what are the formulas for a triangle and a rectangle and a square to find area?
359093
Square and Rectangle= lenght*height\ntriangle=base*height divided by 2
857420
Area of what?\n\nRectangle = b×h\nSquare = l^2\nTrapezium = ½(l+b)h\nEllipse = πlb\nCircle = πr2 \nTriangle = ½b×h
471423
The area of the rectangle is:\n A of rectangle = L X W = 12 x 8 = 96 sq in\n\nThe area of the triangle is:\nA of triangle = 1/2 (A)(B) = 1/2 (A)(32 in) where A is the altitude to the base B of the triangle.\n\nsince both areas are equal:\n96 sq. in = 1/2(A)(32 in)\n96 sq. in = A (16 in)\nA = (96 sq. in)/16 in\nA = 6 inches
426495
It depends on what shape you want to know the area.\n\nSimple shapes have formulas. (such as a circle's area is Pi*r*r)\n\nComplex shapes which have pattern.(you know the equation of lines that form the shape), calculus is used to find the area.\n\nIf a shape is not ordinary. You should draw lines to grid it and count the area square by square.\n\n-Hope this would help-
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.
140062
L x W = Area\n\nCheater
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------------------------------------------------------------------------------------
272507
Draw square ABCD so that A is southwest, B is northwest, C is northeast, D is southeast, and E extends on an imaginary line out from side AD.....\n\nArea of square ABCD = (AB)^2 -- (definition of area of a square)\n\nAB = side of square = short side of triangle ABE\nAE = long side of triangle ABE\nBE = hypotenuse of triangle ABE\n\n(BE)^2 = (AB)^2 + (AE)^2 -- (definition of equation of right triangle; angle A is the right angle of triangle ABE)\n\nThus:\n(AB)^2 = (BE)^2 - (AE)^2
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Surface Area = [Base Area] +\n1/2 × Perimeter × [Side Length]\n\nVolume = 1/3 × [Base Area] × Height\n\nThe Surface Area has two parts: the area of the sides (the Lateral Area) and the area of the base (the Base Area). \n\nThe Base Area depends on the shape, there are different formulas for triangle, square, etc. \n\nBut the Lateral Area is surprisingly simple. Just multiply the perimeter by the side length and divide by 2. This is because the sides are always triangles and the triangle formula is base times height divided by 2
926515
Let x be the length for the square and 14-x for the triangle.\n\nThe area of the square is (x/4)^2 = (x^2)/16.\n\nThe height of the triangle is sqrt(3)(14-x)/6 and the base of the triangle is (14-x)/3, so the area of the triangle is\n\n1/2 bh = [(14-x)/6] [sqrt(3)(14-x)/6] = (14-x)^2 sqrt(3)/36\n\nTotal area A = (x^2)/16 + (14-x)^2 sqrt(3)/36\n\nA = [9x^2 + 4(x^2 - 28x + 196) sqrt(3)] / 144\n\nA = [(9 + 4 sqrt 3) x^2 - (112 sqrt 3) x + (784 sqrt 3)] / 144\n\ndA/dx = [2(9 + 4 sqrt 3) x - (112 sqrt 3)] / 144 = 0\n\n(9 + 4 sqrt 3) x = 56 sqrt 3\n\nx = (56 sqrt 3) / (9 + 4 sqrt 3) = 6.0895 meters  (Answer)\n\nThe area of the square is 2.32 m^2.  The area of the triangle is 3.01 m^2, and the total area is 5.33 m^2.
190089
use heron's formula: \nArea=SQRT(s(s-a)(s-b)(s-c)), \nwhere s=(a+b+c)/2 or perimeter/2. \n\nyou only need to know the sides of the triangles. \nfrom (0,9) to (4,6), the distnace is a=5\nfrom (0,9) to (-1,12) the distance is b=sqrt(10)~3.16\nfrom (4,6) to (-1,12) the distance is c=sqrt(61)~7.81\n\nso, s=(5+3.16+7.81)/2=7.985\n\nplugging these 4 numbers into the formula gives \nSQRT(7.985(7.985-5) (7.985-3.16) (7.985-7.81))=4.486\n\nso the area is approximately 4.486
852685
The area of a rectangle is length * width.  The length is defined in terms of the width.  You know the area, but not the width, so call the width x.  The length is then 2x + 3.\n\nTherefore, you can set up the problem like this:\n\n(2x + 3) * x = 44\n\nMultiply out the x and move the 44 over to get this:\n\n2x^2 + 3x - 44 = 0\n\nThis should look familiar, and you should know how to solve this.  I got:\n\n(2x + 11)(x - 4) = 0\n\nwhich leads to:\n\n2x + 11 = 0           \n2x = -11          \nx = -11/2\n\n x - 4 = 0\n x = 4\n\nNow, you have two answers.  However, only one makes sense, as it is impossible to have a negative width.  Therefore, you can disregard x = -11/2, leaving x = 4 as the width.  You can then substitute x into 2x + 3 to get 2(4) + 3 = 11.\n\nAs a check:\n\n4 * 11 = 44\n\nTherefore, this is a good answer.
533566
h (x) + h (x2) = Area or h(x)=Area if parallelogram is square or rectangle.
170229
Well, a rectangle is not a triangle it is a parallelogram, I think. It has four right angles and two long sided and two short sides. Therefore, the dimensions are as follows: To figure area it is short side * long side. To figure how long all sides are you simply do two short sides plus two long sides. However, I am assuming that this was trick question to see who is paying attention to what a rectangle is.
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Area of a square = L * L\n\nArea of A = x * x ..........(1)\nArea of B = y *y  ..........(2)\n  \n y = x -1               ..........(3)\n\nA = 4 * B              ...........(4)\n\nSubstitute (1) & (2) into (4) gives:\n\nx^2 = 4 y^2           ...........(5)\n\nSub. (3) into (5)\n\nx^2 = 4 (x-1) ^2\n\nSolve:\n\nx^2 = 4( x^2 -2x +1)\nx^2 = 4x^2 -8x +4\nOr\n\n3x^2 - 8x +4 =0\n\nSecond order eq.\n\nx1,x2= [+8 +- Sqr(8^2 - 4*3*4)]/(2*3)\n\nx1,x2 = [8+-4]/6\n\nx1 = 2   &  x2 = 4/6\n\nThen\ny1 = 2-1 =1   & y2 = 4/6 -1 = -2/6 ( illogical )\n\nthen\n\nx= 2  and y = 1
227708
There are two ways to do it...\n\n1.> If you know the base and the height of the triangle:\n\n(Base*Height)/2\n\nso if one of the side was 12 cm., and the perpendicular on it from the point where the other two sides meet is 6 cm., the area is = (12*6)/2 = 36 sq.cm.\n\n2.> If you know all the sides:\n\nHero's formula:\n\nFirst find s (semi-perimeter) as:\n\ns=0.5*(side1+side2+side3)\n\nnow the area= root (s * (s-side1) * (s-side2) * (s-side3))\n\nso if a triangle's sides are 3,4 and 5 cm.:\n\nsemi-perimeter = (3+4+5)/2 = 6\n\narea= root (6(6-3)*(6-4)*(6-5)) = root (6*3*2*1) = root (36) = 6 sq. cm.
981115
that way you can remember the formula with generic terms like L and W then later insert the corresponding numbers.\n\nlength times width equals area\nL * W = A\nl2" * 12" = 144 square inches