = 30 cm, b = 25 cm, h = 25 cm
Total surface area of green house = 2(lb + bh + lh)
= 2(30×25 + 25×25 + 25×30) cm2
= 2(750 + 625 + 750) cm2
= 4250 cm2
(ii) Length of the tape needed = 4(l + b + h)
= 4(30 + 25 + 25) cm
= 4×80 cm = 320 cm
Dimension of bigger box = 25 cm × 20 cm × 5 cm
Total surface area of bigger box = 2(lb + bh + lh)
= 2(25×20 + 20×5 + 25×5) cm2
= 2(500 + 100 + 125) cm2
= 1450 cm2
Dimension of smaller box = 15 cm × 12 cm × 5 cm
Total surface area of smaller box = 2(lb + bh + lh)
= 2(15×12 + 12×5 + 15×5) cm2
= 2(180 + 60 + 75) cm2
= 630 cm2
Total surface area of 250 boxes of each type = 250(1450 + 630)cm2
= 250×2080 cm2 = 520000 cm2
Extra area required = 5/100(1450 + 630) × 250 cm2 = 26000 cm2
Total Cardboard required = 520000 + 26000 cm2 = 546000 cm2
Total cost of cardboard sheet = rs (546000 × 4)/1000 = rs 2184
Dimensions of the box- like structure = 4m × 3m × 2.5
Tarpaulin only required for all the four sides and top.
Thus, Tarpaulin required = 2(l+b)×h + lb
= [2(4+3)×2.5 + 4×3] m2
= (35×12) m2
= 47 m2