How do I calculate the two shorter sides of a right triangle if all I have is the length of the hypotenuse?
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ABOUT THE BINOMIAL THEOREM I AM TEEMING WITH A LOT OF NEWS\n\nWITH MANY CHEERFUL FACTS ABOUT THE SQUARE ON THE HYPOTENEUSE\n\n(W. S. Gilbert, I am The Very Model of a Modern Major General)\n\nI am assuming we are dealing with what we know is a PRIMITIVE Pythagorean triple ie one like (3,4,5) or (5,12,13) or (8,15,17) or (20,21,29) in which there is no common factor between the three sides, NOT one like (6,8,10) which is divisible by 2.\n\nThere are tables published of the smallest triples, in ascending order of hypooteneuse, so you may be able to look it up and not need to calculate it. \n\nhttp://www.math.rutgers.edu/~erowland/tripleslist-long.html\n\nHowever, two points.  \n\nNot every integer can be a hypoteneuse. There are gaps in the table eg after (3,4,5) the next smallesr hypoteneuse is 13 (in (5, 12, 13). There are no primitive triples with a hypoteneuse of 6, 7, 8, 9, 10, 11 or 12. And therefore if you were given a hypoteneuse of 12, there are no integer solutions to it.\n\nSecondly, there are cases where a number is hypoteneuse to more than one triple. 65 is the lowest such number and 85 is the next.\n\n(16, 63, 65) and (33, 56, 65)\n(13, 84, 85) and (36, 77, 85)\n\nThese are known as siblings. Both give correct solutions and you can't say one is right and the other is wrong. Here are a couple of special cases, with more than two solutions:\n\n(47, 1104, 1105)\n(744, 817, 1105)\n(576, 943, 1105) \n(264, 1073, 1105)\n\nThere are EIGHT siblings with a hypoteneuse of 32045, SIXTEEN siblings with a hypoteneuse of 1185665, THIRTY-TWO siblings with a hypoteneuse of 48612265, AND SIXTY FOUR siblings with a hypoteneuse of 2576450045.\n\nHere are rthe eight siblings with a hypoteneuse of 32045:\n\n(2277, 31964, 32045) \n(8283, 30956, 32045)\n(17253, 27004, 32045) \n(21093, 24124, 32045)\n(22244, 23067, 32045)\n(15916, 27813, 32045)\n(6764, 31323 , 32045)\n(716, 32037, 32045)\n\nLacking access to tables, for small numbers, you may be able to work it out, by trial and error, using the following properties\n\n(a) the hypoteneuse must be the biggest side and must be an odd number (to be a primitive triple)\n\n(b) one of the legs will be an odd number and the other an even number\n\n(c) As a result, the difference between the two small sides will always be an odd number. In fact it is either a prime number (eg 1, 7 and 17 feature, but 11, 13 and 19 do not) or a power of a prime number (49 is the lowest such).\n\n(d) the sum of the two legs must be bigger than the hypoteneuse\n\n(e) one and only one of the legs must be divisible by 3\n\n(f)  one and only one of the legs must be divisible by 4\n\n(g)  one and only one side (which might be the hypoteneuse) must be divisible by 5\n\n(h) follows from (e) (f) and (g)\n\n(h) the product of the three sides is always divisible by 60\n\nThus if you knew the hypoteneuse and could see it was not divisible by 5, then it follows the legs must, between them, be divisible BY 3 4 AND 5\n\nSo if you knew the hypoteneuse is 61, it seems obvious to try one of the legs as 60 and see if that works (it does, (11, 60, 61) is a triple. \n\nIn fact you can work out the third member is 11 by adding 60 and 61 to see if it is a perfect square (it is 121) and taking the square root of it, if so. (another property of triples whose two largest sides differ by 1).\n\nGiven that 17 is a hypoteneuse and knowing one of the other sides must be 3 x 4 or 3 x 5 (4 x 5 would be larger than 17) you have a choice of (12, x, 17) where x = 5 or 15 (has to be odd as 12 is even) or (y, 15, 17) where y is 4, 8, 12 or 16)\n\nIt turns out to be (8, 15, 17)\n\nAgain, knowing a property of all triples where the hypoteneuse and the larger leg differ by 2, there is a shortcut, that leads to 8 ... take the average of the two longer sides, multiply by 4 and take the square root. (thus 16 => 64 => 8)\n\nThis holds true for \n(12, 35, 37)\n(16, 63, 65)\n(20, 99, 101) etc\n\nSo it is possible to make intelligent guesses at the
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Try this:\nAssuming the rectangle is centered within the half circle, draw a line from the center of the flat edge of the half circle (should also be the center of one edge of the rectangle) to one of the corners of the rectangle that is touching the edge of the circle.  That line has a length equal to the radius of 10.  By drawing this line you have also formed a 30-60-90 triangle inside the rectangle.  So, the triangle you have just created has a hypotenuse equal to 10 but the sides are unknown.  However, knowing that a 30-60-90 triangle has sides with a relationship of 1, 2, square root of 3, we use the method of similar triangles to  solve for the sides.  The shortest side of the 30-60-90 triangle is thus equal to 5 (this is also equal to the width of the rectangle), the other side is equal to 5*square root of 3 (the length of the rectangle will be two times this value), and the hypotenuse is 10. So the width of the rectangle is 5 and the length is 10*square root 3, so the area of the rectangle is 50*square root 3=86.6
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If the triangle is a right triangle, you can find the other side by Py's Th and then by the formula,(A=1/2bh).\nIf it is not, you should know the angle they make each other or an adjacent angle. Only if, you can find the area by sine laws or cosine laws.
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The Pythagorean Theorem deals with right triangles.  The two sides of the right triangle that are connected to the right angle are its legs, and the third (longest) side is called the hypotenuse.  The theorem says that\n\nleg squared plus other leg squared equals hypotenuse squared\n\nSo if one leg is 3 and the other is 4, the hypotenuse is 5 because 3 squared (9) plus 4 squared (16) = 25, which is 5 squared.
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The 14 foot line is called a secant.  Bisect it, and draw a radius through the point of bisection.  You now have two identical right triangles, with hypotenuses of 12 feet, and known sides of 7 feet.  Use trigonometry to calculate the angle at the center of the circle of either triangle, and then a simple proportion (involving the circle's total circumference) will give you the arc length you want.
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As with many problems, it helps to draw a diagram.  I made a triangle, and then put in each of the altitudes, since they are my only known values.  With these in place, we now have 6 principal right triangles, of which we know the value one leg (the altitude).  In each of these triangles, the hypoteneuse is a complete side of the original triangle, and the angle opposite the known leg is also of the original triangle.  Personally, I chose to work with cosines and tnagents, but you could just as easily have used sines (though I think it might add a step or two).  Remember that the cosine of an angle in a right triangle is merely the ratio of the side opposite the angle to the hypoteneuse.  Don't worry that we don't know the value of any of these angles; I went through the entire problem without ever needing it.\n\nSo, since each angle, in its entirety, is in two of our right triangles, they give us two ratios to compare.  That is:\n\nIf the sides of the original triangle are x,y,z\ncos a1 = 20/x = 15/y\ncos a2 = 20/z = 12/y\ncos a3 = 12/x = 15/z\n\nWith a little cross multiplication and rearrangement of terms, we find that:\n12z = 15x = 20y.  \n\n(Your x, y, z may be in a different order depending on how you labeled your triangle, but if you manipulate your triangle a bit, you should be able to get it to match.  This will keep your coefficients identical to mine.  But even if you don't match it up, in the end you will get the same answer.  That's the beauty of math.)\n\nThe tangents are a little trickier, because they involve a fraction of the original side.  This involves the addition of a new variable on each, such that one portion of the line is "a", and the remainder is x - a; and then b and y - b; c and c - z.  The tangents then follow:\n\ntan a1 = 20/(y - b) = 15/a\ntan a2 = 12/(z - c) = 20/b\ntan a3 = 15/(x - a) = 12/c\n\nRearranging each gives us the three equations:\n20a = 15(y - b)\n12b = 20(z - c)\n15c = 12(x - a)\n\nFrom there, you can solve for either a, b, or c.  It's your choice.  I opted to solve for a.  Substituting the third equation into the second, and then that new equation into the first, we get:\n\n20a = 15(y - (20/12)(z - (12/15)(x - a)))\nwhich reduces to\n8a = 4x + 3y - 5z\n\nNow, side a is in a right triangle with hypoteneuse y and second leg 15, meaning that:\na^2 + 15^2 = y^2   or\na = sqrt(y^2 - 225)\n\nUsing this equation, plus the ratios we received from the cosine functions, we solve for y:\n\n8 * sqrt(y^2 - 225) = (16/3)y + 3y - (25/3)\n\nConveniently, the right half is equal to zero, so dividing by 8 and squaring:\n\ny^2 - 225 = 0\ny^2 = 225\ny = 15\n\nEnter this into our ratio:\n\n15x = 20y\n15x = 20*15\nx = 20\n\nand\n\n12z = 20y\n12z = 20*15\nz = 25\n\nSo finally we have the lengths of all three sides, and adding them together, the perimeter is\n15 + 20 + 25 = 60\n\nThis turns out to be a right triangle, but that is merely happenstance.  This process can be used to solve for any triangle with three known altitudes.\n\nHope this helps.
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If it's a right triangle, use the Pythagorean Theorem.\n\na^2 + b^2 = c^2\n\n\nA and B are the legs (the ones that touch the right angle), C is the hypotenuse which is opposite the right angle.\n\n10^2 + x^2 = 26\n\n100 + x^2 = 676\n\nx^2 = 576\n\nx = 24
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>>>> what is the altitude of a right triangle?\n\nI have several here, and I am at 12 metres above sea level.\nSeriously, you haven’t provided enough information. You need at minimum the length of 2 sides or a side and an angle as well as the orientation to be able to calculate height (altitude).\n\n>>>>>>if the equation is d=2r what is the d and the r what exectly \n>>>>>>>am I finding with this equation \n\nd = diameter. r= radius. You are finding either how far it is from one side of a circle to another (d), or how far it is form the edge of a circle to the centre (r).
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Draw the two points (a,b) and (c,d) on a graph so that the segment connecting them is NOT horizontal or vertical. Then make a right triangle with one vertical leg and one horizontal leg where the connecting segment is the hypotenuse.  \n\nThe length of the horizontal leg is c-a\nThe length of the vertical leg is d-b\n\nSo the length of the hypotenuse, which is the distance from (a,b) to (c,d), is\n\nsquare root of (c-a)^2+(d-b)^2
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Trignometry could be made interestingly easy.\nUnderstand the basics:\nThe longest side of triangle is hypotenuse.\nThe side of the triangle opposite the angle(your angle) is the perpendicular.\nthe remaining side is called the base.\n\nsine theta=perpendicular/hypotenuse\ncosine theta=base/hypotenuse\ntangent theta=sine theta/cos theta=perpendicular/base\n\nAll other formulas are based on these relationships.\nlike sine square theta+cosine square theta=1 etc.
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Insufficient information the included angle or a right angle somewhere is needed.\n\nAssuning these are right angled triangles there are two possibiliies:\n\nEither the left side is a hypotenuse or the right side is a hypotenuse\n\nCase 1;\nBy Pythagoras' Rule\n\n12² = 8² + x²\nSo x² = 80\nx = √80\n= √(16 * 5)\n=4√5\n\nCase 2\nBy Pythagoras' Rule\n\nx² = 8² + 12²\nSo x² = 208\nx = √208\n= √(16 * 13)\n= 4√13\n\nIf neither case is true and the included angle is θ\n\nThen, by the cosine rule\n\nx² = 8² + 12² - 2 * 8 * 12*cosθ\nSo x = √(208 - 16 * 12 * cosθ)\n= √(16 * 13  - 16 * 12 * cosθ)\n=4√(13 - 12 * cosθ)\n\nNOTE: This solution satisfies the inequalities given above as in any triangle 0° < θ < 180°\ncos 0° = 1, cos 180° = -1\nSo 4√(13 - 12 * 1) < x < 4√(13 - 12 * (-1)1)\nie 4√1 < x < 4√25\nwhence 4 < x < 20
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Whether any one likes it or not, the sum of the squares of the sides of a right angled triangle will always like to be equal to the square of the hypotenuse.\n\nPlease, from today say to yourself that I like Pythagoras Theorm. Soon you will begin to like it. It will lead you to many wonders of science
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If you picture your trapezoid with the shorter parallel side and the two legs as sides 4, then the only unknown is the the long parallel side.  That side is at least 4, but with two extra parts on either side (if you drop perpendiculars from the short to the long side, you'd have a rectangle and two triangles).  So the long side is 4 + 2x, where x is the length of those two parts, which is also a leg of a triangle that we created in dropping those perpendiculars.\n\nThe height of the trapezoid is the same as the other leg of the triangle.  We don't know the length of either leg, but we do know that the hypotenuse is 4.  So through the Pythagorean theorem, we have\n\n4^2 = h^2 + x^2\nh^2 = 16 - x^2\nh = sqrt(16 - x^2)\n\nBack to the trapezoid - we know the area of a trapezoid is the average of the bases times the height, so we have\n\nA = (1/2)(4 + 4 + 2x)(sqrt(16 - x^2))\n= (4 + x)(sqrt(16 - x^2))\n= (4 + x)(16 - x^2)^(1/2)\n\nTo find the maximum area, take a derivative, set it to 0, and solve for x:\n\nA' = (4 + x)[(1/2)(16 - x^2)^(-1/2)](-2x) + [(16 - x^2)^(1/2)](1)\n= (-4x - x^2)(16 - x^2)^(-1/2) + (16 - x^2)^(1/2)\n\nIf we set this to 0, then\n0 = (-4x - x^2)(16 - x^2)^(-1/2) + (16 - x^2)^(1/2)\n-(-4x - x^2)(16 - x^2)^(-1/2) = (16 - x^2)^(1/2)\n4x + x^2 = 16 - x^2\n0 = 2x^2 + 4x - 16\n0 = x^2 + 2x - 8\n0 = (x + 4)(x - 2)\nx = -4 or 2\n\nSo your long side has to be 4 + 2(2) = 8 to maximize the area.
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Because the diagonal creates two right triangles you can use the Pythagorean theorem to find the length of the other side:\n\n5^2 +x^2 = 9^2\n25 + x^2 = 81\nx^2 = 56\nsqrt of 56 = 7.48\n\nNow that you have the length of the other side you can use the formula    area = length x width:\n\n5 x 7.48 = 37.4 feet
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We first find the most obvious altitude (the altitude to the base 6).  That is h = √(8² - 3²) = √55. (Pythagorean Theorem)\n\nThus, the area is \nA = bh/2 = 6√55/2 = 3√55\n\nNow, any combination of base and altitude will yield the same area, thus,\nbh/2 = 3√55\n\nWe need to find the length of the altitude to the base 8, so\n8h/2 = 3√55\n4h = 3√55\nh = 3/4 √55\n\nSince we know that 3/4 √55 < √55, then the length of the shortest altitude of the triangle is 3/4 √55.\n\n^_^
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There are six ratios of the three sides of a right triangle. These ratios have been identified by name throughout the centuries. In the USA they are known as sine, cosine, tangent, cotangent, secant, and cosecant. Your calculator has buttons for three of these: [SIN], [COS], and [TAN]. Remember them with the non-word SohCahToa. \n\nThe tangent ratio of an angle is simply the ratio of the opposite side to the adjacent side of a right triangle. \n\nIf you know the angle, type \n\n[TAN] > angle > [ENTER].\n \nThe answer will be the tangent (ratio of the opposite side to the adjacent side).\n\nIf you know the opposite and adjacent sides and need to find the angle, type \n\n[2nd] > [TAN] > opposite > [/] > adjacent > [ENTER]. \n\nThe answer given will be the desired angle.\n\nCaution: \n(1) Different calculators require these buttons to be pushed in a different order.\n(2) If your angle is in degrees, then make sure your calculator's mode is in degrees.\n(3) Don't get the numbers backwards or you'll find the cotangent of the angle and not the tangent.
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The hypotenuse is closely associated with the right triangle. In the right triangle, the hypotenuse is the longest side. The Pythagorean Theorem is a^2 + b^2 = c^2, which is (leg^2 + leg^2 = hypotenuse^2). a and b are both legs of the triangle and c is the hypotenuse