the missing side of a triangle?
596750
The Pythagorean theorem only works with right triangles.  Try using SIN, COS or TAN.
792000
I suppose the triangle is right-angled.\nLet the unknown side be 'x'\nIn that case,\n5^2 + x^2 = [9*(sqrt3)]^2\n25 + x^2 = 243\nx^2 = 243 - 25\nx^2 = 218\nx = sqrt(218)\n   = 14.7648230602334(approx)
1084262
use pythagoras theorm,if left hand side is equal to right hand side then it is a right triangle.\n\nsince 8 is the biggest no here that means the length of hypotenues is 8. \n\nlet side 1 be a=4\nlet side 2 be b=4 sqrt(3)\nlet hypotenuese be c=8\n\naccording to pythagoras theorm,\n\n(a)*(a)+(b)*(b)=(c)*(c)\n\nL.H.S\n\n=4*4+4sqrt(3)*4sqrt(3)\n=16+48\n=64\n\nR.H.S\n=8*8\n=64\n\nsince L.H.S=R.H.S\n\ntherefore, it is a right triangle\n\nhope u understand it\nif u hv any prob u can contact me
985667
They taught us SOHCAHTOA back in OL trig, but I never really felt the need for it, I mean just picture a right angled triangle and the longest side is the hypotenuse,h. Take an angle and the side opposite it is h.sin x , where x is the angle. The other side is h.cos x
1139471
Yeah where's angle 1 and 2.\nBut remember the angle opposite the longest side has the greastest measurement.
1028326
1 - A is enough because with the lengths of 2 sides you can use pythagoras' theorem to work out the length of the other
967287
The triangle will have 2 sides of length x, and the remaining side will be length P-2x making the parameter x + x + P - 2x = P.\nBegin by drawing a line from the vertex where the two sides of length x meet, to the center of the side which is length P-2x. The line should be perpendicular to the side P-2x thereby splitting the triangle in half, into 2 right triangles. One of the right triangles has hypotenuse = x and one leg = (1/2)(p-2x). So the remaining leg has length S.R. (x^2 - (p/2 -x)^2). We calculate the area of this right triangle, and double it since there are two triangles:\nArea = (2)(1/2){(1/2)(p-2x)S.R.(x^2 - (p/2-x)^2)\n         = (1/2)(p-2x)S.R.{px - (p^2)/4}\n         = (1/4)(p-2x)S.R.(4px-p^2)\nNote that your given answer is wrong - it has an x^2 where it should have a p^2....
776798
Every side of this triangle has a pair of ends and all the three pairs are cyclic. ie, if AB,BC & CA are the three sides of the given triangle , then the side AB has the ends (4,5),(1,1), BC has the ends  (1,1)and (-2,5) & CA has the pair of ends (4,5),(-2,5).\nBy simple inspection of these pairs we can find that the points C and A are at the same height from the abcissa hence parallel to it. By this we can conclude easily that the length of CA=6 units.\nFurther, to be an isocelous triangle in this case the point B should be on the mid line of CA, and it is normal to CA. The mid point of CA is three units away from its ends C&A. As CA is parallel to absissa we can clearly conclude that any point between (1,0)&(1,5) will be the vertex of an isosceles triangle. Here the point B is (1,1) on the said line. Thus proved that the given points form an isosceles triangle.\nWe alredy know that CA = 6\nAnd the other sides are equal to sqrt(4-1)^2 +(5-1)^2 = 5 units\nNow, we can find the area of this triangle by Heron's Formula \nie sqrt[8(8-6)(8-5)^2] =12 sq units
544028
You have to have a right triangle.\nThen you figure out which is side is opposite or adjacent from the angle you need.
726169
It is possible only when the given triangle is isosceles or equilateral. It's also possible for right-angled triangles but only when the given angle isn't 90 degrees.
1002935
Its the Pythagorean Theorem. It applies to right angle triangles. \nIt states that a^2 + b^2 = c^2\nWhere a and b represent the lengths of legs of the triangle and c is the Hypotenuse.\nThe Hypotenuse is the side that is opposite the right angle (also the longest side). Whereas the legs are the two sides that meet at the right angle.
506764
well, id need a pic to explain it...but, you can draw two lines from the midpoint of one of the sides, and then you have 4 triangles...the big one, and three little ones inside...
785502
This is a right isosceles triangle.  It's right because the Pythagorean Theorem works and it's isosceles because 2 sides are equal.
219885
triangle- 3side\nquadrilateral- 4side\npentagon- 5side\nhexagon- 6side\nheptagon- 7side\noctagon- 8side\nnonagon- 9side\ndecagon- 10side\ndodecagon- 12side\nicosagon- 20side
1074757
The problem is that 30, 31, and 63 are not the sides of a valid triangle.  For any triangle with sides a,b,c, you must have\na + b > c\nb + c > a\nc + a > b\nBut you have a + b < c, so the triangle isn't valid.
392786
you may be dealing with "right" triangles, but you don't identify the angle that is 90, so your question is bout impossible to answer. this type of question usually involves  \na squared + b squared = c squared (a^2 + b^2 = c^2) where a and b are the 'legs' of the triangle in question. (the two 'legs' of a right triangle make the 90 degree corner) :)
991928
Assuming you mean a regular hexagon, then with the 30-60-90 relationship, you should get that half the length of a side is 2.\n\nSo, the area of that yellow triangle in the link below is\n.5 * 2 * 2 * sqrt(3) = 2 * sqrt(3)\n\nThere are 12 triangles, so the area is 24 * sqrt (3).