an isosceles  triangle has fixed perimeter P ( so P is a constant).if x is the length of one of the two equal?
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....
764538
For any given triangle ABC with sides a, b and c.\n\nCos C = (a^2 + b^2 - c^2)/2ab  This is the cosine rule.\n\nSin C =sqrt(1 - Cos^2 C). You may need to adjust the sign of the sign to match the triangle.
906875
The trick is turning the word problem into equations.\n(Key: length = l; width = w)\n\nThey tell you the length is 3 feet more than twice the width, so:\nl = 2w + 3\n\nThen they tell you the perimeter (which is twice the width and twice the length) is 36, so:\n2l + 2w = 36\n\nYou should be able to take it from there.
788933
the value of pi is always the same because the ratio between circumference and diameter will remain constant
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
804994
Let the Length of the side of the rectangle be x and the length of the side of the triangle is L\n\nThen when the rectangle is inscribed it is symmetrically placed (otherwise you would not get a rectangle). So the distance from its base vertices to the vertices of the triangle are each ½(L - x)\n\nThus the width of the rectangle = ½(L - x) tan 60°\n= √(3)/2 * (L - x)\n\nThus the area of the rectangle A = length * width\n= x * √(3)/2 * (L - x)\n= √(3)/2 * (Lx - x²)\n\ndA/dx = √(3)/2 * (L - 2x)\n= 0 for stationary points\nWhence L - 2x = 0 ie x = ½L\nd²A/dx² = -√(3) < 0 for all x Thus the curve is always concave down So the stationary point is a maximum.\n\nThus the length of the rectangle is ½L and the width is √(3)L/4\n(and its area is √(3)L²/8)
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
442290
1. pythagoros theorem applies to any right angled triangle.\n\n2. The geometric mean of a set of positive data is defined as the nth root of the product of all the members of the set, where n is the number of members.\n\n3. in an isosceles triangle 2 sides and 2 angles are equal\n\n4.hypotenuse is opp. to the right angle.it is the largest side.\n\n5.true
783069
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.
1015708
Do you mean a "fixed perimeter?"
976501
You mean "one less."\n\nCall the base x.\nThen each leg is 2x-1.\nAnd the perimeter is x + (2x-1) + (2x-1), which equals 18.\n\nJust solve for x and you can calculate the lengths of the sides.
551409
you can cross a line from one of them (A) equal (just like the line in the down ) and have the triangle just equal (B) and also for C . then yu have them in a line and we know that a line is 180 .
331358
it's the Pythagorean Theorem\n\nIn any right triangle, the area of the square whose side is the hypotenuse (the side of a right triangle opposite the right angle) is equal to the sum of the areas of the squares on the other two sides.
915466
Let a and b the lengths of the sides of the rectangle (with a the side with the semi-circle).\nThe perimeter is 2b+a(pi/2+1)=16, and the area ab+pi/8*a^2. Substituting b in the area (from the perimeter equation), the area is 8a-a^2*(1/2+pi/8). \nSetting the derivative to 0, we have a=32/(4+pi). Then b comes from the perimeter constraint as b=8*(1-(2+pi)/(4+pi)).\nNumerically this gives a=4.48, b=2.24 and an area of 17.92
552905
The AREA of an isosceles triangle is found by taking 1/2 the sum of the long and short parallel sides, times the height. One way to compare this with the parallelogram is to take two copies, turn one 180 degrees and glue them along the equal (nonparallel) edges. \nYou now have a larger parallelogram - with one side composed of the sum of the original short and long parallel edges. Use the usual formula for the area of this parallelogram, then divide by 2 (since it is composed of two equal copies).
343312
perimeter = 2 (side1 + side2)\narea = side1 * side2
324878
I'm assuming you are asking for a formula to calculate the area when you just know the base (x) and not the height.  But, you need to know the height, even if the triangle is a right triangle.  To see this, imagine you had a right triangle standing on it's shortest side.  That shortest side would be your base, x.  Now stretch the triangle taller and taller.  The base length still stays the same, but the area inside the triangle grows and grows.  (And the triangle is still a right triange.) So, unless you know the height, you can't know the area.\n\nNow, what I'm guessing, is that your triangle might actually be not only a right triangle, but an isosceles triangle too, meaning that 2 of the sides are the same length.  And because it's a right triangle too, that means the two shorter sides (that connect at the right angle) are the same length.  So, your base and your height are the same length.\n\nIf this is true...\n\nlet x = length of base\nsince triangle is isosceles and right, height also = x\n\nSo the traditional formula becomes:\n1/2 BH = 1/2 x*x\n(sorry, can't figure out how to write "x squared" right now)\n\nHope that helps...