Area of Triangle = 1/2 Base x Height. Its a theorem or postulate?
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The area of a parallelogram is area = base x height. Now, drawing a diagonal of the parallelogram splits the shape into two congruent triangles. Therefore, each triangle will have an area equal to (1/2)base x height.\n\nThis is true for any triangle; that is, any triangle has a corresponding parallelogram associated with it, which can be cut in half.
42660
One. \n\nThis relation is really just Pythagoras' theorem, hiding in a new form. Recall:\n\nH^2 = A^2 + B^2 \n\nThat is, the square of the hypotenuse H is the sum of the squares of the other two sides A and B, for a right triangle with sides A, B and hypotenuse H.  \n\nOf course, A=H*cosx and B=H*sinx for some angle x, so that \n\nH^2= (H^2)*(sinx*sinx + cosx*cosx)\n\nleading to the result:\n\n1=sinx*sinx + cosx*cosx\n\nThis result also is nicely connected to the simple fact that \n\nx - x = 0 \n\nbecause \n\nexp(i*x)=cosx + i*sinx and\nexp(-i*x)=cosx -i*sinx\n\nwhere i=sqrt(-1). \n\nThen multiplying these two, \n\nsinx*sinx + cosx*cosx=exp(i*x)*exp(-i*x)=\nexp(i*[x-x])=exp(0)=1
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er that formula is false and you cant prove it. i can disprove it for ya.\nyou have a trapezoid with legs a and b, and c and d being the bases. you know that a formula for a trapezoid is the average of the bases times the height: (h)(b1+b2)/2. (c+d)/2 is already the average of the bases, since c = b1 and d = b2. now you have to multiply it by the height. if the formula you're supposed to prove is true, then (a+b)/2 has to be equal to the height. how can that be possible? that will never happen unless you have BOTH of the of the legs are perpendicular to the bases. think about it. if they arent perpendicular to the bases, no matter how you draw the trapezoid, the average of the non parallel legs (a+b)/2 will always be greater than the height.
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.............6\n........______\n....../.|...........|&#92;\n...../..|...........|..&#92; 2\n..../_|_____|__&#92;\n\n\nThe triangle on the right is a 30-60-90, so the shorter side must be 2/1 = 1.  So the length of the bottom part is 6+1+1 or 8.  The height must be 1√3, so the area is 1/2(6+8)√3 = 7√3.\n\nGopal screwed up when he did 2*2cos60.  That's 2*2*1/2 = 2, not 4.
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Using Pappus Guldinous Thheorem: A=2*pi*y*L\narea of angular section A1=2*pi*(r/3)*(square root(r^2+l^2))\n                        A1=2*pi*9/3*sq root(9^2+12^2) \n                        A1=90*pi   \narea of circular base section A2=2*pi*(r/2)*r=pi*r^2 [which proves the theorem]         \n                              A2=pi*9^2=81*pi  \nso total area A= A1+A2=171*pi=537.21 sq ft.
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height times the base area. divided by three.
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The volume of a cylinder equals the (area of the base)*height = π r2h
142078
Don't you mean the Pythagorean Theorem?  That theorem is a basic tenent of Geometry that states: In any right triangle, the square of the length of the hypotenuse (side opposite the right angle) is equal to the sum of the squares of the other two sides.   A^2 + B^2 = C^2.
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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
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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.
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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.
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Sure, how about a triangle with a base of 3cm and a height of 4cm?\n\nA = 1/2 b * h\n6 = 1/2 b * h\n12 = b * h\n\nSo any numbers where b * h = 12 will work...\n1 cm x 12 cm\n2 cm x 6 cm\n3 cm x 4 cm\n0.5 cm x 24 cm\n1.2 cm x 10 cm\n2.4 cm x 5 cm\netc.\n\nAnd you can have a right triangle, or an acute triangle or an obtuse triangle... doesn't matter, as long as the base and height multiply to give 12...
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Oh c'mon! Smiddy hasn't had geometry for 20 years! but this is basic. Pythagorean theorem states that on a triangle A-squared plus B-squared = C-squared where C is the hypotenuse. The hypotenuse of a triangle is the diagonal bit if it's two halves of a square and we know by definition that on a square A and B are equal. Therefore 2X squared = 8-squared or 2x^2 = 64\n\nx^2 = 32\nx = sq.rt 32\n\nthe area of the original square is x times x so you're back at 32 square centimeters
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From your question, we know the 3 points of the triangle are:\n(0,0), (0,y) and (x, y).\n\nThis points us to the fact that the right angle is at point (0,y).\n\nHence, area A of triangle is given by the product of base and height divided by 2. Therefore, we have:\n\n Area A of triangle = 1/2 [base.height]\n                    = 1/2 y.x\n\nWe are told y = x2, therefore:\n Area A of triangle = 1/2 y.x\n                    = 1/2 x3 (read as half x-cube)
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In any right triangle, the area of the square whose side is the hypotenuse (the side of the triangle opposite the right angle) is equal to the sum of the areas of the squares on the other two sides.\n\na^2 + b^2 = c^2
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you can express the area as the sum of the areas of the 4 inner triangles which are each 1/2 base x height.  Each base is an edge of the quadrilateral.  The height can be determined in 2 ways and you will need both to solve the system of equations:\nfor example: to determine the height of APD, recognizwe that the height line splits APD into 2 right triangles.    Call the point where the height line intersects AD point E. The edge AD will be split between the 2 tringles proportionally:  AE = 24/69  AD and DE = 45/69 AD.  Now you can determine the height in terms of AD using pythagorean theorum.  Go from there...\n\nTry this site...
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no, it won't..\n\nHOWEVER there are some other theorems through which you can calculate the length of a certain side of a triangle given the lengths of the other the sides.\n\nLikewise it is also possible to calculate (through analytical geometry) the area of any triangle given the lengths of all sides.\n\nSo, practically, given the lengths of at least two sides of a triangle, you can calculate the length of the third side and then calculate the total area of the triangle.\n\n(8th grade mathematics)