What is the length of an altitude of an equilateral triangle with side lengths 8 square root of 3?
459122
the side length is 2\nso the altitude should be 3^(1/2)\ngood luck
595452
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.
930340
PROBLEM 1:\n\nsqrt(x^3) = 8\n\nSquare both sides:\nx^3 = 64\n\nNow take the cuberoot of both sides:\nx = cuberoot(64)\n\nSolve:\nx = 4\n\nPROBLEM 2:\n\ncuberoot(x^2) =4 \n\nCube both sides:\nx^2 = 4^3\n\nSimplify:\nx^2 = 64\n\nTake the square root of both sides:\nx = ±8
991928
Assuming you mean a regular hexagon, then with the 30-60-90 relationship, you should get that half the length of a side is 2.\n\nSo, the area of that yellow triangle in the link below is\n.5 * 2 * 2 * sqrt(3) = 2 * sqrt(3)\n\nThere are 12 triangles, so the area is 24 * sqrt (3).
881588
First, the isosceles triangle has sides of 10, 12, 12.  You can find the angle by law of cosines or, as I prefer, just divide the triangle into two right triangles.  The short side is 5 and the hypotnuese is 12 so the angle at G is arcsin(5/12) and the total angle FGA is twice that (since we split the original triangle in two) <FGA = 2*arcsin(5/12) = 49.249 deg.\n\nIf you make a pyramid, The radius of the hexagon (10 cm), the 12 cm side and the height form a right triangle so the height is:\n\nh = sqrt(12^2 - 10^2) = sqrt(44) = 6.633\n\nFinally, for the area, we need the height of the isosceles triangle which is the tjhird side of the right triangle in the first part. This height is:\n\nc = sqrt(12^2 - 5^2) = sqrt(119) = 10.909\n\nSo the area is: A = 3*10*sqrt(119) = 327.261
995285
The line going from one corner to it's opposite in a square becomes the hypoteneuse of two right triangles.  Try it out, draw a square and slice it from one corner to its opposite and you'll see what I'm talking about.\n\nNow take the length of one side and call that 'a'.  Take the side next to it and call it 'b', then do a simple equasion;\n\na squared plus b squared = c squared\n(or  a'2 + b'2 = c'2)\n\nwhere c is the hypoteneuse, or the distance from one corner to it's opposite.  If you break the equasion down you can get;\n\nc = square root (a'2 + b'2)\n\nIf the figure is a perfect square (all four sides being of equal length) and not a rhombus, trapezoid or paralelogram, then c will always be 1.414213562 times the length of a side.\n\nPS: 1.414213562 is the square root of 2, as in the square root (1'2 +1'2)\n\nDuckie
616025
Pretend that:\nx= length of base\n\nUsing the information given:\n4x-8= length of one of the legs\n\n\nProcedure:\n2(4x-8) + x = 29 ---> (there are 2 legs- so you x2)\n8x-16 + x= 29\n9x-16= 29\n9x= 29+16\n9x= 45\nx=5\n\nResult:\nSince x=5 and 4x-8 is one of legs of the triangle, by simplying applying the x=5 into 4x-8, you can get your length of one of the equal legs.\n\n4x-8= 4(5)-8= 20-8= 12\n\nANSWER: D) 12 cm
811567
do u mean angle or side? for side length you use the pithagorean theorem... a^2+b^2=c^2 ... yopu hace c, c=16, and a, a=9\nso ... 9^2+b^2=16^2......   b^2=16^2-9^2===175\nb= square root of 175 =13.2\nI
899919
16.72 inches squared.\n\n\nDivide an equilateral triangle into two equal right triangles (meaning, draw the height).  The right triangles will have angles of 30, 60, 90, right (an equilateral triangle has 3 60 degree angles)?\n\nThe ratio of the sides in such a right triangle (the smallest leg being a) is a:a(sq(3)):2a, where 2a is the hypotenuse and a(sq(3)) is the other leg.\n\nThe "other leg" in this case is the height.  So a, half of the length of the side of the whole equilateral triangle, is 5.382/sq(3).\n\nMultiply this by 2 to get the whole length of the side.  We have our base and height.\n\n\nThe area of an equilateral triangle is 1/2bh.  b=2(5.382)/sq(3)\nh=5.382\n\nPlug those in and you get 16.72.
1021808
Ok, you've got a good one here.\nFor a right triangle, height and base are the short sides; and area of the triange is (bh)/2. Since we know the area, and the length of the sides, we can plug it in to an equation, and find x. So...\nA = (bh)/2\n60 = ((5x)(3x-1))/2\n120 = (5x)(3x-1)\n120 = 15x^2 - 5x\n15x^2 - 5x -120 = 0\nDivide both sides by 5 to get\n3x^2 - x -24 = 0\nOk, there's a number of ways to do this, but the simplest way is to list the combinations of integers that will multiply to get -24 (the constant in the above equation). That gives us the combinations of (1, -24; 2, -12; 3, -8; 4, -6; 6, -4; 8, -3; 12, -2; and 24, -1). Now, we need to figure out the one combination that, when the second number is multiplied by three (the coefficient of x^2), you can add it to the first number to become (-1). It happens to be (8, -3) {-3*3=-9, -9+8 = -1}. There; you've factored the equation, leaving you with...\n(3x+8)(x-3)=0\nSince multiplication rules say that when ab=0, either a=0 or b=0, we can test by setting both numbers equal to zero; giving us...\n3x+8 = 0 ; x-3 = 0\nx = -8/3; x = 3\nSince x is an absolute length, and lengths can't be negative, we can disregard (-8/3) for now. That gives us x=3, and putting that back into the original lengths gives us sides of 15 and 8. To find the hypotenuse, we use Pythagorean Theorum: a^2 + b^2 = c^2, substituting in the known lengths gives us:\n15^2 + 8^2 = c^2\n225 + 64 = 289  = c^2\nc = √289 = 17\nSo, the hypotenuse of your triangle is 17.
748059
First use the Pythagorean Theorem.\n\nThe hypotenuse is 20. One of the legs has length x. Call the length of the other leg y.\n\nx^2 + y^2 = 20^2\nx^2 + y^2 = 400\n\nNow solve for y in terms of x.\n\ny^2 = 400 - x^2\ny = sqrt(400 - x^2)\n\nNow, since the triangle is right, the two legs are the altitude (or height) and the base.\n\nA = (1/2)*b*h\nA = (1/2)*x*y\nA = (1/2)*x*sqrt(400-x^2)\n\nThere's the area in terms of x!
778352
.............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.
503124
first split the triangle in half. \nfind the height of the triangle (with the 12cm side as base)\n\n7^2 - 6^2 = (height)^2\nheight = 3.61\n\nUse formula\n\nArea = (0.5) (height)(base)\n     = (0.5) (3.61)(12)\n     = 21.66 cm squared
981773
In a 45-45-90 triangle, there is an important ratio. leg-leg-hypotenuse = 1-1-square root of two.\n\nTo find the length of the leg if the hypotenuse is given...\ndivide the hypotenuse by square root of two.\nFor example... if the hypotenuse were 5, the leg would be 5/square root of 2 = about 3.5
865929
since it is right angle triangle, the addition of square of two arms will be equal to the square of hypotenuse.\n\nNow consider the smaller leg is 'x' inches, thus the other leg would be 'x+14' inches and hypotenuse would be 'x+16' inches.\n\nFrom the property of right angle triangle, we have now\n\n(x+16)^2 = x^2 + (x+14)^2\n\nor, x^2+32x+256=x^2+x^2+28x+196\nor, -x^2+4x+60=0\nor, x^2-4x-60=0\nor x^2-10x+6x-60=0\nor x(x-10)+6(x-10)=0\nor (x-10)(x+6)=0\n\nso x can have 2 values -6 and 10. since it is a distance it can not be negative, thus the shortest length of the right angle triangle is 10 inches. The other arm would 24 inches and the hypotenuse will be 26 inches.... (answer is 10,24,26)\n\ncheers:)
436114
well, you could be an architect and be using a 30-60-90 triangle and you needed to know a particular side of it. you can use x times the square root of 3, x being the shorter leg, to find the length of the longer leg.
912921
altitude = height = 552km\nAngle of dep = 39.7\n\nTangent of Angle = Alt/ Radius\n\nRadius = Altitude/ Tan(39.7) = 552/tan(39.7) I think... p.s not really sure