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The area of the four walls of a hall is 660 sq m the length is twice the breadth and the height is 11 m find the area of the ceiling?
Wiki User
∙ 10y ago
The combined area of all four walls is the height (h) times the perimeter (p) of the hall. Since the height is given as 11 m, we can calculate the perimeter...
p = 660 m2 ÷ 11 m = 60 m.
The perimeter is two times the length (l) plus two times the breadth (b).
p = 2l + 2b
The length is given as twice the breadth.
l = 2b
Substituting 2b for l in the previous equation gives us...
p = 2(2b) + 2b = 4b + 2b = 6b
We had determined that the perimeter is also 60 m, so...
6b = 60 m
Dividing each side of the equation by 6 gives us...
b = 10 m
Substituting 10 m for b in the equation for l above gives us...
l = 2(10 m) = 20 m
The area of the ceiling is...
a = l x b
Finally, substituting the values of l and b yields...
a = 20 m x 10 m = (20 x 10) x (m x m) = 200 m2
Of course, the last step in solving any word problem is to confirm that we found what was asked for (in this case, the area of the ceiling), which we did. Also, one advantage that word problems have over regular math problems is that they give us the benefit of what I call the ridiculous test. Ask yourself whether or not the magnitude and/or units of your answer is ridiculous for what was asked for (200 m2 is a reasonable ceiling area for a hall).
Wiki User
∙ 10y ago
Wiki User
∙ 10y agoThe combined area of all four walls is the height (h) times the perimeter (p) of the hall. Since the height is given as 11 m, we can calculate the perimeter...
p = 660 m2
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