area of the isoceles triangle???
443635
You can use Heron's formula if you know the length of all three sides. \nHeron's Formula:\nLet s be the semiperimeter (which is half the perimeter) and a,b,c be the length of the sides, then the area is\nA = sqrt[s*(s-a)*(s-b)*(s-c)]\nperimeter = 4+4+6 = 14\ns = 14/2 = 7 (because s is one-half the perimeter)\nA = sqrt[7(7-4)(7-4)(7-6)]\n= sqrt(7*3*3*1)\n= sqrt(63)\n= 3*sqrt(7) which is approximately 7.93725\n\nYou can also use the Pythagorean Theorem to find the height.
669681
1.  Area of rt triangle whose leg is 6ft and whose\n    hypotenuse is 10ft?\n\n    Area = 24 sq. ft.  \n\n  Explanation:\n\n    First we need to find the other leg's length...\n    Using Pythagorean Theorem,\n    \n    c**2=a**2+b**2 \n       \n    where c is the hypotenuse, a is one leg, and \n    b is the other leg.\n\n    So, 100 = 36 + b**2;  b**2 = 100-36 = 64; therefore b=8.\n    Now, Area = 1/2(base*height) since the base is one leg\n    (we'll say it is a) and the height is the other leg \n    (therefore we'll name it b) \n    the Area = 1/2(8*6) = 1/2(48)= 24 sq. ft.\n\n2.   The hypotenuse is approx. 4.9 meters.\n    \n  Explanation:\n     \n     Using Pythagorean Theorem: c**2=a**2+b**2 we find that\n     c**2 = (3.5)(3.5) + (3.5)(3.5)\n     c**2 = 12.25 + 12.25\n     c**2 = 24.25\n     c = sq. rt. (24.25)\n     c = 4.9244 meters\n     hypotenuse is approx. 4.9 meters.\n\n3. hypotenuse is approx. 20 inches long.\n\n  Explanation: Area = 1/2(x)**2 \n               where x is length of one leg.\n\n     x**2 = 2*Area;\n     x**2 = 2*98\n     x**2 = 196\n     x = sq. rt. (196)\n     x = 14 inches\n\n     Now, to find length of hypotenuse, we use the\n     Pythagorean Theorem:\n\n     c**2 = a**2 + b**2; since a=b this can be rewritten so\n     c**2 = 2(a)**2\n     c**2 = 2(14)**2 -- Remember 14 is the length of a leg.\n     c**2 = 2(196)\n     c**2 = 392\n     c    = sq. rt. (392)\n     c = 19.7989\n     c is approx. 20 inches long.\n\n\n4.   The researcher will climb approx. 10 meters \n     to reach the crack.\n\n   Explanation:\n      \n      Using the Pythagorean Theorem: c**2 = a**2 + b**2\n      If we set a = 24, then c**2 = (24)**2 + b**2,\n      but we also know that c + b = 36 \n      which means c = 36 - b.\n\n      So to solve for this we need to set\n      (36-b)**2 = (24)**2 + b**2\n      1296-72b + b**2 = (24)**2 + b**2\n      1296 - 72b = (24)**2\n      1296 - 72b = 576\n      -72b = 576 - 1296\n      -72b = -720\n         b = -720/-72\n         b = 10 \n\n      And now we have our 10 meters the researcher \n      must climb to reach the crack in the tree.\n\n\n6.    Approximately a 42 inch stick is the longest stick\n      that will fit in a box that is 24 in X 30 in X 18 in.\n\n  Explanation:\n\n      First use Pythagorean Theorem to find the diagonal\n      inside the rectangle of 24 in X 30 in.\n\n      c**2 = (24)**2 + (30)**2\n      c**2 = 576 + 900\n      c**2 = 1476\n      c = approx. 38.418 in. -- we will call it 38 in.\n\n      Now, use Pythagorean Theorem to find the diagonal\n      inside the "new and invisible" rectangle that is \n      38 in. X 18 in.\n\n      c**2 = (38)**2 + (18)**2\n      c**2 = 1476 + 324\n      c**2 = 1800\n      c = sq. rt. (1800)\n      c is approx. 42.43 in.  \n      \n      We will call "c" 42 inches long.
422511
if u kno the base and height.. then area= (base x height)/2\n\nif the triangle is scalar,\nthen :\n\nfirst find s, using hero's formula\n\ns= (a+b+c)/2   .... where a,b,c are the sides of triangle.\n\nthen , area = squareroot(s x (s-a) x (s-b) x (s-c)
921006
Area = πab where a and b are the semi-major and semi-minor axis length
204855
Looks to me like you are trying to figure out a "PERIMETER" rather than an "AREA".  If so, you need to ADD the lengths of all the sides together.  If you ARE looking for an area, it is simply Length x Width (for a rectangle or square), 1/2 the base x height (for a triangle), or Pi x radius squared (Pi is approximately =3.1415926) for a circle.  If the "shape" of the object you are trying to find the area of is irregular, then you must divide it up into these familiar shapes first, find EACH individual area, then add those areas together for the total area.
890633
1. total area of a box of sides a,b,c is 2(ab+bc+ca)\na=4,b=3,c=6 => surface area = 108 sq.ft\n..\n2.Lateral area = pi*radius*slant ht\npi=3.1415,radius=2m(4m/2), slant ht=15m\nans is 188.49sq.metres\n..\n3.   volume of pyramid = (1/3)* height * base area\n= (1/3)* (6) * (6*6)\n= 72 cu.metre\n..\n4. lateral area of cylinder = 2*pi* radius* height\n=2*3.1415*12*60\n=4523.76cu. mm\n..\n5. total surface area = outer surface + inner surface + top surface + bottom surface\n= 2*pi*8*height + 2*pi*5*height + 2 times ( pi*(8-5)^2)\n[the last term is the sum of top and bottom surfaces which are circular rings]\nheight is not given :(
143817
The first area to be defined was the obvious one of a rectangle.  The number of squares of unit size that would fill up a rectangle is clearly the number of such squares that fit in a row across one dimension of the rectangle, times the number of such rows needed.  Since each square has a length and width of one unit, it's easy to see that the length x width = area of a rectangle.\n\nA rectangle can be divided into two right triangles of equal size.  Hence, the area of a right triangle is 1/2 x length x width, only in triangles we call those things base and height.  Now, any arbitrary triangle can be broken up into two right triangles.  Doing so, it's easy to show that the area of *any* triangle is 1/2 x base x height, where the height is defined as the perpendicular distance from the line containing the base to the opposite vertex.  Similar techniques using triangles can be used to find the area of other regular polygons. \n\nThe area of things like circles can be computed from integral calculus, although I suspect there is probably a geometric method to find the area of a circle as well.\n\nThe area of figures bounded by more complex functions than straight lines and circles is found using integral calculus.
1081188
Here's simply why:\n\n2 isosceles triangles are not always similar. The common angle for an iso triangle is 45-45-90. Other isosceles triangles have angles 20-20-140, or 70-70-40, etc (2 angles are the same and all three add up to 180).\n\nFor equilateral triangles, they are always similar. All equalleral triangles have the angles 60-60-60, no matter what.
572270
triangle area = 1/2(height * width)\n\ncircle area = pi * radius^2
460116
It depends what it looks like.\nWithout seeing the actual picture I cannot find the area.
708770
angle cdb = ancle cbd   (since traingle cbd is an isosceles)\n\nangle bcd =  180  -  (2*65)   =   50\n\nangle  acb =   180 - 68 - 30 =  82\n\n\n\nUsing sine rule,  (considering traingle  abc)\n\n(ab/sin 82) =  (bc / sin 68)\n\ngiven: ab= 5  ,  therefore  bc = 4.68   (2 decimal places)\n\n\n\n\nusing sine rule again,   (for triangle bcd)\n\n(bd/sin 50) =  (bc / sin 65)\n\nbc= 4.68   ,  therefore  bd = 3.96
196800
Surface area. Seriously. 8D
680192
The area of a triangle is 1/2 the base time the height.\n\nA=1/2bh\nA=1/2(10)(4)\nA=(5)(4)\nA=20 units^2
833189
Area = 5 square meters\nBase = 2x\nArea of any triangle = (Base*Height)/2\n5 = (2x*Height)/2\n5 = x*Height (The 2's cancel each other out)\nx*Height = 5\nHeight = 5/x(Divide both sides by 'x')\n\nI hope that helped
619584
Do you mean inside area of triangle ?\n\nArea= 1/2 (base)(height)
175713
Your question is too complicated to appreciate the math in it, which is supposed to be the real beauty in the first place. Anyway, I have tried figuring it out and come with equations. I do not even want to try and begin substitution and end up with more complications. I just thought once you saw the relation, you would throw those 10 points my way.\n\nAssuming a to be sides of the squares of the prism and b to be the length of the rectangles, we know that the sum of all dimensions is 14a+5b. Now I do not even want to get started why you brought it the story of the equilateral triangle. You can just say the sum  of the dimensions is 61. Anyway, since any side of the trianle, as it is equiangular hence equilateral, is twice 14a+5b and 122, the sum of the dimensions will be 61. So your first equation is 14a+5b=61\n\nYou know the volume of the prism is the length X width X heiht = a x a x b = (a^2)b This is equal to the surface area. The surface area is the sum of all the areas on the prism and that is a^2+4ab. (How about you calculate how that came up?)\n\nSo there is your second equation, (a^2)b = a^2+4ab So then from 14a+5b=61, express a in terms of b or the other way around and plug it into the second equation and experience brutal complications and figure out what the variables are. I think you know it is too easy to calculate each dimension and area after that.
636579
Okay, I got something really different, but it might be wrong, I used tangents and stuff to figure it out, and I assumed a dodecagon had 12 sides, but I might be wrong. Anyways, I got 38.4 for the perimeter, and 115.2 for the area.