. A cubical box has each edge 10 cm and another cuboidal box is 12.5cm long, 10 cm wide and 8 cm high (i) Which box has the greater lateral surface area and by how much? (ii) Which box has the smaller total surface area and by how much?
A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30cm long, 25 cm wide and 25 cm high.
(i)What is the area of the glass?
(ii)How much of tape is needed for all the 12 edges?
Answers
Answer:
q1
From the question statement, we have
Edge of a cube = 10cm
Length, l = 12.5 cm
Breadth, b = 10cm
Height, h = 8 cm
(i) Find the lateral surface area for both the figures
Lateral surface area of cubical box = 4 (edge)2
= 4(10)2
= 400 cm2 …(1)
Lateral surface area of cuboidal box = 2[lh+bh]
= [2(12.5×8+10×8)]
= (2×180) = 360
Therefore, Lateral surface area of cuboidal box is 360 cm2. …(2)
From (1) and (2), lateral surface area of the cubical box is more than the lateral surface area of the cuboidal box. The difference between both the lateral surfaces is, 40 cm2.
(Lateral surface area of cubical box – Lateral surface area of cuboidal box=400cm2–360cm2 = 40 cm2)
(ii) Find the total surface area for both the figures
The total surface area of the cubical box = 6(edge)2 = 6(10 cm)2 = 600 cm2…(3)
Total surface area of cuboidal box
= 2[lh+bh+lb]
= [2(12.5×8+10×8+12.5×100)]
= 610
This implies, Total surface area of cuboidal box is 610 cm2..(4)
From (3) and (4), the total surface area of the cubical box is smaller than that of the cuboidal box. And their difference is 10cm2.
Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm2
q2
Length of greenhouse, say l = 30cm
Breadth of greenhouse, say b = 25 cm
Height of greenhouse, say h = 25 cm
(i) Total surface area of greenhouse = Area of the glass = 2[lb+lh+bh]
= [2(30×25+30×25+25×25)]
= [2(750+750+625)]
= (2×2125) = 4250
Total surface area of the glass is 4250 cm2
(ii)
SEE THE ATTACHMENT TOO
From figure, tape is required along sides AB, BC, CD, DA, EF, FG, GH, HE AH, BE, DG, and CF.
Total length of tape = 4(l+b+h)
= [4(30+25+25)] (after substituting the values)
= 320
Therefore, 320 cm tape is required for all the 12 edges.
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