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there are so many tricks like that. working in a drawing office similar to a bookies in that there are alwys numbers flying about and people shouting out sums cos they can't find their calculator. I've managed to work out loads of little tricks, that make you seem a lot smarter than you really are :)
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i think Q2 is a trick. If there is rope attached to 3 of the corners, and all 3 lengths of the triangle are different, the field cannot possibly be square as stated in the question. ???????
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Why not?
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because in a square, all the sides have to be the same length, surely by using 3 corner posts, the enclosed shape has to be a right angle triangle and would have to include 2 edges of the field. so all 3 sides of the triangle cannot be different.
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I think you've misunderstood the question.
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Q4 is 24 days NAP
i dont think so, i think youve misunderstood me anyway back to work........ |
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"the enclosed shape has to be a right angle triangle"
No. The shape enclosed by two edges and two separate ropes (the third rope being from the "middle corner" of the three to the junction of the two other ropes) does not have to be a triangle at all. |
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Draw 4 dots making a square. Connect the left side and the top side. You have a backwards 7. Now draw another dot near the bottom right dot inside the square and think of that as the bull's location. Now draw a line to that"bull" dot from bottom left dot. Draw another Line to that dot from top left dot to "bull" dot. Draw a third line to that "bull" dot from top right dot. You have created a wonky kite shape inside the field.
One of those lines is 30m, one is 40, one is 50. Two sides of the kite are equal lengths. Still no idea how to work it out from that though. |
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I think you need to break either the shape or the other part of the square up into right-angled triangles (two right-angled triangles and a rectangle for each part) then use some combination of trigonometry and simultaneous equations. However, having said that, I've not actually tried to do it beyond drawing a picture similar to the one you've just described.
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this is "cecil rhodes" all over again
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I think the bull is a red herring.....
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........and I think the answer is 2500m sq. The 30 and 40m ropes are irrelevant
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Bul question has me stumped.
Unless you can prove mathematically, that its possible for 3 corners af 3,4 and 5 to be joined. If its possible to have 50m pinned in one corner, 40m pinned in opposite, and then 30m pinned between them, it will create a triangle with 50*40*64ish 64/ sqrt(2) = 45.25 (length of one side) which equals an area of 2048 But I have no idea how you would prove that the 30m would reach the bull... Perhaps someone more clever than me could tell ;) |
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i'm sad, I actually went home from the office just to try and work this out! Can't get any work done from thinking about it!
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i think you might need to combine the sin rule and the cosine rule for non right angled triangles and then solve simultaneously to try and find the angles
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Ive got
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I like your thinking, boggle.
Does no one know the 'true' answer? This really bugging me now :) |
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i dont like it cause it uses more than 3 of the corner posts. cant better it myself though
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if it loops back on itself, fk, its possible to only use 3 posts.
It looks like thats the winner, and I can finally start some work today. |
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i don't think that is the winner. There is a situation v.similar to the 3200 answer given earlier that will work. just don't now how to prove it
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okthis is the last im committing my mind to this now. if all 3 ropes arejoined, making 1 rope 120m long, and said rope is placed tight around 3 posts making a right angle triangle, isosceles triangle, then the area of the square (ie twice the area of the triangle) is 1235.32 sqm to 2dp. this may not be the answer on the paper but it fits the criteria of the question and the maths is right so i rest on that.
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Great band - sorry don't know any answers.
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