Da T.

asked • 01/16/16

Maths question need answer urgently please help much aprpreciated

A radio ham places an aerial mast where it gives the best reception on the roof of his rectangular garage. He then fixes wire supports from the top of the mast to each corner of the roof. The lengths of two opposite supports are seven metres and four metres and the length of one of the others is 1 metre. Find the length of the remaining support.

1 Expert Answer

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David W. answered • 01/16/16

Tutor
4.7 (90)

Experienced Prof

Don L.

tutor
Hi David and Da, please help me resolve a problem I am having with this question.
 
The three given supports, 7 meters, 4 meters, and 1 meter meet at a common point somewhere on the roof of the garage. 
 
Let the 7 meter and 1 meter supports be on one side of the garage meeting at opposite ends of that side. If I drop a perpendicular line from the point where they meet, to the side they are on, I can create two right triangles with the 7 meter support as the hypotenuse of one of the triangles and the 1 meter support the other hypotenuse. From this information, I know the maximum length of the side they are on will be less than 8 meters ( 7 times the cosine of some angle is less than 7 and 1 times the cosine of some angle is less than 1.)
 
Similarly, the side with the 4 meter support and the 1 meter support will be less than 5 meters in length.
 
If we let the maximum length be 8 meters and the maximum width be 5 meters, the diagonal of the rectangle from the point where the 7 meter support is anchored to the point where the 4 meter support is anchored will be less than 9.5 meters in length.
 
If this information is correct, I am not sure about the equation 7 + 4 = 1 + x. x would be longer than the diagonal of the garage.
 
Can you help me resolve this?
 
Don
 
 
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01/16/16

Da T.

Thank you so much David i can't thank you enough, i kept asking my family members and they kept telling me that the question was missing information thanks so much:)
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01/17/16

Don L.

tutor
Hi David and Da, thanks for taking the time to add to your explanation. I believe you have one solution. There are two depending on which corner the 7-meter support is attached. If it is attached to the same side as the 4-meter support, your answer is correct. If it is attached to the same side as the 1-meter support, there is a different solution.
 
Here is the other solution, the two right triangles are 1, y, 7 and 4, y, x (note, y would represent one side of the rectangle.)
 
Solve the 1, y, 7 right triangle for y gives y2 = 48.
 
Solve the 4, y, x right triangle for x gives x2 = 64, x = 8 meters.
 
David thanks your explanation.
 
 
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01/17/16

Gotz H.

When I draw three circles with a common centre and the given radii (1,4 and 7) then the corners of the garage will be somewhere on these circles. (The height of the mast is assumed to be near 0) I then simply draw any rectangle L1 x W1 (red) that has one point on each of my three circles. The distance from the fourth point back to the centre of the circles is the length of the last support (x). If I change the assumed rectangle to L2 x W3 (green) then x changes. I therefore believe that there is no mathematical solution to this problem because one side of the garage must be given in order to calculate the length of the last support (x). Please correct me if I’m wrong.
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02/14/16

Gotz H.

Just checked my drawing and it was inaccurate. x remains in fact constant. The equation is actually pretty simple when we consider the mast being placed on a corner of the garage. In that case the height is 1m. The garage is then
√(42 – 12) = √15 wide and √(72 – 12) = √48 long. X is therefore
√(42-12 + 72-12 + 12) = √64 = 8m in length.
The same works when we consider that the mast has no height at all and the supports are flat: Assuming that the mast sits on one side of the roof, the garage roof would then be (4+1)m = 5m wide and 72 - 12 = √48m long. The remaining support would therefore be √(72 - 12  + 42) = √64 = 8m long.
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02/14/16

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