Challenge for challengers. ABC is a triangle with X a point on AB and Y a point on AC such that XY is parellel
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a) AX = 2.25 cm.\n\nb) area AXY upon area ABC = 0.5625\n\nc) area AXY upon area trapezium XYCB = 0.4375
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1) Assuming an aritmetic sequence\n\nbn = b1 + (n-1)d \nletting b3=a1 = 20, then b7=a5 = 32, and b25 = b23\na5 = a1 + (5-1)d\n32 = 20 + (5-1)d\n32 = 20 + 4d\n12 = 4d\n3 = d\n\nb25 = a23 = 20 + (23-1)*3\nb25 = 20 + 22*3\nb25 = 20 + 66\nb25 = 86\n=====================================================\n2)\na75 = a1 + (75-1)d; d=-3\n\na75 = 5 + (75-1)*-3\na75 = 5 + 74*-3\na75 = 5 + -222\na75 = -217\n\nSn = n/2(a1+an)\nS75 = 75/2(5 + -217)\nS75 = 75/2(-212)\nS75 = -7950\n===================================================\n3) ????\n\na5 = 5*5 - 4 = 21\na15 = 5*15 - 4 = 71\na25 = 5*25 - 4 = 121\n\nOR MAYBE\n\na1 = 5*1 - 4\na1 = 5 - 4\na1 = 1\na30 = 5*30 - 4\na30 = 150 - 4\na30 = 146\n\nSn = n/2(a1+an)\nS30 = 30/2(1 + 146)\nS30 = 30/2(147)\nS30 = 2205\n================================================\n4)\n\nrth term is nCr-1m^n-(r-1)k^r-1\n\n5th term =\n4C4m^4-(5-1)k^5-1\n4C4 = 1\n\n5th term = k^4
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A triangle in which all 3 sides are of different lenghts, with different angles.
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1 - A is enough because with the lengths of 2 sides you can use pythagoras' theorem to work out the length of the other
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either A, B, or AB is possible
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In this case A, B, and C are all equal.  And yes, A will always equal C.
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The midpoint of A and B is (15+7)/2 = 11.\nIf C = 11 then AC =BC.\nTherefore AC < BC  for 11<C<15
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a) "Suppose Triangle ABC and Triangle DEF are congruent."\nb) "Suppose there exists a line segment with at least two right bisectors."\nc) "Suppose there is an isosceles triangle for which the angles opposite two of its congruent sides are not congruent."\nd) "Suppose there is a line segment with at least two midpoints."\ne) "Suppose there are two intersecting lines such that opposite angles are not congruent."\nf) "Suppose AB is not equal to CD."
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So you need to prove that each element (x,y) E br(xo, yo) has a neighbourhood contained in br(xo, yo),\n\nTake a point (xa,ya) E br(xo,yo) and let d = r - ||(xa,ya) - (xo,yo)||.  d is positive since ||(xa,ya) - (xo,yo)|| < r.\nThen the neighbourhood bd(xa,ya) satisfies the condition because for any point (x,y) in bd(xa,ya) we have\n||(x,y) - (xo,yo)||\n<= ||(x,y) - (xa,ya)|| + ||(xa,ya) - (xo,yo)|| (triangle inequality)\n< d + ||(xa,ya) - (xo,yo)|| (because (x,y) E bd(xa,ya))\n= r.\n\nIn plain English: If the point (xa,ya) has distance d from the boundary of the set, then a neighbouroud of (xa,ya) whose elements are all less than d away from (xa,ya) will be contained in the set as well.
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the base of a triangle times the height times 2\n\nb*h*2
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A n B\n(A intersect B)
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Wow, that last person used way too much stuff.\n\nDraw in one diagonal of parallelogram ABCD, say BD.  You have congruent angles ABD and BDC, ADB and DBC because they are alternate interior angles and BD is congruent to itself.  The two triangles are congruent (ASA) which makes the parts you want congruent by CPCTC.
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Its been a long time since I took Geometry but maybe I can give you a small pointer.\n\nTry dividing the rectangle into 3 segments lengthwise and 3 segments widthwise\n\nA--B--C--D\nE--F--G--H\nJ---K--L--M\nN--O--P--Q\n\nDraw the diagonal through NKGD\n\nProve that the segments NK, KG, and DG are equal (prove that the triangles are the same)\n\nNow draw the segments from the vertex Q to the mid point of AD and AN.  Prove that the mid point of AD is the same as the mid point of BC and the same for AN and EJ.\n\nNow we have to prove that the line segment we drew in the last step instersect points K and G. You can probably do this with similar triangles (the triangle NQ and mid point of EJ is similar to JK and mid point of EJ therefore the point K is in the triangle)\n\nSo if you prove that the lines drawn from Q to the midpoints intersect K and G and we proved that NK KG and GD are congruent we have proved the original hypothesis.\n\nYou can turn this gabably gook into a real mathematical proof.\n\nGood Luck
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|\n  | \n  |  A\n  |\n  |_______\n......B\n\nThis is what "A" perpendicular to "B" looks like.  I hope this helps!
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There are 3 solutions for ABCDEFGHJ -\n\n2 6 7 8 3 5 1 4 9\n2 6 7 9 5 3 1 4 8\n6 7 8 4 1 3 2 5 9
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a^2 + b^2 = c^2 where c is the hypotonuse of the triangle.\n               :&#92;\n          a  : &#92; c\n               :__&#92;\n                b\n\na and b are interchangable but the key is the c must be the site opposite the right angle.\n\nThen you just plug in the numbers if you are given a or b\nsay a=2 and c=3\n\nthen 2^2 + b^2 = 3^2\n4 + b^2 = 9\nb^2 = 9-5 = 4\nb = square root of 4 or 2\n\nIf you are given a and b it is somewhat easier\na=3 b=4\n3^2 + 4^2 = c^2\n9+16 = c^2\n25 = c^2\nc = square root of 25 \nc = 5
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you are talking about "Pascalles (not sure about spelling) Triangle." It goes 1,11,121,1331 and so on.