Do the diagonals bisect the 90 degree angles to create 45 degree angles in a rectangle?\n\nOr is that only in squares?
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The answer to your question is NO!  That is assuming a rectangle that fails to pass the test as a square, i.e. two sides of one length and two sides of another length.\n\nSince you have a difference of opinions in preceding answers, may I suggest an experiment.\n\nDraw a rectangle that is 1/2 inch wide and 11 inches long.  Now draw a diagonal in that rectangle.  It should be obvious that you are NOT bisecting the angles.  \n\nIf you are trying to prove this one way or the other, look to isosolese triangles with a 90 degree vertex.  If an isosolese triangle has a 90 degree angle the base can only be opposite that angle.  By definition an isosolese triangle has equal sides and equal base angles.  That means when you apply the Pythagorean theorem the square of the sides must equal each other.  So after you have drawn the diagonal in your rectangle, that diagonal must be opposite a 90 degree corner.  That means the length of the rectangle is one side of the triangle with the width being the other.  Since the length and width of the rectangle are not equal, the sides of the triangle are not equal.  Since the sides of the triangle are not equal, the triangle is not isosolese.  Since the triangle is not isosolese, the base angles are not equal.  And since all the angles of a triangle must equal 180 degrees and one angle is known to be 90 degrees, and the other two angles are known to be unequal neither of the other two angles can equal 45 degrees.\n\nThis is a bit involved, I hope you can follow the logic.
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One of the most simple requirements for tiling: if a polygon tiles a plane, then in each of polygon vertexes on the plane, the sum of adjacent angles should equal 360 degrees (one of the angles may be a side of another polygon, making up 180 degrees). \n\nFor example, imagine or draw a "house-shaped" pentagon, that is a square with a right-angled triangle on top (right angle pointing up). It has two angles of 90 degrees at the bottom, two 135 degree angles where "walls" and "roof" meet, and a 90 degree angle at the top ("roof" tip). You can tile them side-by-side and then tile the same pentagons on top, flipped upside down. At the "tip of the roof", you will see that two 135 degree angles and one 90 degree angle meet. 135+135+90=360. If it would not be the case, there would be a crack in the tiling.\n\nHowever, some polygons have such angle values that it is impossible to make a sum of angles equal to 360 or 180 degrees. For example, consider regular pentagons. They have angles of 108 degrees. However you add up 108 degree angles, you cannot make up 360 or 180.
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ok: i hope you sketched this out: forget for a minute the solid's diagonal is 26cm. let's use the ratios as actual units. so the one side with a 3*4 rectangle has a 5unit diagonal (from Pythagoras).\nnow-picture the right triangle with the two adjacent sides to the right angle, one side is 5 and one is 12 (that's the solid's diagonal). using OUR units for now, THAT diagonal is 13units (Pythagorean again). luckily, 26 is 2 * 13, so hopefully the measurements are 6,8,24, would such a solid yield a diagonal of 26?
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I would think that the area is proportional to the *square* of their corresponding sides.\n\nIf you had a rectangle that was 3 x 4 and the corresponding rectangle was 6 x 8, the ratio of their sides is 1 : 2.  But their areas are 1 : 4 (2²).
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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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Short Answer:\nThe diagonal line or hypotenuse is not a straight line.  It bends either way thus both figures are NOT triangles.\n\nLong answer:\n\nThe two diagrams are not perfect triangles; they are quadrilaterals. Why is that significant? Look at the "diagonal line" on each of the diagrams. You may be able to see, if you look closely, that the supposed diagonal line is not really a straight line. Rather, it's two connected segments (two straight line segments) with a slight angle to it where the two segments meet.\n\nThe top diagram bends in a convex manner, so that there is slightly more area contained below the diagonal. In the bottom diagram, the bend is in the opposite direction.\n\nTo help you see this better, look at the green traingle in the top diagram. It's dimensions are 2 units high, and 5 units long. The Red triangle is 3 units high. In order for the two line segments that make up the combined diagonal to be a straight line, the Red triangle would therefore need to be 7.5 units long (2/5 = 3/x; therefore x=7.5). Instead, it's 8 units long. If that Red Triangle were only a half a unit shorter along the base, then the diagram would all line up, and we wouldn't be trying to solve this problem!\n\nThe four pieces that make up the upper diagram "appear to fit exactly" on a triangle that is half of a rectangle 13 units long and 5 units high. (Count the squares: 13 across, 5 down.) If "appearances are correct", then that would imply that the total area of the four pieces is half of 13x5, or half of 65. So you would think that the total area of these four pieces is 32.5 square units. Let's total them up and see:\n\nIn the top diagram, you can see that the two irregular pieces combine to make a 3x5 rectangle. So these two pieces are 15 square units (you can count the squares, as well! One is 7 squares, the other is 8.) That's part of our total.\n\nThe Red Triangle is half of a 3x8 rectangle. So it's 12 square units. So add 12 to our total.\n\nThe Green Triangle is half of a 2x5 rectangle. So it's 5 square units. So add 5 to our total.\n\nTotal Area of the four pieces is 15+12+5 square units, or 32 square units. So the top quadrilateral is covering only 32 squares out of the 32.5 square units that a triangle would cover. It LOOKS LIKE it covers 32.5 squares, but really it only covers 32!\n\nLet's check the bottom diagram. Obviously the four pieces only cover 32 square units, using the same math as above. There's a 33rd square which is empty.\n\nIf you take the top diagram, and rotate it 180 degrees, so that it fits perfectly on the bottom diagram, you'll see that the red triangles are butted against each other, as well as the green traingles. Therefore, if the top diagram has a slight angle caused by the line segments, the bottom diagram has a slight angle exactly complimentary to the top one. Therefore, if the top diagram undershoots the 32.5 square units by just a half of a square, then the bottom one OVERshoots it by half a square.\n\nAdd up the two half squares, and you have the whole hole.\n\nFour pieces that cover 32 squares, plus four pieces that cover 32 squares, plus one empty square equals 65 squares, which is the area of rectangle 13 units long and 5 units high.\n\n(7+8+12+5) + (7+8+12+5) + 1 = 13 x 5\n\nClear as mud, eh?
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360 degrees because only zero degree or 360 degrees places the angle in the same quadrant. if u are asking multiples of then the answer would be a finite number i.e. 360 degrees.\n\nexplaination:\ntake a circle and hold its radius at the circumference, now move it round the circle with its center point fixed. it will come at the same place only after 360 degrees, or it will substend the same angle as it was when u started the movement.
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I think so, actually...\n\n1. Draw 2 perp. lines.\n2. Draw 2 hypotenuses to each base such that the triangles made are 45-45-90 triangles (extend the bases as well, if you need to).\n3.  You have just made an isoceles triangle since the base angles are congruent.\n4.  The line that is perp. to the base is now the perpendicular bisector because the triangle is isoceles.\n\nViola!
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There are 88 or 44 depending on how you define what a unique diagonal is.  ( Sorry, not trying to be "egghead-y" here, just trying to give you a complete explanation )\n\nANSWER OF 88:\n\nLet's look at an n-sided figure.  For each point (where two sides meet), you can't draw a diagonal from the point to the same point itself, nor from the point to the two points next to it on either side.  But, you can draw a diagonal to every other point, or to n-3 points.  You can do this for each of the n points.\n\nSo, in an n-sided figure, you can have n(n-3) diagonals.\n\nSo, in an 11-sided figure, you can have 11(11-3) diagonals = 11*8 diagonals = 88 diagonals.\n\nNotice this answer assumes that the diagonal from point A to point B is different than the diagonal from point B to point A!\n\nANSWER OF 44:\n\nTo eliminate such "duplicates", we need to look at it this way for an n-sided figure:\n\nPoint 1 has n-3 diagonals (same reasoning as above)\nPoint 2 has n-3 diagonals (same reasoning as above)\nPoint 3 has n-4 diagonals (since 1 of its diagonals was already counted under Point 1)\nPoint 4 has n-5 diagonals (since 2 of its diagonals have already been counted)\n... etc ... till we get to ....\nPoint n-2 has 1 diagonal\nPoint n-1 has 0 diagonals\nPoint n has 0 diagonals\n\nSo an 11-sided figure would have 44 diagonals.\n\nThe general formula being n(n-3)/2.\n\nIf you try this general formula for something that you can easily draw and count the diagonals of, such as a square or a pentagon, you will see it holds true.  (A square has 2 diagonals = 4(4-3)/2 = 4/2 = 2.  A pentagon has 5 diagonals = 5(5-3)/2 = 10/2 = 5.)\n\n~ ♥ ~
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Best to plot this on a piece of graph paper; "+i" is  to the right' , "+j" is upwards, "-i" is to the left, "-j" is downwards, all with respect to the origin.\n\n1. Draw the vectors on the graph paper. "u" is in the second quadrant (i is neg and j is pos). "v" is in the fourth quadrant, with (i is pos, j neg) . There are two answers to this question, since you could go from u to v or from u to v in the opposite direction. I always take the angle from zero to "u" and add it to the angle from 0 to "v".  For "u", the tangent of the angle to the vertical axis is 1/4, then angle = 14%. So the angle from zero degrees to the vert axis = 90, and "u"s angle from zero degrees is 90 + 14 = 104 degrees. For "v", the tangent of the angle to the horizontal axis is 4/3 = 1.333.. and the angle = 53 degrees. Therefore the angle from "u" to "v" is 104 + 53 = 157 degrees. (the other answer is 360 degrees - 157 degrees = 203 degrees if you rotate in the other direction).\n\n2. Again use graph paper. 240 degrees is 30 degrees less that 270 degrees (the vertical axis, down) draw a line 10 graph units long. It will be in the third quadrant where i and j will both be negative. At this point you can use a polar to rectangular converter, or do it by hand. By hand: the i side of the triangle formed with the down axis can be found by saying sin(angle) = i/10, and cos(angle) = j/10. The angle is 30 degrees (from vert downwards). so sin(30) = 0.5, and, cos(30) = 0.866; thus i= 0.5 and j = 0.866 (ignoring pos/neg). Since the vector is in the third quadrant, both i and j are negative.\nAnswer = -0.5i -0.866j\n\nHope this helps!
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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.
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Simple answer: Any non-square rectangle can have a line drawn within it parallel to the short side that partitions it into a square and a smaller rectangle. If the long side of the rectangle is a and the short side is b, the new rectangle will have dimensions of b x (a-b). If this is not obvious to you, remember that one side of the rectangle is the side of the square (which has dimensions b x b), and the other is the part of the long side which is not occupied by the square (thus having a length equal to the length of the long side minus the length of the square). \n\nNow, some rectangles, when so partitioned, have the property that the new rectangle formed by the partition will have the same ratio between the lengths of its sides as the original rectangle. That is, a/b = b/(a-b). Such rectangles are called golden rectangles, and the ratio of the sides of a golden rectangle is called the golden ratio.\n\nThe golden ratio is easily shown to be both irrational and algebraic. Irrationality is proven by contradiction: note that all rational numbers can be represented in lowest terms - that is, as the ratio of two integers that have no common factors other than 1. However, the golden ratio cannot be represented in lowest terms, since if some ratio a/b were a representation in lowest terms, b/(a-b) would be a valid representation in even lower terms, which contradicts our statement that a/b is in lowest terms. Q.E.D.\n\nComputing the golden ratio is done by solving the equation we provided: a/b = b/(a-b). We arbitrarily choose b to be 1 - we could let b be anything, but letting b be 1 makes the ratio a/b much easier to compute, since it will simply be equal to a. Thus we have a=1/(a-1). Multiplying, we get a(a-1)=1 and from that a² - a = 1 Putting this in standard form, we have a² - a - 1 = 0, which shows that the golden ratio is algebraic, since our formula for computing it IS a polynomial - this stands in contrast to many other fundamental constants like pi and e, which cannot be expressed as a root of any polynomial of any degree (that is, they are non-algebraic).\n\nCompleting the square then yeilds:\n\na² - a - 1 = 0\na² - a + 1/4  = 5/4\n(a - 1/2)² = 5/4\na - 1/2 = ±√(5)/2\na = (1 ± √5)/2\n\nHowever, since the golden ratio is a ratio of lengths, and all lengths are positive, we know that it cannot be (1 - √5)/2, since that is negative. Thus the golden ratio is (1 + √5)/2 exactly.
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Discounting rotations and mirror images, there is only one way to make a magic square using the digits 1 through 9.  \n\nFirst, we can prove the 5 has to go in the middle:\n\nLet the numbers in the top row from left to right be a, b, and c.  Let the numbers in the middle row from left to right be d, e, and f.  (Note e is the middle box.)  Let the numbers in the bottom row from left to right be g, h, and i.     \n\nWe know a + b + c = 15 and g + h + i = 15\n\nWe also know a + e + i = 15, b + e + h = 15, and c + e + g =15.  Therefore, (a + e + i) + (b + e + h) + (c + e + g) = 15 * 3 = 45.  We can rearrange this to (a + b + c) + (g + h + i) + 3e = 45.  Since a + b + c = 15 and g + h + i = 15, that reduces to 15 + 15 + 3e = 45, or 3e = 15.  Therefore e (the middle box) must equal 5.   \n\nKnowing the middle box is 5, try 9 in a corner and you'll notice a problem - the 6, 7 and 8 can't all be placed in the square without the row or column with the 9 exceeding 15.  So, the 9 has to occupy a side box.  Since we are discounting rotations and mirror images, we can arbitrarily select box b for the 9.  This forces the placement of the 1 in box h.   \n\nWith the 9 and 1 now placed, the 8 and 6 have to go in the row with the 1.  So, we'll put the 8 in box g and the 6 in box i.  (We still haven't done anything to create a second square if rotations and mirror images are discounted.)  That forces the placement of the final four numbers.
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The interior and exterior angle add up to 180 degrees; so far the smallest would be 180 - 110 = 70 degrees.  But since there are 5 angles in a pentagon, we have to first make sure that the unknown angle of the pentagon is not larger than 110 (making it's supplement, the exterior angle, smaller than 70 degrees.\n\nDraw a pentagon on your scratch paper (it doesn't have to be perfect) and pick one of the vertices (the "corners" where the sides meet).  If you draw a line from that vertex to all of the other vertices, you see that you've divided the pentagon into 3 triangles.  Since each triangle has 180 degrees, the whole pentagon has angles that add up to 3 x 180 or 540 degrees.  From the four angles you've been told (85 + 90 + 95 + 110 = 380) you can figure out that the fifth angle has to be 160 degrees (540 - 380 = 160).  So the exterior angle that matches up with that angle is 180 - 160 = 20 degrees and choice B. is correct!
351328
Assume the rectangle has dimensions a x b\nDraw a radius to one of the verticies of the rectangle.\nThen drop a perpendicular from the center of the circle to the midpoint of one of the sides of the rectangle.\nThis forms a right triangle with sides a/2, b/2 and hypotenuse 5.\n(see picture: http://mudandmuck.com/str2/optimization.JPG )\n\nSo (a/2)^2 + (b/2)^2 = 5^2 = 25\nor a^2/4 + b^2/4 = 25\na^2 + b^2 = 100\nand b = √(100 - a^2).\n\nNow we want to maximize the area of the rectangle..\nThe area y = ab = a√(100 - a^2)\n\nTake the derivative with respect to a and set = 0.\n\ndy/da =  √(100 - a^2) +  a * (1/2) * (100 - a^2)^(-1/2) * (-2a)\n\ndy/da = √(100 - a^2) - a^2/√(100 - a^2)  = 0\n\nmultiply both sides by √(100 - a^2) to get:\n\n100 - a^2 - a^2 = 0\n2a^2 = 100\na^2 = 50\na = √50 = 5√2\nb = √(100 - a^2) = √50 = 5√2\n\nSo the rectangle is indeed a square of side length  5√2
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Use graph paper and graph out this parallelogram..\nA(-6,0) B(0,10) C(6,0) D(0,-10)\nAngle A = ang C (definition of parallelogram)\nAngle B = ang D (def. of para)\nThe diagonals bisect the angles(given)\nThe diagonals bisect each other(def. of Para)\nUsing side-angle-side you can prove all the triangles formed by the diagonals are congruent. So that means all sides of parallelogram are equal and you have a rhombus.
123024
You have to draw the lines and mark the points.\nIf Cp has a 90 degree angle with the line you just draw and crosses it right in the middle, then you have graphic proof that it is the perpendicular (which means it has a 90 degree angle) bisector (which means that it splits another line in two halfs).