How do I calculate the area of a triangle with only the length of the sides?
602085
You use Hero's formula. Add together the sides and divide by two (also called the semiperimeter), and let that value be S. Arbitrarily pick sides to be A, B, and C, then do the following:\n\nArea= Square root (S*(S-A)*(S-B)*(S-C))\nAnd you'll have the area.\n\nIncidentally, Brahmagupta's formula works the same way for quadrilaterals that can be inscribed in a circle, only you use A=square root (S-A)*(S-B)*(S-C)*(S-D). S remains the semiperimeter.
850777
Area of figure = Area of equilateral triangle + Area of arc AB + Area of arc BC.\n------------------------------------------------------------------------------------\nNow, to find the area of an equilateral triangle, use the formula\n\nA = ((s^2)/4)*(sqrt3) where s is the length of the side. Here, S= 7, so now solve the equation.\n\n((s^2)/4)*(sqrt3)\n=((7^2)/4)*(sqrt3)\n=(49/4)*(sqrt3)\n=12.25*sqrt3\n------------------------------------------------------------------------------------\nTo find the area of the arc, you need to see angle ABC as a central angle for a circle. Basically, see ABC as a sector of a circle where AB is the radius. Each angle in an equilateral triangle is 60 degrees.\n\nIn order to find the area of a sector, use the formula\n\na = (area of circle)*(d/360) where d is the number of degrees. Here, d equals 60. The area of a circle is (pi)(r^2) where r is the radius of a circle. Here, r equals 7. Now solve.\n\na = ((pi)(r^2))*(d/360)\na = ((pi)(7^2))*(60/260)\na = ((pi)(49))*(1/6)\na = 49pi * (1/6)\na = (49(pi))/6\n\nThe area of the sector is (49pi)/6. This area, however, includes the equilateral triangle. So, in order to find the area of the arc, we need to subtract the area of the triangle from the area of the sector.\n\na = ((49pi)/6) - 12.25(sqrt3)\nSimplify this on paper to get\na = (98pi - 147sqrt3)/12\n\nThe area of arc AB equals the area of arc BC. So, to get the total area of the two arcs, multiply by two to get\n\narea of two arcs = (98pi - 147sqrt3)/6\n------------------------------------------------------------------------------------\nNow add the area of the triangle and the two sectors.\n\nA = (12.25pi * sqrt3) + (98pi - 147sqrt3)/6\nSimplify this to get a total area of the figure...\n\nA = (196pi - 147sqrt3) / 12\n------------------------------------------------------------------------------------
1184743
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.
522898
Imagine you unraveled the cylinder, the middle would turn into a rectangle, right? so to find the surface area, take the height of the cylinder and multiply it by the circumference of the circles at the bases, then add in the area of the circles. Circumference is 2(pi)r and area is (pi)r^2.
89149
Take the two extremes of the length of the sides.  The shortest a side can be is 1cm.  With a length of 1cm, the width would be 71 cm (1+1+71+71=144).  This is an area of 71 (1 x 71=71), which is the smallest possible.  The largest possible area is of a square with perimeter 144, which is each side being 36cm.  This is an area of 1296.  So, the difference is 1296-71=1225.
722736
the angles are just unknown if i get you correctly.\nthere are sin, cos, and tan proportions you can work out to find the rest. i think you only need three of the measurements to work out the rest of the triangle, like 2 sides and an angle or 2 angles and a side.  not positive.
598737
I wish I remembered terms for this stuff.\nFind the length of the base, divide that by two.\nUse this number to find the length of a line drawn from the point of the triangle to the base. Congrats, that number is the height.\nTriangle ABC.\nLength from point A to C is 3.\nFrom A to B is 4.\nSo B to C is 5.\nTake the base, we'll say line AC, and divide that by 2, so you get 1.5.\nUse that number, now, as the length from A to D.\nSo, 1.5 squared plus 4 squared is 18.25. Square root of 18.25 is 4.27, so your height is 4.27.\nPythagorean Theorem.\nA^2 + B^2 = C^2
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....
549797
enclose the ireegular shape with a rectangle and find the empty spaces. then subtract that with the area of the enclosed rectangle to get ur answer.
604203
to find the surface area of a pyramid take 1/2 of the perimeter of the base times the slant height and add on the area of the base.  First lets find the area of the base with 3 sides of 17.32 each. You need more information for this, unless you have been taught some simple trig functions. To get the area of the base we need to know the height of this base(if it were standing up).If I use the tangent function I find that the height of the triangle is approximately 15 inches. So the area of the base is 1/2 times 17.32 times 15. I did the tangent of 60 degrees = the height of the triangle divided by 1/2 the side of the triangle.  I hope this helps solve the rest of the problem.
975485
H*W= area\nheight times width\n\none side has a height of 8 and its width is 9 thus the area would be 72
824775
The Area of a triangle is half the area of a square. So take one verticy and take a 90degree line from that up to the other corner.\n\nOh cr@p I'm starting to forget it all now. Too long ago.\n\nIt's something like that.
471779
We use the formula  area = (1/2)bc.sin(A)\n\nThe formula above can be "proven" easily by definition that a triangle's area is just 1/2 x "base" x "height".\nNote in above, base = b. and height = c sinA. \nIts one way of looking at it. Draw a diagram.\n\nWe'll use that and some basic trigonometry to prove "Hero's formula".  \n\nAnd sin(A) = 2 sin(A/2).cos(A/2)    \n\nSo area = bc.sin(A/2).cos(A/2)\n\nNext we find expressions for cos(A/2) and sin(A/2) in terms of \ns, a, b, c, where a, b, c are the sides of the triangle and s is \nthe semi-perimeter.\n\nSo  s = (a+b+c)/2\n\n                                    \ncos(A) \n\n=   2cos^2(A/2) - 1 \n\n= [ b^2 + c^2 - a^2 ] / [2bc]\n\nSo,\n                                  \n2cos^2(A/2) \n\n= 1 + [ b^2 + c^2 - a^2 ] / [2bc]\n\n= [(b+c)^2 - a^2 ]/[2bc] = [(b+c-a)(b+c+a)]/[2bc]\n\n= [(2s-2a)2s ]/[2bc]                          \n\nSo,\n              \ncos^2(A/2) =  [s(s-a)] / [bc]\n\n                                   b^2 + c^2 - a^2\nNext, \n\ncos(A) \n\n= 1 - 2 sin^2(A/2) \n\n= [b^2 + c^2 - a^2] / [2bc]\n\nSo,\n\n2sin^2(A/2) = 1 - [b^2 + c^2 - a^2] / [2bc]\n\n2sin^2(A/2) = [a^2 - (b-c)^2] / [2bc]\n\n2sin^2(A/2) = [(a-b+c)(a+b-c)] / [2bc]\n\n2sin^2(A/2) = [(2s-2b)(2s-2c)] / [2bc]\n \nHence,                                 \n                                    \nsin^2(A/2)  = [(s-b)(s-c)] / [bc] \n\n                                    \nWe can now return to the formula for the area of the triangle using the above:\n\n(1/2)bc.sin(A) = bc.sin(A/2).cos(A/2) \n\n(1/2)bc.sin(A) = [bc.sqrt[s(s-a)(s-b(s-c)]/[bc]\n\n(1/2)bc.sin(A) = sqrt[s(s-a)(s-b)(s-c)]\n\nDone.\n\nNote : The dots "." mean nothing just "spacing".\nRegards
579191
Area of a triangle:\n\nA = 1/2 (B)(A)\nwhere B = base which could be any of the three sides\nand A = altitude (or height), which is the perpendicular distance to the base B from the opposite vertex.
730817
Define the lengths of the smallest triangle:\nAP = ½ AC\nAQ = ½ AB\n\nSquare the equations:\nAP^2 = ¼ AC^2\nAQ^2 = ¼ AB^2\n\nPythagorean theorem for smallest triangle:\nAP^2 + AQ^2 = PQ^2\n\nSubstitute:\n ¼ AC^2 + ¼ AB^2= PQ^2\n\nMultiply by 4:\nAB^2 + AC^2 = 4PQ^2\n\nPythagorean theorem for triangles ABP and ACQ:\nAP^2 + AB^2 = BP^2\nAQ^2 + AC^2 = CQ^2\n\nAdd the two equations:\nBP^2 + CQ^2 = AP^2 + AQ^2 + AC^2 + AB^2 \n\nSubstitute:\nBP^2 + CQ^2 = PQ^2 + AC^2 + AB^2\n\nSubstitute:\nBP^2 + CQ^2 = PQ^2 + 4PQ^2\n\nTherefore:\nBP^2 + CQ^2 = 5PQ^2\n\n\nIf you get stumped with this type problem, sometimes it helps to start at the end and work backwards...
184797
find the angle between two sides.
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.