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A homeowner wishes to insulate her attic with fiberglass insulation to conserve energy. The insulation comes in 40-cm wide roll...Asked by Micha
A homeowner wishes to insulate her attic with fiberglass insulation to
conserve energy. The insulation comes in 40-cm wide rolls that are cut to fit between
the rafters in the attic. If the roof is 6 m from peak to eave and the attic space
is 2 m high at the peak, how long does each of the pieces of insulation need to be?
Round to the nearest tenth.
conserve energy. The insulation comes in 40-cm wide rolls that are cut to fit between
the rafters in the attic. If the roof is 6 m from peak to eave and the attic space
is 2 m high at the peak, how long does each of the pieces of insulation need to be?
Round to the nearest tenth.
Answers
Answered by
Micha
I am sorry, How would I set this problem up?
Answered by
Ms. Sue
First, you have a right-angle triangle; its height is 2 m and its hypotenuse is 6 m. To find the base, use the Pythagorean theorem: a^2 + b^2 = c^2. This gives you half the length of a piece of insulation. Double that number to find the total length of each piece.
If you post your answer, we'll be glad to check it.
If you post your answer, we'll be glad to check it.
Answered by
Micha
Okay,If I understand this is what I come up with.
square root of 6^2-2^2=x^2
square root of 36-4=x^2
square root of 32=x^2
x=4 square root of 2*2= x=8 square root of 2= approx. 11.3 m
square root of 6^2-2^2=x^2
square root of 36-4=x^2
square root of 32=x^2
x=4 square root of 2*2= x=8 square root of 2= approx. 11.3 m
Answered by
Ms. Sue
Right! You've got it! :-)
Answered by
Micha
Thanks! :)
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