JUST ANSWER T OR F...\n1.All diagonals from a single vertex of a hexagon are drawn. Four triangles will result.\n2.a PERSON WHO LOOKS AT A SEQUENCE OF 14, 12, 10, 8 AND DESIDES THAT THE NEXT TWO TERMS ARE 6 AND 4 IS DOING DEDUCTIVE THINKING?\n3. If "if p,  then q" is the conditional statement, then "f q, then p" is the converse?\n4.a square is a rhombus?\n5.an equilateral triangle is an isosceles triangle.?\n6.the legs of a trapiziod are parallel?\n7.the compliment of an acute angle is in obtuse angle?\n8.Vertical angles can never be complimentary?
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I think you're asking too much to request that we do your homework for you. If you ask a single question, perhaps someone would get you started so that you could figure out the answer for yourself. No one wants to deprive you of an education and the pride that comes with learning and doing for yourself.
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Given the parallelogram ABCD with angle A supplementary to angle B, and angle C supplementary to angle B.\n\nSince angle A supp to angle B, then AD parallel to BC because the interior angles on the same side of the transversal are supplementary.\n\nSimilarly, since angle C supp to angle B, then AB parallel to DC because the interior angles on the same side of the transversal are supplementary.\n\nTherefore ABCD is a parallelogram by definition.\n\nQ.E.D.
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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.
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Draw a picture - a square with corners (0,0), (2,0), (0,2), (2,2), showing the range of possible values for x and y.\nNow, what part of that square has the difference < 1? Its a hexagon to be precise; its the entire square except for a little triangle at the top left, and a little triangle at the bottom right. The two sloping sides go from (0,1) to (1,2) and (1,0) to (2,1).\nThose triangles each have area 1*1/2 = 0.5, so area 1 combined.\nSo the total area left is 4-1 = 3, so the answer is 3/4.
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PART (a)\nUsing Pythagoras’ Theorem in triangle ABC,\nAB^2 + BC^2 = AC^2 = 85 [since AC = sqrt(85)]\nBoth sides AB and BC has to be positive integers.\nBy trial we take AB = 6, BC = 7 [since 6^2 + 7^2 = 36 + 49 = 85]\n\nAgain, using Pythagoras’ Theorem in triangle CDE,\nCD^2 + DE^2 = CE^2 = 85 [since CE = AC = sqrt(85)]\nBoth sides CD and DE has to be positive integers.\nBy trial we take CD = 2, DE = 9 [since 2^2 + 9^2 = 4 + 81 = 85]\n\nThus we have the length of four sides of the hexagon ABCDEF,\nAB = 6, BC = 7, CD = 2, DE = 9.\n\nNow, let us concentrate on triangle AFE.\nIn any triangle, the sum of any two sides is always greater than the third side.\nSo, in triangle AFE, AF + FE > AE\ni.e. AF + FE > sqrt(85) = 9.22 [since AE = AC sqrt(85)]\n\nAs length of 4 sides of the hexagon are fixed so for minimum perimeter the sides of triangle AFE has got to be minimum. As AE = sqrt(85) is also fixed so the other two sides has to be minimum and hence their sum has to be minimum i.e. (AF + FE) has to be minimum. But AF + FE > 9.22 and also both AF and FE has to be positive integers, \nso (AF + FE)min = 10.\n\nNow, 10 = 1+9 = 2+8 = 3+7 = 4+6 = 5+5 \n\nSides AF and FE cannot attain any pair of these values (1,9; 2,8; 3,7; 4,6; 5,5) as length of each side of the hexagon has got to be different positive integers and we already have the nos. 2, 6, 7, 9 as length of 4 sides.\n \nSo, AF + FE = 11 [next positive integer to 10]\nAgain 11 = 1 + 10 = 2 + 9 = 3 + 8 = 4 + 7 = 5 + 6\n\nThus, AF and FE can attain 2 pairs of values, either 1 and 10 or 3 and 8. \nSo the minimum value of (AF + FE) of possible values of AF and FE is 11.\nHence the minimum value of perimeter of the hexagon is \nAB + BC + CD + DE + EF + (FE + AF)\n= 6 + 7 + 2 + 9 +11\n= 35 \n\nPART (b)\nTo find a minimum value of AC, AC has to be less than present value sqrt(85). Further there must be two pairs of positive integers whose sum equals AC^2 (which is less than 85). So those positive integers has got to be in the range 1 through 9. The squares of these positive integers are presented below. One cannot find two pairs from these numbers such that the sum of each pair is equal to any number less than 85. \n1 4 9 16 25 36 49 64 81\nSo the minimum value of AC is sqrt(85) and the corresponding perimeter of the hexagon is 35.\n\nTo get the figure of this hexagon give me your email id so that I can send it as attachment.
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The figure we are talking about is an isosceles trapezoid. The lower base is equal to the two diagonals and the upper base and the two sides are all equal in length.\n\nCall the trapezoid ABCD with AB being the lower base and cd bbeing the upper base. The two sides which are both equal to the upper base are BC and AD. Also notte that AB parallel to CD.\n\nThe diagonals are AC and BD. \n\nIt turns out that the two base angles are bisected by each of the diagonals.Since the base angles are equal then they are each divided into two equal angles. Call each of them angle 1. also By the parallel postulate, angle CDB and angle must also be equal to angle 1.\n\nNow for the base to be the same length as the diagonals, triangle\nABC and Triangle must both be isosceles. This will be true only if angle ADB=.twice angle 1 and angle ACB = twice angle 1.\n\nTherefore the two base angles add up to 4 angle 1s and the agles C and D are both = 3 angle ones. Thus the four vertex angles add up to 10 angle 1s. Since the total of all angles in a quadrilateral is 360 degrees, the angle 1 must 360/10 =36 degrees.\n\nThus each base angle = 2 angle 1s = 2*36 =72 degrees and each of the upper angles must be equal to 3 angle 1s =3*36=108 degrees.\n\nIn summary angle A and angle B = 72 degrees and\nangles C and D both = 108 degrees.\n\nHope this was not too confusing. It will help if you draw a picture.
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Your first Question:\n\nNow,\n\nImagine a Triangle with vertices ABC. \nAny line from A to BC is AD. D is the point of intersection on BC.\n\nLet E and F be the midpoints of Sides AB anc AC respectively.\n\nJoin E and F to meet AD in G.\n\nNow for the proof:\nTake Triangles AGE and ADB these two are similar.\n\nReason being that EF would be parallel to BC from Mid Point theorem. Therefore, the two triangles will be similar as Angle E and Angle B are equal and Angle A is common. \n\nNow take Triangle ADB, E is the mid point of AB and EG is parallel to BD there for from Converse of Midpoint Theorem G becomes the Midpoint of AD.\n\nHence proved.
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The question apparently is "How many diagonals does a polygon with n sides have?"  \n\nThe first formula is n(n-3)/2.  \nOne way to see this is to notice that you can draw (n-3) diagonals from every vertex of the polygon. This is because there are (n-1) other vertices, but two of them are adjacent vertices and so don't count towards making diagonals. This seems to give n(n-2) diagonals, but this way of counting counts every diagonal twice since each diagonal connects to two vertices, so you have to divide by 2.\n\nBy a recursive formula, we mean a way of expressing the answer for n vertices in terms of the answer for n-1 vertices. Suppose, for example, that you already know the answer for a polygon with n-1 vertices. Now if you add another vertex between two of the vertices of the original polygon, then all the diagonals of the original polygon will still be diagonals of the new polygon, and so will the side joining the two vertices that you added a new vertex between, and so will the line segments joining the new vertex to all the other vertices of the original polygon. (Got all that?) So if we let diag(n) be the number of diagonals for a polygon with n sides, we get the formula:\n\n        diag(n) = diag(n-1) + n - 3 + 1 or\n\n                  diag(n-1) + n - 2\n\nThe first formula is better, since it actually gives you the answer.  \nBut sometimes it's easier to get a recursive formula first and use \nthat to get an explicit formula (your first formula is an explicit one \nsince you only need the number of vertices in the polygon to get the number of diagonals in that polygon).  This is called "solving the recursion."  Sometimes a recursive formula is the best you can do because there simply is no explicit formula.\n\nI hope this helps.
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well, geometry proofs become very easy when you get used to them. what you have to do in a proof is to first state what is given by the diagram or figure, and then look at the properties that the figure has. for example if you are looking at a rectangle and they tell you to prove it is a rectangle and they show you its diagonal you should put  that the diagonal is congruent to itself becuase of the reflexive property, and if you were given to sides more you should put that the quadrilateral is a recyangle becuase of he SSS theorem (Side-Side-Side) if you need more help just send me an email i am in 10th grade and i am stuidying geometry\n\nGood luck!!
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Take the average between the side of an inscribed square\nand that of an inscribed hexagon. With a hexagon you\nget an isosceles triangle with base as a side and a 60\ndegree angle at the apex, leaving 120 degrees to be split\nbetween the equal pair which are then also each 60 degrees.\nSo you have an equilateral triangle whose base is equal\nto the radius of the circle and of course the base is a side\nwhich has a length of 1. \n  Now inscribe a square. There are now 4 isosceles\ntriangles with a 90 degree angle at the apex and a pair\nof 45's at the base. Draw an altitude from the apex, \nperpendicular to the base, thus bisecting the 90 at\nthe apex making another 90, 45, 45 triangle whose\nside is 1/2 the side of that of the square. The hypotenuse\nis a radius of the circle so the side of the last mentioned\ntriangle is 1/2^2 (.707). This makes the side of the\nsquare 1.414. So averaging 1 and 1.414 you get 1.207.\n Using trig we find the real answer to be 1.1755.\nSo the percentage of error is 2.67 percent from 1.207.\nI wouldn't win an engineering prize with that kind of\naccuracy, but I think I could build a pretty good\ncatapult.
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64 proofs\nexplanation\nProof #1\nThis is probably the most famous of all proofs of the Pythagorean proposition. It's the first of Euclid's two proofs (I.47). The underlying configuration became known under a variety of names, the Bride's Chair likely being the most popular.\n\n    \n\nThe proof has been illustrated by an award winning Java applet written by Jim Morey. I include it on a separate page with Jim's kind permission.\n\nFirst of all, ABF = AEC by SAS. This is because, AE = AB, AF = AC, and\n\n  BAF = BAC + CAF = CAB + BAE = CAE.\n \n\nABF has base AF and the altitude from B equal to AC. Its area therefore equals half that of square on the side AC. On the other hand, AEC has AE and the altitude from C equal to AM, where M is the point of intersection of AB with the line CL parallel to AE. Thus the area of AEC equals half that of the rectangle AELM. Which says that the area AC2 of the square on side AC equals the area of the rectangle AELM.\n\nSimilarly, the are BC2 of the square on side BC equals that of rectangle BMLD. Finally, the two rectangles AELM and BMLD make up the square on the hypotenuse AB.\n\nThe configuration at hand admits numerous variations. B. F. Yanney and J. A. Calderhead (Am Math Monthly, v.4, n 6/7, (1987), 168-170 published several proofs based on the following diagrams\n\n   \n\n\n\nProof #2\n    \n\nWe start with two squares with sides a and b, respectively, placed side by side. The total area of the two squares is a2+b2.\n\n    \n\nThe construction did not start with a triangle but now we draw two of them, both with sides a and b and hypotenuse c. Note that the segment common to the two squares has been removed. At this point we therefore have two triangles and a strange looking shape.\n\n    \n\nAs a last step, we rotate the triangles 90o, each around its top vertex. The right one is rotated clockwise whereas the left triangle is rotated counterclockwise. Obviously the resulting shape is a square with the side c and area c2.\n\n(A variant of this proof is found in an extant manuscript by Thâbit ibn Qurra located in the library of Aya Sofya Musium in Turkey, registered under the number 4832. [R. Shloming, Thâbit ibn Qurra and the Pythagorean Theorem, Mathematics Teacher 63 (Oct., 1970), 519-528]. ibn Qurra's diagram is similar to that in proof #27. The proof itself starts with noting the presence of four equal right triangles surrounding a strangenly looking shape as in the current proof #2. These four triangles correspond in pairs to the starting and ending positions of the rotated triangles in the current proof. This same configuration could be observed in a proof by tesselation.)\n\n\n\nProof #3\n    \n\nNow we start with four copies of the same triangle. Three of these have been rotated 90o, 180o, and 270o, respectively. Each has area ab/2. Let's put them together without additional rotations so that they form a square with side c.\n\n    \n\nThe square has a square hole with the side (a-b). Summing up its area (a-b)2 and 2ab, the area of the four triangles (4·ab/2), we get\n\n  c2 = (a-b)2+2ab \n  = a2-2ab+b2+2ab \n  = a2+b2 \n \n\n\n\nProof #4\n    \n\nThe fourth approach starts with the same four triangles, except that, this time, they combine to form a square with the side (a+b) and a hole with the side c. We can compute the area of the big square in two ways. Thus\n\n  (a + b)2 = 4·ab/2 + c2  \n\nsimplifying which we get the needed identity.\n\n\n\nProof #5\n\nThis proof, discovered by President J.A. Garfield in 1876 [Pappas], is a variation on the previous one. But this time we draw no squares at all. The key now is the formula for the area of a trapezoid - half sum of the bases times the altitude - (a+b)/2·(a+b). Looking at the picture another way, this also can be computed as the sum of areas of the three triangles - ab/2 + ab/2 + c·c/2. As before, simplifications yield a2+b2=c2.\n\nTwo copies of the same trapezoid can be combined in two ways by attac
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let \nc=hypotenuse\na,b = legs\n\ntherefore:\nc=25\na=2b-10\n\nform pythagorean theorem;\n\na^2+b^2 = c^2\n(2b - 10)^2 + b^2 = 25^2\n4b^2 - 40b + 100 + b^2 = 625\n5b^2 - 40b - 525 = 0\n\ndividing both sides by 5\nb^2 - 8b - 105 = 0\n\nfactoring the left side\n\n(b - 15)(b + 7) = 0\n\nb + 7 = 0\nb = -7 absurd (no side can be negative)\n\nb - 15 = 0\nb = 15\n\nsolve for a:\n\na = 2b - 10\na = 2*15 - 10\na = 20\n\ntherefor the two legs of the right triangle are\n20 cm and 15 cm\n\nchecking:\nc^2 = 20^2 + 15^2\nc^2 = 400 + 225\nc^2 = 625\nc = 25 cm.\n\n\nthat's it
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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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Unfortunately, the people above have misused geometry.\n\nThe way that you'd solve this in a geometry class would involved quite a bit of trigonometry. Here's a way to solve it without using as much. (I worked it out both ways and both methods give the same result. I also checked the result.)\n\nI'm assuming that your octagon is regular.\n\nDraw a regular octagon. Call the length of each side a.\n\nNow use vertical and horizontal lines to draw gridlines on the octogon. This divides the octagon into four triangles, four rectangles, and a square in the middle.\n\nThe square has area a^2..\n\nThe triangles are 45-45-90 triangles having hypotenuse a. This means that both sides are a / square root of 2. This means that each triangle has area:\n\n1/2 * (a / sqr(2)) * (a / sqr(2)) = a^2 / 4.\n\nThe rectangles have one side that is a side of the octagon (a) and one side that is a leg of the triangle (a / sqr(2)).\n\nThis means that the rectangles have area:\n\na * (a / sqr(2)) = a^2 / sqr(2).\n\nThe octagon's area is equal to the sum of the areas of its regions:\n\n4 triangles + 4 rectangles + 1 square\n\n4*(a^2 / 4) + 4*(a^2 / sqr(2)) + a^2 = \n\na^2 + 2sqr(2)*a^2 + a^2 = \n\na^2 (2 + 2sqr(2)) = 64.\n\nSolve for a, and you get approximately 3.6407 feet.\n\nThat is to say, an octagon having a side length of 3.6407 feet has the same area as a square having a side length of 8 feet.\n\nThis seems contradictory until you realize that an regular octagon isn't made up of just squares and triangles. If they were, then the "diagonal" sides would be longer than the "horizontal" and "vertical" sides.
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Prove that if in an isosceles triangle a line segment does one of the following things it also does the other two.\n\n      A\n       ^\n    /  |  &#92;\n   -------\nB    D    C\n\nAD goes from the vertex to the midpoint of the base \nAD is perpendicular to BC \nAD bisects /A \n\n\nConverse: If the same line does any two of the previous three things, then the triangle is isosceles.
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THE ANSWER TO THE CONGRUENT TRIANGLES IS NO\n\nTwo triangles are congruent if:\n\nall 6 pairs of corresponding angles and sides are congruent. \nCorresponding Triangles Corresponding Congruent Angles Corresponding Congruent Sides\n\nfactor = 1\n\na/f = b/d = c/e = 1\n \nDABC @ DDEF <A @ <D \n<B @ <E\n\n<C @ <F\n AB @ DE \nBC @ EF\n\nAC @ DF\n \n\n  \n\n\n\nThe factor for congruent triangles is 1. Remember this fact for future problem solving activities.\n\n \n\nIf all 3 pairs of corresponding sides in two triangles are the same, then the triangles are congruent.\n\nMove the corners A, B or C of the triangle above. Watch how DDEF changes.  Since the 3 pairs of corresponding sides of the two triangles are equal each time you manipulate the applet, the SSS (Side, Side, Side) Congruence Relation proves each pair of triangles are congruent.
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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."