ABCD is rectangle.Digonal BD is drawn.A line from C is drawn to meet at F on AB.The point at which they inters
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How about finishing the question so we can help
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How about (a+c)/(b+d)  ? more interesting!!\n\nbtw this only work when a/b != c/d
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The faster way to solve this questions is using graphical method. Draw the respective triangles (choosing appropriate scale) and you will get answer.
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1 - D\n2 - B\n3 - C\n4 - C
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Hint: the three side are equal long.\n\nTh
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considering both y values are -2 for A and C....\n\nwe know that a straight line exists between these two points, which means that it HAS to be 1 of the 4 sides.\n\ngiven that, the coordinates or K and L HAVE to be (3, infinity (EXCEPT -2)) and (-4, infinity (EXCEPT -2))
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make a common denominator\n\na/b - c/d = a*d/b*d - c*b/d*b = (a*d-c*b)/b*d
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by using a compass, draw two archs of the same degree to intersect each other on each end.  each point of intersection will be your point A & B use a straight edge to connect. Now you have a proven straight line.
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The answer is (D).\n\nThe perimeter is 9x (just add up the 4 sides 2x+3x+3x+x to determine this) -- which eliminates (B) and (C) as a possible answer.\n\nThe area is 6x^2 -- which eliminates (A) as a possible answer, leaving only (D).\n\nLooks like you have a trapezoid, where the base is 2x, the sides are 3x each, and the top is x.  Draw straight lines down from your top to your base, so that your trapezoid is in 3 pieces:  a right triangle with base = x/2 and hypotenuse 3x, a rectangle x by 2x, and another right triangle with base = x/2 and hypotenuse 3x.  The area of the inner rectangle is 3x^2, and you can put the two triangles together to get another x by 2x rectangle for another 3x^2 area, or a total area of 6x^2.\n\nI hope that helped.\n\n~ ♥ ~
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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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The longest line is BC.  (See the correctly drawn diagram in the attached link.)\n\nThis relies on a rule of triangles that the longest side is *always* opposite the biggest angle.\n\nLook at the two upper triangles (ABF and FED).  Their biggest angle is on the outside making their longest edges part of the inner triangle (BFD).\n\nHowever, for that triangle, the longest side is BD which is longer than any of the other sides so far.  So far we have eliminated all other sides above as shorter than BD.\n\nThat gets us to the bottom triangle whose longest side is BC (longer than BD).\n\nThus the longest line is BC... \n\nAs you noted, the picture is *definitely* not drawn to scale.\n\nJust for fun, I decided to take the measurements and actually draw the image *to scale*.  The resulting picture is attached as a link.  Hopefully now it is more apparent that BC is the longest side, as proven previously.\n\nIn the future, I'd look out for angles like BDC that are deliberately drawn far from their correct value.  That was the first clue that BC might be the longest side...\n\nAlso, as others have noted, until you are given at least one measurement, you can't determine the actual length, just the relative length.  For example, BC is about 4% longer than BD...
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Since A lies on the x-axis, the only ratio that would make sense would be AA /AB = 0/AB =0\n\nHowever from your answer I think you were asking: "If I draw a line segment connecting the y-intercept(call it point C) to point B, what is the ratio of CA/AB?"\n\nThe equation to the line including A and B is simply y = x-3. So the y-intercept is -3. This gives the 3 points C=(0,-3),A=(3,0),B=(4,1) The ratio of the y values below to above the x-axis is -3:1 just as you got. Of course the ratio of the lengths CA / AB = 3:1
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Any point of the vertical line B, which is 90 degrees from a line A and going through A's midpoint, is of equal distance from the end points of the line A.\n\nPractical Image\nThink of a cross. The horizontal line A of the cross is divided into equal parts by the vertical line B of the cross. Then any special point on the vertical line B is equally distant from both ends of A.\n\nIt often helps to make a drawing.\nBest wishes.\nCordially, India.Magica
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Plot the three boundary lines, shade the correct sides; the answers occur at the vertices of the shaded area (a triangle).  Just plug in these vertices to C = 4x + y and see what the highest and lowest outcome is.
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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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since each one gives you a *-----* i would say all of them are line segments.\n\nHowever since D. is the only one with an indefinite beginning and end and not a possible end or beginning, i would definately have to go with \n\nD.) -5 <= x <= 5\n\nThat is the only one that has a defined beginning and end and not a undefined end and/or beginning.
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A _______________ B\n... &#92;.................. ...............&#92;\n..... &#92;.................. ...............&#92;\n....... &#92;_______________&#92;\n...... D.................. ...............C\n\nLet ABCD be a quadrilateral with AB || DC and AB = DC\n\nDraw diagonal AC\n\nNow in Δs ABC, CDA\nAB = DC (Data)\n<BAC = <DCA ( Equal alternate angles, AB||DC)\nAC is common\nTherefore ΔABC ≡ ΔCDA (SAS property)\n\nTherefore < BCA = < DAC (Corresponding angles of congruent triangles are equal)\n\nTherefore BC||AD (alternate angles are equal)\n\nSo ABCD is a parallelogram as opposite sides are parallel