If the vertcies of square "A" are the midpoints of ... ?
920664
area a to b is 1/2..\n\nhow? assume a side of the inner square as 1 so that its area would be also 1, by pythagorean theorem, half of the side of square b is equal to squareroot of 1/2, so the side of b is 2*squareroot(1/2), that makes the area to be equal to 2, getting the ratio of areas, that is 1:2 n_n
741830
the ratio is 1:2 or 1/2\n\noh, by the way it goes like this:\n\nassuming the length of each side of square A is a, then the area is a^2.\n\nsince, the vertices of the smaller square (A) is the midpoint of the sides of the big square(B), using the diagonal of square A, we can find the sides of the bigger square.\n\nThe diagonal is a*squareroot of 2, in which when we square to get the area ===>  gives us 2a^2.\nThus, the ratio is between Square A' area (a^2) and B's area (2a^2) is 1:2 or 1/2
1066144
Without the knowledge of where point e is, this problem is insolvable.\n\nI suspect, however, from a brief diagramming session, that you will need to solve for the length of ab using the Pythagorean Theorem, and set up similar triangles.
937024
Straight B's would be a 3.0, A's would be a 4.0.\n\nThat's about 1/2 B's and A's.
774873
False for all quadrilaterals.\nBut all squares and isosceles trapezoids can be inscribed in a circle.
152263
A is on the x-axis (horizontally), while B is on y-axis (vertically).
758662
For example, a square is inscribed if the square is inside a circle.  The four vertices of the square are touching the circle..\n\nThe square will be circumscribed if it will be outside the circle.  In this case the circle will be touching the square in four points on the sides.
624413
yep\n\n*** Edit - for you idiot types, if a quadrangle has four sides that are not congruent, then it is NOT a square. A square has 4 right angles and 4 sides of equal length, ALWAYS.
1130245
I used to be a square but I'm all _right_ now :-)\n\nSeriously, you should do your own homework.
681882
I can't see your image, but if it is like any of these other pictures (see link), the problem is with the slopes of the triangles.  They may appear the same, but they aren't and that is where you lose a fraction of a square.  These small differences eventually add up to a full square if you rearrange things differently.\n\nThe key is in the slopes.\n\nI wish I could see the picture because I like area paradox puzzles like these.\n\n-----\n\nI'm back and can see your picture now... it is just like the first paradox in my link.\n\nThe answer, as I said is the slopes of the triangles.  One triangle has a rise of 2 for a run of 5 (slope = 0.40) and the other has a rise of 3 for a run of 8 (slope = 0.375).  So despite how it looks in the picture, that is not a straight line along the top.\n\nIn the top case, the slope causes a slight bend inward on that line.  In the bottom case, the slope causes a slight bend outward on that line.  If you take the difference between these two you get 1 sq. unit which accounts for the "missing" square.  The second link shows this visually.
566727
I'm afraid that with the given information, this would be impossible to determine. The only thing one could conclude from this information is that this polygon has an even number of sides. \n\nIt could be a square with sides of 3cm each\nIt could be a hexagon with sides of 2cm each\nIt could be an octagon with sides of 1.5cm each\nIt could be a decagon with sides of 1.2cm each\nIt could be a 12-gon with sides of 1cm each\n\nThe list is infinite unless you are given more specific information, such as the length of each side or the angles within the shape.
921006
Area = πab where a and b are the semi-major and semi-minor axis length
285703
Yes, if the circles are congruent.
570600
Ok, first just so you get the figure right, draw dots n name them like i have n at the places like m showing them below\n      A                B\n      \n    E                   F     \n\n   D                      C             G\n\nNow, join AB, D-C-G, EF , A-E-D and B-F-C  also.. join\n A-F-G .. GET IT?? and you'll have AE equal to ED as well as BF equal to CF as the median of the trapezium joins the mid=points of the non-parallel sides...\n\nNOW IN triangle BFA and triangle CFG ,you have\nBF congruent to CF ----   Proved tht F midpoint of BC\nangle BFA congruent to angle CFG--vertically opposite angles\n and as AB is parallel to CD (opposite sides of a trapezoid) it is also parallel to CG ( construction was to join D,C and G , remember??\n\ntherefore, abgle ABF congruent to angle FCG --- alternate angles for parallel lines AB and CG ..\n\nSO YOU NOW HAVE triangle ABF congruent to triangle CFG --- Angle-Side-Angle Test of congruency...\n\ntherefore, AF congruent to FG --- corresponding sides for congruent triangles\n\nNOW, IN THE BIG TRIANGLE ADG,\n  E is the midpoint of side AD - by construction\n  F is the midpoint of AG  --- Proved AF congruent to FG\n \ntherefore... EF parallel to DG --- By the Midpoint Theorem which states that-- A line passing through the midpoints of any two sides of the triangle is parallel to the third side (and has half the length of the third side )\n\nSO WE'VE NOW PROVED EF // DG\n .\n. . , EF IS ALSO // TO DC --- D,C AND F LIE ON THE SAME LINE.. I.E. THEY ARE COLLINEAR\n\n SO, EF//DC\nBUT DC IS PARALLEL TO AB ---- OPPOSITE SIDES OF THE TRAPEZIUM ABCD\n \n .\n. . , EF IS ALSO PARALLEL TO AB\n\n .\n. ., WE CAN PROVE THAT ----> AB // EF // DC  (REASON- both AB and DC are parallel to the same linE EF)  \n                     \n\n .\n. . WE HAVE PROVED THAT THE MEDIAN OF THE TRAPEZIUM (in this case GE)  IS PARALLEL TO BOTH THE BASES (i.e AB and CD )
549388
Yes.\nAny square is a rhombus.\nSome rhombuses are squares (the ones that have all angles of 90 degrees).
1116258
Yes, in that case, the circle is called a point.
726789
What are the co-ordinates of the other point? just use the midppint formula for a segment. All the best!\n\nLife's Lovely! Love & Live Life!