How do you find the perimeter of a triangle with variables?
952527
These aren't angles, so it won't equal 180.  These are the measurements of each side, but given with variables.\n\nPerimeter = side A + side B + side C\n\nYou just add the terms.\nPerimeter = (x + 3) + (2x + 1) + 6\n\nNow group your x terms and your unit terms:\nPerimeter = (x + 2x) + (3 + 1 + 6)\n\nNow simplify:\nPerimeter = 3x + 10\n\nThere's your perimeter --> 3x + 10\n\nBONUS INFO:\nAnd if they had continued the question and said the perimeter was 40, you should be able to figure out the sides...\n\n40 = 3x + 10\n3x = 30\nx = 10.\n\nSide A = x + 3 = 13\nSide B = 2x + 1 = 21\nSide C = 6\n\nFor your question though, the perimeter is:\n3x + 10
766258
The length of the wire must be (2x5)+(2x7)=24cm.  So each side of the triangle must be 24/3=8cm. long.
136459
Find the factors of 56 (or all the possible measures of sides that would equal an area of 56). \n2 and 28\n4 and 14\n7 and 8\n\nThen add 2l + 2w to get the perimeters.\n2(2) + 2(28) = 60\n2(4) + 2(14) = 38\n2(7) + 2(8) = 30\n\nSo the least perimeter would be 30 cm.
1108873
Here's the process you would use to solve the problem.\n\nFirst, you need to know that an isosceles triangle has two equal sides.\n\nTherefore, you have two variables in the problem\n                    x = length of leg (there are two of these)\n                    y = length of base\n\nWe know that "the length of one of the equal lengths of an isosceles triangle is 8 cm les than 4 times the length of the base."  Therefore, we can write "x" in terms of "y":\n\n                    x = 4y - 8\n\nFurthermore, we know that the perimeter is 29 cm.  Now we can set up an algebraic expression to solve for "x."\n\n2x + y = 29   (initial equation)\n2(4y-8) + y = 29 (substitution)\ny = 5 (solved)\n\nWe have the length of the base ("y").  Now we can solve for the length of one of the equal legs.  Remember from above that\n\nx = 4y - 8\nx = 4(5) - 8\nx = 12\n\nSolved!\n\nIf you are still confused, draw an isosceles triangle to visualize the problem.  Label the base "y" and the two legs "x."
1118220
Perimeter = 60 = 2L + 2w\nw = (60 - 2L)/2 = 30 - L\n\nArea = 216 = Lw\n216 = L (30-L)\n216 = 30L - L^2\nL^2 - 30L + 216 = 0\n(L + 6) (L - 36) = 0\nL = -6, 36\nLength cannot be negative\nTotal Length = 36\neach length = 36/2 = 18\n\n60 = 2(18) + 2w\n60-36 = 2w\n24= 2w\nw = 12\n\nTo check:\nA = LW\nA = (12)(18) =216\n\nSo, 12ft x 18 ft
976501
You mean "one less."\n\nCall the base x.\nThen each leg is 2x-1.\nAnd the perimeter is x + (2x-1) + (2x-1), which equals 18.\n\nJust solve for x and you can calculate the lengths of the sides.
764538
For any given triangle ABC with sides a, b and c.\n\nCos C = (a^2 + b^2 - c^2)/2ab  This is the cosine rule.\n\nSin C =sqrt(1 - Cos^2 C). You may need to adjust the sign of the sign to match the triangle.
303035
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.
943161
The faster way to solve this questions is using graphical method. Draw the respective triangles (choosing appropriate scale) and you will get answer.
897796
Let L be the length\nLet W be the width\nWe know the perimeter is:\n2(L + W) = 300\nSolving for L:\nL + W = 150\nL = 150 - W\n\nNow write a formula for the area:\nArea = L * W\nArea = (150 - W) * W\nArea = -W² + 150W\n\nYou can see this is the formula for a parabola.  I'll simplify things by expressing it in terms of x:\nf(x) = -x² + 150x\n\nBut it isn't in vertex form:\nf(x) = a(x - h) + k\n\nTo get it in that form, first pull out the -1 to get x²:\nf(x) = -1(x² - 150x)\n\nNow take the coefficient on the x term (-150), take half of it (-75) and square it (5625). Add and subtract this:\nf(x) = -1(x² - 150x + 5625 - 5625)\nNow you can write this as a square:\nf(x) = -1[ (x - 75)(x - 75) - 5625)\nSimplify:\nf(x) = -1(x - 75)² + 5625\n\nYou have a downward facing parabola, with a maximum vertex at the point (h, k) or (75, 5625)\n\nSo the maximum is when the width is 75. When you solve for length you get that the length is also 75. So the maximum area for the patio is a square with sides of length 75 feet.  The total area will be 5,625 sq. feet, a maximum.
967287
The triangle will have 2 sides of length x, and the remaining side will be length P-2x making the parameter x + x + P - 2x = P.\nBegin by drawing a line from the vertex where the two sides of length x meet, to the center of the side which is length P-2x. The line should be perpendicular to the side P-2x thereby splitting the triangle in half, into 2 right triangles. One of the right triangles has hypotenuse = x and one leg = (1/2)(p-2x). So the remaining leg has length S.R. (x^2 - (p/2 -x)^2). We calculate the area of this right triangle, and double it since there are two triangles:\nArea = (2)(1/2){(1/2)(p-2x)S.R.(x^2 - (p/2-x)^2)\n         = (1/2)(p-2x)S.R.{px - (p^2)/4}\n         = (1/4)(p-2x)S.R.(4px-p^2)\nNote that your given answer is wrong - it has an x^2 where it should have a p^2....
771384
perimeter = 2*L + 2*W\n58 = 2*L + 2*12.\nCan you solve for L?
838640
The perimeter is the measure of all of the sides added together.  If you have a square, the sides are all equal.  So, 3+3+3+3= 12 OR 3x4=12.  The answer is D.
434240
Generally, midpoints divide the lengths into half. So then, the whole perimeter of LMN is 1/2 of ABC. So the ratio is 1/2.\n\nYou can pick two CORRESPONDING sides, and their ratio is still 1/2. To find the area, there's a formula:\n\n(Side 1/Side 2)^2\n\n(One length of LMN/One length of ABC)^2\n= (1/2)^2\n= (1^2)/(2^2)\n= (1)/(2*2)\n= 1/4\n\nNote: one length of LMN doesn't have to be 1 and another length of ABC doesn't have to be 2. They can share common multiples, but you end up simplifying to ratio 1/2.\n\nratio of perimeter of triangles LMN to ABC is 1/2\nratio of areas of triangles LMN to ABC is 1/4
48612
Let r be the radius of the circle and s the length of the square's sides.\ncircle perimeter=2*pi*r, square perimeter=4*s\n2*pi*r=4*s => s=pi*r/2\ncircle area=pi*r^2, square area=s^2\ns^2=(pi*r/2)^2=pi^2*r^2/4\nBut pi<4 => pi^2*r^2/4<pi*r^2 => square area < circle area
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.
528659
This is simple algebra. I'll use P for perimeter, and L and W for length and width.\n\nWe know that 2L + 2W = P = 94. Also, 2W + 3 = L.\n\nSo, put these equations together and solve.\n\nL - 2W = 3\n2L + 2W = 94\n\nAfter solving, you get L = 97/3 and W = 44/3.