whAT IS THE SQUARE ROOT OF AN ISOSOLECES TRIANGLE?
82067
something to do with sex
71806
In an isosceles right triangle, sides a and b are equal and c is the hypotenuse.  Pythagorean's Theorum says that a^2 plus b^2 = c^2 for all right triangles.  here, a and b are equal, so you can say 2(a^2)=c^2.  If you want to solve for c, you find that c = a(rad2)
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
785502
This is a right isosceles triangle.  It's right because the Pythagorean Theorem works and it's isosceles because 2 sides are equal.
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True\nAn isosceles triangle is defined as a triangle which has two equal angles.  It doesn't say it has to be exactly two.
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)
737743
The positive square root is 11.  The cube root is about 4.946 (it's not an integer).
882750
Listen to the first guy.\nThe distance on the y axis is 1 ((-5)-(-4)) and on the x axis it is 3 ((-3)-0).\nThis gives you a right angles triangle with sides of 3 & 1. \nUnder Pythagoras theoem the sum of the squares is equal to the square of the hypoteneuse so\n1^2 + 3^2 = 10.\nTherefore the distance is the square root of 10.
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Heron's formula for triangles:\na = √s(s-a)(s-b)(s-c)\ns is the semiperimeter or 1/2 (4+4+4)\ns = 6\na = √6(2)(2)(2)\na = √3×2²×2²\na = 2×2×√3\na = 4×1.7\na = 6.8\n\n\nJose, you calculated the area of the right triangle, not the entire equilateral triangle you started with.
803140
OK.. First of all it's an isosceles triangle.. thus the base is divided by the vertical line into two equal halves of 3 and 3 each.. \nNow just look at one triangle.. say the left one.. it is a right traingle (the vertical line is perpendicular to the base) having base of 3 and hypotenuse of 21. \nThus if the vertical height is 'h'     => h^2 + 3^2 = 21^2   (^2 means square.. ie power 2) \nsolving for h = (441 -9)^(1/2) = 20.78  (use a calculator)\nNow for theta..         Cos a = Base / hypotenuse \n                                 cos a = 3/21 = 1/7\n                                 a= cos^-1 (1/7)          (Cosine inverse)\n                                 a = 1.43 radians         (Use a calculator :-) )\nGuess thats it !!
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5.196 (3 times root 3)\nDive the triangle into 3 equal triangles.  Then divide each one of those triangles into 2 triangles.  The hypotnuse of each triangle has a length of 2, and by trig. (it is a 30-60-90 triangle) the base has a length of the square root of 3 and the height is 1. Area of a triangle =(base times height)/2=(radical 3)/2.  There are 6 smaller triangles total, so multiply the area by 6=total area=3 times radical 3 (5.196)\nHope this helps, it would be a lot easier if I had a picture that I could post and it would make more sense.
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The sides of the initial right isosceles triangle are 1 and 1.  Therefore the area = 1 x 1 x 1/2 = 1/2.\n\nThe sides of the enlarged triangle are 8 and 8.  Therefore the area = 8 x 8 x 1/2 = 32\n\nDilation:\n\nAssuming the preimage is the smaller of the 2 images:\nThe are of the preimage = 1 therefore the sides of the preimage square is 1 by 1.\n\nIf the scale factor is 1/4 , then the sides of the image square is 4 by 4 therefore the area = 16 and the perimeter = 16 (4+4+4+4)\n\nInversely if the preimage is the larger of the 2 images, the size of the image sqaure is 1/4 by 1/4 and the area = 1/16 and the perimeter = 1 (1/4+1/4+1/4+1/4)
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let the base of the isosceles triangle be x\nheight of the triangle=h\ngiven the radius of the circle is 4\n(4-h)^2+(x/2)^2=4^2\n16-8h+h^2+x^2/4=16\n4h^2-32h+x^2=0\nx=rt(32-4h^2)\narea=1/2*x*h\n=1/2*h[2rt(8-h^2)]\ndA/dh=0 for max\nafter finding h sub to get x
978895
Can you use a square root symbol?  If so, sqrt(9) + 9 - 9
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30 squared is 900.\n\n1/2 of that is 450\n\nthe square root of that is 21.213, so this is the length of the side.  this works because the the diagonal of a square gives us an isoceles right triangle, and the rule for calculating the sides of a right triangle are \na2 + b2 = c2
542378
Assuming you are using degrees, the answer is 9. If you're looking for the square root of that, it would be 3. \n\nIf you want to figure it out yourself, remember that cot is 1/tan, cot^2 = 1 / tan^2.
545467
The lenghth of the missing side is 20 in. because an isosceles triangle has only 2 alike sides so you multiply 26.5 X 2 and get 53in and then subtract 53in. from 73in. and get the 20in.