where can i find a quadrilateral with two pairs of congruent sides and four right angles?
143210
fancy description of a rectangle.\n\nIf it had four congruent sides, it would be a square.
306739
4 times the length of one side
844185
square-ish in shape\ncome in different sizes\nmany varieties
107799
a right angle triangle can be complmentary and so can 2 acute triangles, these two can also be congruent.
667352
octagons have 8\npolygons with "x" sides will have "x" axes of symmetry
745336
All parallelograms have opposite sides congruent, but if all four sides are congruent, it's a rhombus.
891869
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
1105920
They add up to 360 degrees. (If you walked around the quadrilateral you would end up facing the same way as when you started, hence you went through 360 degrees. For a re-entrant or crossed-over quadrilateral, make that 360k degrees)
880763
The coordinates of the vertices of the 3 rectangles, by symmetry, would be of the form, (0, +- 1, +-x), (+-1, +- x, 0), and (+-x, 0, +-1), where x is unknown.  (0,1,x), (0,1,-x), and (1,x,0) forms one equilateral triangle. Therefore we know that 4x^2 = 1+(x-1)^2 +x^2 .  This reduces to x^2 + x -1 = 0.  Solving for x, we have x =  1/2(-1 + sqrt(5)).   From this, we can arrive at the ratio of rectangle sides at 1/2(1 + sqrt(5)), which is the golden ratio.
375748
All squares are rectangles,\nall parallelograms are quadrilaterls\nand all squares are parallelograms
959978
use pythagorean thereom a^2 + b^2 = c^2. (c being teh longest side or the hypotenuse) the two legs are a and b. so you have the sq rt of 169. (sqrt of 12^2 + 5^2) which is 13
1159510
the four angles add up to 360 degrees
436643
If the shape is a quadrilateral, the measures of its opposite angles WOULD add up to 180 degrees.  This is because the measure of an inscribed angle (an angle with vertex on the circle) is half the measure of its intercepted arc.  Between the two opposite angles, their combined arc is the whole circle which measures 360 degrees.  So the sum of the inscribed angles is half of that, or 180.
763972
Let there be a right-angled triangle.\nnow, \nwe know that, \n(2n^2+2n+1)^2=4n^4 + 4n^2 + 1 + 2(4n^3 + 2n+ 2n^2)\n=>(2n^2+2n+1)^2=4n^4 + 4n^2 + 1 + 8n^3 + 4n+ 4n^2\n=>(2n^2+2n+1)^2=(4n^4 + 4n^2 + 8n^3) + (1 + 4n+ 4n^2)\n=>(2n^2+2n+1)^2=(2n^2 + 2n)^2 + (2n+1)^2\nThis gives solutions\n(3,4,5); (5,12,13); (7,24,25); (9,40,41)........................\nthe above relation is true for all n belonging to natural numbers\nthis way you can find solution for any pair of sides the 3rd side been given.\nput the given side equal to one in equation and find positive n by solving the quadratic and find all the sides.\nEg:If hyputenuse is given as 5\nSolve 2n^2 + 2n + 1= 5\n=>n=1 therefore other sides are 3 and 4\n\nChoose this as best, if correct.
1021945
The first five squares are easy:\n1, 4, 9, 16, 25.\n\nRectangular:\n2, 6, 12, 20, 30\n\nTriangular:\n1, 3, 6, 10, 15\n\nPentagonals are:\n1, 5, 12, 22, 35
1096150
To determine if it is a parallelogram:\nIf a quadrilateral has one pair of opposite sides which are parallel and equal in length, then the figure is a parallelogram.\n\nTo determine if it is a rectangle:\nIf a parallelogram has two pairs of opposite sides which are parallel and equal in length, then the figure is a rectangle.\n\nGraph the points and determine the distance between the points, or, in your case, two points have an X of 0 and two have X of 3\n\nBetween the (0, 2) and (0, 6), there is a distance of 4\nBetween the (3, -2) and (3,2) there is a distance of 4\n\nBetween the (0,6) and (3, 2) there is a distance of 3 horizontally.\nBetween the (0,2) and (3, -2) there is a distance of 3 horizontally.\n\nYour figure has two pairs of opposite sides that are equal in length and parallel.  You have a rectangle.
33049
Start with a square. All sides being equal and all internal angles being equal (90 degrees), there are four equal sides, each of which contain only two points --> hence, for N = 4, here are four couples of points which do not contain the other two non-coupled points. If they did contain the other point(s), it would not be a square and thus N would not equal 4. Which isn't true.