a yard need to be enclosed by a house on one side and two fences forming a right triangle on the other, if the
113932
Yep, it's the Pythagorean theorem question.\n\nA^2=B^2+C^2\n\nA is always the hypotenuse.\n\n20^2 = 400\n\n400 = 144 + B^2\n256 = B^2\n16 = B\n\nSo the length of the last side is 16 yards.\n\nHence, the total length is A + B.\n\n12 plus 16 is 28 yards total of fencing.\n\nThe area of an enclosed yard is one yard squared.\n\nBut if you're curious as to what the area is of the triangle created by the house and the fencing, then it's 1/2(B*C)\n\n1/2*192= 96yd^2 is the area of the triangle.
1028326
1 - A is enough because with the lengths of 2 sides you can use pythagoras' theorem to work out the length of the other
706050
Ok... the first thing to realize is that you are finding the maximum area with the least amount of materials.  To do that you should put the shed at a corner of the rectange, that way two sides of the shed act as part of the fence.  Assume that the shed is not "usable area" for the shep pen.  Accordingly, you need to solve the problem twice, once with using the shed as part of one side, and once using the shed as a corner.\n\nOk, the first thing you need to do is show that the maximum area of a rectangle occurs when the lenght and width are equal.  That is the only part that needs any type of equation analysis, and it really doesn't require calculus.\n\nA=L*W\nAssuming that the width is at most as long as the L, then the equation goes to:\nA=L*(x*L), where x a percentage, ranging from 0 to 1.\nClearly, the area (A) will be maximum when x=1.  That shows that in your problem, the pen will have the maximum area when the length and width of the rectangles are equal.\n\nSo, you know that the fence is 160m., and you have 4 sides, s1, s2, s3 and s4.\nSo:\n160m=s1+s2+s3+s4\n\nNow, if you use the shed as part of one side, namely s4, then the perimeter equation for the shed is:\n180m=s1+s2+s3+s4+20m\n\nSince the maximum area occurs when the lenght and width are equal:\ns1=s2=s3=s4-20m, so substituting s2, s3 and s4 for s1 in the permiter equation, you get:\n4s1+20m=180m\nSolve for s1,\ns1=45m (call this the length)\ns4=45-20=25m\nin this case, the area will be S1*S1, since the lenght and width of the pen will be equal.  \nSo the area is 45*45=2025 sq. m.\n\nNow, put the shed in a corner of the sheep pen.\nSo in our example, we'll make the shed part of s1 and s4.\n\nso the primeter equation will be:\ns1+20m+s2+s3+s4+20m=200m\nNow, s1=s2-20, s3=s2 and s4=s2-20\nSo substituting for s1, s2 and s4, you get:\ns2-20m+s2+s3+s4-20m+40m=200m\nsolve for s2,\ns2=50m=s3 and s1=30m (s2-20m)\nSo max area is 50m*50m=2500 sq. m\nBut, now you need to reduce the area by the area of the shed (400 sq. m), so final usable area is:\n2100 sq. m which is more than using the shed as part of only 1 side.
111898
Fences on the border.
637271
It is hardly impulsive and 370 miles isn't nearly long enough.  Staffing may go up in the short term but will eventually be reduced in favor of pet projects.  At that point, the fence will remain.  It is much more cost effective than many of the other elements.  It is called a "95%" fence, a triple fence with sensors cameras and trenches along either side.  You go to the places they plan to put it and tell me if you see any curb appeal there now.
298695
In a house. It will protect you from the outside....sometimes
763822
Divide the right angle into three equal parts, or three 30-degree angles.\n\nSounds like homework to me (which, hey, I'm cool with!), as if you owned this parcel this is NOT the way you'd want to divide it for real -- you wouldn't want three neighbors all bunched up like that at the confluence of the former right angle.\n\n.
786241
Defence seperates my yard from my neighbor's.
122782
the fence  built on your property is your the fentce touching the property of your neighbor isnt urs anymore.
304879
It's long, it doesn't have a fence it's just open terrain!  That is what the bill is supposed to correct, but for some reason criminals don't want us to build a fence because it would be harder for their criminal families to come over as well!
394622
A triangle in which all 3 sides are of different lenghts, with different angles.
63969
A hexagon has 6 sides.
644975
its always a good idea to have a fence around any kind of pool, if it is on your property and some kids decide one day to go for a dip and something happens to them, then you are responsible for whatever may happen... better safe than sorry
751913
its happening on both sides of the fence. i wish this fence could be torn down
1115391
I think your teacher is looking for choice b., but technically it's choice a., because any triangle can be split into right triangles by drawing appropriate altitudes and then using the theorem for those smaller right triangles.
709642
No, a yard is a yard is a yard.
903586
A trapezoid can be constructed so it has only 2 right angles.