Asked by Marylouise
A painter needs to cover a triangular region 63 m by 66 m by 73 m.A can of paint covers 70 square meters. How many cans will be needed?
Answers
Answered by
mathhelper
by Heron's Formula
s = (1/2)perimeter
= (1/2)(63+66+73) = 101
s-a = 101-63 = 38
s-b = 101-66 = 35
s-c = 101-73 = 28
area = √(s(s-a)(s-b)(s-c) )
= √(101*38*35*28) = appr 1939.39 m^2
or, find the acute angle opposite side 63
63^2 = 66^2 + 73^2 - 2(66)(73)cos A
cosA = .59319...
angle A = 53.616135
area = (1/2)(66)(73)sin53.616135 = 1939.39 m^2, same as above
number of cans = 1939.39/70 = 27.71 cans
the painter will need 28 cans, (how do you buy .71 of a can ?)
s = (1/2)perimeter
= (1/2)(63+66+73) = 101
s-a = 101-63 = 38
s-b = 101-66 = 35
s-c = 101-73 = 28
area = √(s(s-a)(s-b)(s-c) )
= √(101*38*35*28) = appr 1939.39 m^2
or, find the acute angle opposite side 63
63^2 = 66^2 + 73^2 - 2(66)(73)cos A
cosA = .59319...
angle A = 53.616135
area = (1/2)(66)(73)sin53.616135 = 1939.39 m^2, same as above
number of cans = 1939.39/70 = 27.71 cans
the painter will need 28 cans, (how do you buy .71 of a can ?)
Answered by
Maggi Margaret
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Naomi
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