What is the length of an isosceles triangle with congruent sides of length 26 1/2 and perimeter 73 in.?
545467
The lenghth of the missing side is 20 in. because an isosceles triangle has only 2 alike sides so you multiply 26.5 X 2 and get 53in and then subtract 53in. from 73in. and get the 20in.
890883
1)One leg of a right triangle is 12 cm long. The diference between the other two sides is 2 cm.\nFind the sides of the triangle. (Pythagorean Theorem)\n\nWhat we can take from this problem is that one LEG is 12 cm long, while the other LEG is unknown (we will call it X) and the HYPOTENUSE -- the longest side of a right triangle -- is also unknown (we will designate it X+2, since it differs 2 cm from the unknown leg).\n\nSetting up Pythagorean Theorem as always:\n\n "Leg squared" + "Leg squared" = "Hypotenuse squared"\n\n  12^2 + x^2 = (x+2)^2\n\nExpanding:\n\n  144 + x^2 = x^2 + 4x + 4\n\nCollecting like terms:\n\n  4x = 140\n\nDivide by 4:\n\n   x = 35\n\nThe unknown leg is 35 cm, while the hypotenuse is 37.  Thus, we can also call the triplet (12,35,37) a "Pythagorean Triple" (of which there are infinitely many!).\n\n\n2) The hypotenuse of a right triangle is 58 cm long. The diference between the other two sides\nis 2 cm. Find the sides of the triangle. (Pythagorean Theorem\n\nIn this case, we know the hypotenuse, but do not know the lengths of the legs.  For convenience, we will again use the expressions X and (X+2) for each side's length.\n\nBy P.T.:\n\n x^2 + (x+2)^2 = 58^2\n\nExpanding:\n\n x^2 + x^2 + 4x + 4 = 3364\n\nCollect like terms:\n\n 2 x^2 + 4x - 3360 = 0\n\nDivide by the GCF (2):\n\n  x^2 + 2x - 1680 = 0\n\nCan we factor this trinomial easily?  We could, if we knew integer factors of -1680 (one positive, one negative) that combine to make 2.\n\nLuckily, -1680 can be factored into -2*2*2*2*3*5*7, and these can be grouped into two sets:\n      2*3*7 = 42\n -2*2*2*5 = -40\n\nFactor:\n\n (x+42) (x-40) = 0\n\nAlthough two solutions come from these factors (-42 and 40), we may only consider the positive answer, since lengths of triangle's legs are always positive!\n\n Therefore, x=40 (Leg 1), and Leg 2 is 42 cm.
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
942453
From symmetry, you have all the angles\n69 for two, and 180-69= 111 for the other two.\nUse the law of cosines: for triangle abc with opposing angles A,B,C we have the length of the 3rd side c for given a,b from:\n\nc = sqrt(a^2 +b^2 -2abcos(C))\n\nIn your case: a=4, b=6 and C = 69 so\n\nBD = sqrt(36+16-48(cos(69)) = 5.9\n\nFor C = 111 we have:\n\nAC = sqrt(36+16-48(cos(111))= 8.32
875314
Fisrt divide the regular octagon into 8 isosceles triangle. The resulting vertex angle would be 45 deg.\nDivide the isoscles into congruent right triangles with an angle of 22.5 deg, 67.5 deg and 90 deg.\nThe length of the sides opposite 22.5 deg would b equal to\nhypotenuse (radius of the circle) times sin 22.5 deg.l\nThat would be 6.046398231 cm.\nDoubling this value would result to the side of the octagon,\nand that is 12.09279646.\nMultiplying the side by 8, we get the final answer or the perimeter of the regular octagon inscribed in the circle.\nthe answer is\n96.7423717 cm.\n\nYou my use this formula for any regular octagon inscribed in a circle with radius x.\nP = 16x sine 22.5\nContact me if this is wrong.
906875
The trick is turning the word problem into equations.\n(Key: length = l; width = w)\n\nThey tell you the length is 3 feet more than twice the width, so:\nl = 2w + 3\n\nThen they tell you the perimeter (which is twice the width and twice the length) is 36, so:\n2l + 2w = 36\n\nYou should be able to take it from there.
473359
DEepnds on the shape. If it's a quadrilateral then it's 25 m. If it's a triangle, then it's 15 m. But that's only if all the shapes are regular (all sides and angles the same).
563121
a right angled isosceles triangle\n\nplease give me 10 points now
1074757
The problem is that 30, 31, and 63 are not the sides of a valid triangle.  For any triangle with sides a,b,c, you must have\na + b > c\nb + c > a\nc + a > b\nBut you have a + b < c, so the triangle isn't valid.
621583
What type of triangle is it?\nIf it is an equilateral triangle, all sides will be the same length, so p=(1.25 X 3)m\n\nHope this helps
204855
Looks to me like you are trying to figure out a "PERIMETER" rather than an "AREA".  If so, you need to ADD the lengths of all the sides together.  If you ARE looking for an area, it is simply Length x Width (for a rectangle or square), 1/2 the base x height (for a triangle), or Pi x radius squared (Pi is approximately =3.1415926) for a circle.  If the "shape" of the object you are trying to find the area of is irregular, then you must divide it up into these familiar shapes first, find EACH individual area, then add those areas together for the total area.
684692
Question 20906\nSubmitted on 5/28/2002\n\nFind the area of a nonagon inscribed in a circle of radius 24cm.\n\nSolution 20906\n\nThe nonagon can be divided into nine congruent isosceles triangles. \nThe lengths of the sides are 24 cm and the vertex angle is 360/9 = 40.\nThe base of one triangle is\nB = 2 * 24 * sin (20)\nThe height of the triangle is\nh = 24 * cos (20)\nSo the area of one triangle is\nA = (1/2) * 2 * 24 * sin (20) * 24 * cos (20)\nA = 24^2 sin(20) cos(20)\nAnd the area of the nonagon is\n9 * 24^2 sin (20) cos (20)\n= 1666.1\n\n\nSubstitute in your 1 meter for the 24 centimeters and you'll get your answer.  I believe the answer should come out to somewhere around 3.353 m^2
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
354025
the perimeter of a figure is the distance around the edge of the figure.  so in this case, the triangle is 6 in + 8 in + 10in = 24in.
915466
Let a and b the lengths of the sides of the rectangle (with a the side with the semi-circle).\nThe perimeter is 2b+a(pi/2+1)=16, and the area ab+pi/8*a^2. Substituting b in the area (from the perimeter equation), the area is 8a-a^2*(1/2+pi/8). \nSetting the derivative to 0, we have a=32/(4+pi). Then b comes from the perimeter constraint as b=8*(1-(2+pi)/(4+pi)).\nNumerically this gives a=4.48, b=2.24 and an area of 17.92
1023518
Use the Law of Sines.\n\nsin A/a = sin B/b\n(sin 36)/24 = sin B/47\nsin B = (47 sin 36)/24\nsin B = 1.151079452\nAt this point you have a problem--it is impossible for the sine of an angle to have a value greater than 1.\n\nSo either you copied the problem wrong, or you are being asked to recognize that a triangle cannot be formed with these dimensions.
865033
okay. so. there's a special set of rules for a 30-60-90 triangle.\n\nthe sides are proportional to each other by the ratio \n\nx : v/( 3)x : 2x\n\nwhere the v/ is the square root.\n\nthat means that the greatest angle corresponds to the greatest length.\n\nbecause the longest side is 7.5 meters, that means that is 2x.\n\ndivide by two for x, and that is the 30* length. (because its the smallest angle, therefore is opposite the smallest length)