Question 78337: Hi.. need some help with this word problem that is part of a math chapter on Applications of the Pythagorean Theorem... thanks!
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. The roof is 6 m from peak to eave and the attic space is 2 m high at the peak.
For the home described in this example, if the roof is 7 m from peak to eave and the attic space is 3 m high at the peak, how long does each of the pieces of insulation need to be? Round to the nearest tenth.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 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. The roof is 6 m from peak to eave and the attic space is 2 m high at the peak.
For the home described in this example, if the roof is 7 m from peak to eave and the attic space is 3 m high at the peak, how long does each of the pieces of insulation need to be? Round to the nearest tenth.
:
This is just a "find the side of right triangle" problem:
The roof length is the hypotenuse, one side is the height to the peak;
We need to find the horizontal side:
:
Remember a^2 + b^ = c^2, let the horizontal side = a
:
a^2 + 3^2 = 7^2
:
a^2 + 9 = 49
:
a^2 = 49 - 9
:
a = sqrt(40)
:
a = 6.3 meters
:
Later as I thought about it, I realized that it would be 2 * 6.3 = 12.6 meters
for a conventional roof ceiling. Sorry for the screw up
|
|
|
