of a right triangle has a hypotenuse (slant) of square root of 52, how do you find the lengths  of the legs?
966570
Is it isosceles (with the legs the same length)?\n\nThen you can solve using:\na² + b² = c²\n\nc² = sqrt(52)² = 52\n\nThus a² + b² = 52\n\nIf a = b, then a² = 26\na = b = sqrt(26)\n\nOtherwise, if you don't have some other information, you can't uniquely solve for the length of the legs.  The legs could have lengths from nearly 0 to nearly sqrt(52).
847510
Several ways. The first that comes to mind is to figure the area with Heron's formula (area = sqrt(s(s - a)(s - b)(s - c)), where s = (a + b + c)/2). The area can also be figured as 1/2 base times altitude, so you know the altitudes of the triangle. The altitudes give you right triangles where you know the hypotenuse (a side of the triangle), and a leg, so you can compute the trig functions of all the angles. Then the angles can be computed as arcsin or arccos of the angle (arcsin(altitude/side))
247800
it is basically:\na2 + b2 = c2   -->  (A squared)+(B squared)=(C squared)\n  \n\nif you have a right triangle.you have three sides.the two legs.and the hypotenus(sp?).a and b are the two legs and c is the hypotenus.\n\nin an equation.if you are given a and b.you can square each of them and then add the two values together.the will give you c squared.to find c you then need to take the square root of the value(a squared + b squared).\n\nif you have c and a leg(example:a).you square c.and then subtract a squared from that value(c squared).that will give you b squared.to get b you just get the square root of c squared - a squared.\n\n\nhope that helps.
185495
The Pythagorean Theorem can only be applied to Right Triangles only.\n\nIn a right triangle, the side opposite the 90 degree angle is called the HYPOTENUSE. The other two sides are called LEGS.\n\nTHEOREM:\nThe square of the hypotenuse is equal to the sum of the squares of the legs. \n\nEQUATION: Supposing,\nc = hypotenuse\na = first leg\nb = second leg\n\nThen:\n\nc^2 = a^2 + b^2\n\nThis is the conventional way of assigning the letters to the sides of the right triangle. However, if other letters will be assigned, then determine which letter will represent the hypotenuse so that its square can be placed on the left side of the equation. The others will be placed on the right side.
969211
just clarifying first answer... \n\nlet   a = 1st leg\nand b = 2nd leg\n       c = hypotenuse\n\na= 2 + 2b -->> one leg(a) is two inches(2+) more than twice the other leg(2b)\n\nsubstitute a\n[(2+2b)^2] + [b^2] = [13^2]\n[4+4b+(4b^2)] + [b^2} = 169\n5b^2 + 8b + 4-169 = 0\n5b^2 + 8b - 165 = 0\n  A         B     C\nusing quadratic formula which is -B plus or minus the square root of B^2 - 4AC all over 2A\n\n-8 plus/minus squareroot of 64 - [4(5)(-165)] all over 2x5\n\n-8 plus/minus squareroot of 58 all over 10\n\n[-8 + 58] / 10 = 5\n\nor \n\n[-8-58] /10 = -6.6\n\nnegative sides do not exist so one side (b) is 5\n\ngoing back to a= 2 + 2b \na = 2+10\na = 12\n\ntherefore\na = 12,\nb = 5\nc = 13
567965
means that the base is 4 cm rite?\nok, u obtained the base, then now u need the height.\nHow to find the height?,\nUse trigo!. U know what izzit, well, i assume u know.\n\nFirst split the triangle in half. \n            /|&#92;\n           / | &#92;\n          /__|__&#92; \ntake one side only,\n            /|\n           / |\n          /__|\n\nthe slant side is 4 cm rite?,\nand to find the height,\n\nsin 60 = opposite/hypotenus\nopposite = (hypotenus)sin 60,\nopposite = 3.46cm.\n(oppossite is the height, hypotenus is one of the side of the triangle)\n\nU have the height!\nUse triangle formula\nArea = 0.5(heigth)(base)\n     = 6.93cm(squared)\n\nOR, for easier way,\nuse trigo formulae for triangle area\n\nArea = 0.5*a*b*sin C\n(C is the angle of any two sides while a and b is both sides making the angle)\n\n     = O.5 (4) (4) (sin 60)\n     = 6.93 cm squared
882750
Listen to the first guy.\nThe distance on the y axis is 1 ((-5)-(-4)) and on the x axis it is 3 ((-3)-0).\nThis gives you a right angles triangle with sides of 3 & 1. \nUnder Pythagoras theoem the sum of the squares is equal to the square of the hypoteneuse so\n1^2 + 3^2 = 10.\nTherefore the distance is the square root of 10.
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.
737368
let one leg be x\nhypotenus = 3x\n\nother leg = sqrt(9x^2-x^2) = sqrt(8) x = 2 sqrt(2) x\nsum = x+3x+2xsqrt(2) \n= 4x + 2xsqrt(2) \n3x is integer and x(4+2sqrt(2)) is between 6 and 8\nso 3x(4+2sqrt(2)) is between 18 and 24\n\nso 3x is between 18/(4+2sqrt(2)) and 24/(4+23sqrt(2))\nbetween 18/6.8and 24/6.8\nonly possible value is 3\nhypotenuos = 3
793698
Start by drawing a picture of this.  Draw a right triangle.  I put the right angle at the bottom left.  Then draw an arbitrary line from the upper left vertex down to the base.  This separtes the large right triangle into a smaller right triangle and what I will call the "outer triangle."  This gives you the 1000 Jim moved; it's to the right of where the line intersects the base.  I called this leg a.  I drew in the angles I was given, and labeled all of the line segments.  The hypotenuse of the large right triangle is b, the hypotenuse of the smaller right triangle (also the only unlabeled leg of the outer triangle) is c, and the vertical leg of the right triangle is d.  d is the height of the escarpment.  I named the angles after the sides they face.  So the 10 degree angle is C, the 32.5 degree angle is D, and angle A is opposite leg a, and is an acute angle in the outer triangle.  Since I was given angle D, I could find its supplement, B, which is an obtuse angle in the outer triangle, opposite leg b.  B = 180 - D = 180 - 32.5 = 147.5.  Since the sum of angles in a triangle is 180, and I know C and B, I can also find A = 180 - B - C = 180 - 147.5 - 10 = 22.5.\n\nNow that I know every angle in the outer triangle and the length of one leg, a, I can use the law of sines.  The law of sines says that a / sin(A) = b / sin(B).  Rearranging, b = a*sin(B) / sin(A) = 1000*sin(147.5) / sin(22.5).  Now look at the larger right triangle.  You know the length of the hypotenuse, b, and you know the angle C opposite the unknown d (I called it C because it is also opposite leg c in the outer triangle.).  By trigonometry, d = b*sin(C) = b*sin(10), where we use b from above.
776061
The theory is this.. \n\nThe square of the length of the hypotenuse is the sum of the squares of the adjacent sides.\n\nIn this case you have two built in triangles you can solve... the front and the side of thebox.  you need to solve one of these to find the hypotenuse from the top of the box to the bottom opposite corner... (the diagonal line from the top rear corner to the bottom front corner)  this is the longest stick possible that you can fit in a box of these dimensions\n\nso that being said.. the side of the box measures 3 x 5\npythagorean theorem states (a^2+b^2=c^2)\ntherefore 3^2+5^2=c^2   --> 9+25=c^2  so C=5.8 ( square root of 34)\n\nnext we solve the "compound" triangle.  using the "C" value we calculated as one of the sides in the new triangle.\n\n5.8^2 + 4^2 = C^2   --> 34+16= C^2  --> 50=C^2\nC=7.07 (square root of 50)\n\nSo the answer is no  only a stick 7.07 feet long will fit
46479
Oh boy. I'm not sure that i can explain why the rules are what they are, but i know what the rules are, and i can tell you:\n\n1. 45-45-90\n -The two legs are equal\nLeg 1=a\nLeg 2=b\na=b\n -The hypotenuse is 2 times the square root of a (or b, since they are equal)\n\n2.30-60-90\n -The key to this is the small leg, "a". The hypotenuse is equal to 2a (2 times "a").\n -The large leg is equal to 3 times the square root of the small leg, "a".\n\nHope this helps!\n\n-incarnadyne49\n\nThe first guy is wrong, i'm sure!!! I would know, i am a calc student, and access to a geometry textbook!!
791994
First distance: Pythagorean Theorem is a squared + b squared = c squared.  So (2*2) + (3*3) = ???  Then find the square root of ??? \n\nThen the bearing: We first need to find the measure of angle A.  Since we know the length of all 3 sides, any of the SohCahToa formulas work.  Tangent is easiest here.  tan(A) = opposite/adjacent (length of opposite leg divided by length of adjacent leg).  tan(A) = 3/2.  Use the tan-1 key (probably will require shift key).  \n\nWe know the first leg of the trip is south which represents 180 degrees.  So add the found angle to 180 degrees.  (Draw it out, it should make sense.)
352673
Make right triangles with the height as one leg, a radius (segment from center of base to one of its vertices) as the other leg, and lateral edge as the hypotenuse.  The radius will equal the length of the base edge as it's a side of an equilateral triangle.  So 10^2 + 15^2 will equal lateral edge squared.\n\nSlant height would be the hypotenuse of a right triangle with legs of the pyramid's height, and the apothem (distance from center of base to midpoint of a side.)  You can get the apothem from Pythagorean theorem from the right triangle formed by half a base edge, a radius, and the apothem.  Then use the apothem in another Pythagorean theorem to get slant height.\n\nLateral area = 6 times the area of any one triangle.  Each triangle's area is 1/2 times 10 times slant height.\n\nVolume = 1/3 (area of base)(height).  The area of the base is 1/2 (apothem)(perimeter).\n\nThose are the steps.  Hope you can figure them out from this typing!
803140
OK.. First of all it's an isosceles triangle.. thus the base is divided by the vertical line into two equal halves of 3 and 3 each.. \nNow just look at one triangle.. say the left one.. it is a right traingle (the vertical line is perpendicular to the base) having base of 3 and hypotenuse of 21. \nThus if the vertical height is 'h'     => h^2 + 3^2 = 21^2   (^2 means square.. ie power 2) \nsolving for h = (441 -9)^(1/2) = 20.78  (use a calculator)\nNow for theta..         Cos a = Base / hypotenuse \n                                 cos a = 3/21 = 1/7\n                                 a= cos^-1 (1/7)          (Cosine inverse)\n                                 a = 1.43 radians         (Use a calculator :-) )\nGuess thats it !!
365246
mathematician's approach is the only correct approach in this thread. The square root function does not distribute over addition/subtraction. Consider Sqrt[1 + 1] vs. Sqrt[1] + Sqrt[1].\nIf you're more of a geometer than an algebraist, you will get the same answer from considering the right triangle with leg of length Sqrt[1-4x^2], implying hypotenuse of length 1 and the other leg of length 2x. Your expression is then 2tan(t) where t is the angle opposite the leg of length Sqrt[1-4x^2]. Also, x is obviously (1/2)cos(t), so dx = -(1/2)sin(t) dt.
520768
First you need to find the slope of the line. Use the info from the slope (rise over run) to make a right triangle with hypotenuse is the line. Use inverse tangent on info from slope to find angle.