creat 5 triangles with the letter H and 2 lines?
927347
Draw lines from the top of the vertical lines in the H to the opposite lower corners. There you go.\n\nIn response to the comment below:\nBTW\nI guess the middle line of the H would have to be not directly centered between the lines.  It has to be a little highter or lower so the 2 lines do not intersect on the center line.  Then you come out with 5 individual triangles and not triangles within triangles.
331507
sorry hunny you need at least 2 side lengths in order for us to help you figure it out... plus once you have that the formula for figuring it out is using the pathagorean theorem! a2+b2=c2 (s = squared) say for instance... the 2 sides of your triangle are 7 and 3inches. 3>2+7>2=c>2\nwhich would turn into 9+49=c>2~~~~~~~ then 58=c>2 then to get c by itself(which is needed) you would take the square root of 58! your hypotenuse would equal 7.6 or rounded, 8!\nnow if you use my steps you should be able to do your problem! I'm a 10th grader in geometry and it's the most lovely subject!
703485
1) 9 * (t + 1)\n2) 8 * (v + 7)\n3) 25 * (5 - h)
780884
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.
120928
ok, let the triangle be ABC with sides AB, BC, AC.\ndraw a typical line, get too point on in with the distance of BC, name them B and C. then draw an arc with the center of B and the length of AC and another one eith the center of C and the length of AB. if they had a common point, it would be A and your triangle is there, but if not, there is no triangle with those three sides.
239138
well you could split up the thing into triangles. you know what the base is (1m) and the height is unknown, so lets call it h for now. so for each side, the area of the triangle is h/2 (bh/2 and b=1 so h/2). so for every 2 sides, the area is h. so the total area is 18h. \n\nso to find h, you have to use some trig. the equation to find the internal angles is (n-2)*180/n. cuz the sum of the internal angles is (n-2)*180. and then to find each angle, you have to divide by number of angles (which is same as number of sides). so in this case 34*180/36 = 170 degrees each angle. so now for each side, if you make a triangle with the center of the polygon, u should notice that the triangle is isosceles and the base angle is half of the side angle. \nas in the bottom 2 angles are equal and they are equal to 170/2. which is um 85. you can try this with any polygon if you dont believe it. try a square, (n-2)*180 = 2*180 = 360 is the sum of the internal angles. so each angle is 360/4 = 90 degrees. so if u connect each vertex to the center of the circle, you get 4 isosceles triangles and the base angles are 45 degrees each. \nso you have 85 degrees as the angle. so now if you split the isosceles triangle into 2 right angle triangles, you have 3 sides in each right angle triangle: the h, 0.5m and the line from the vertex to center. so you have the angle 85 degrees and you need h and you have the adjacent side = 0.5m. so you can use tan 85 = h/0.5 and you get h. \nso the total area is 18h. \n\ni really hope you understand that. its kinda hard to explain without drawing it out and showing you. if you need more help, then message me or somethin. later
389484
Draw a triangle with the letter "M" INSIDE and the triangle should cut the top 2 vertices and the 3 bottom vertices.\n\nyou should get 8 triangles and the last 1 is the big triangle with "M" inside.
701484
The answer is 632 triangles!\n\nCheck out the website below for confirmation and a list of other shapes and the number of triangles formed by their vertices.
1070589
Assuming you mean as 4 shapes equivalent to the original L shape, the answer is given in the attached image.
460943
You can't do it in three.  You need four lines to complete.  Click below for the solution.
791300
Use the FOIL method; First, Outer, Inner, Last.\n\nFirst, n * n\n\nOuter, n * 5\n\nInner, n * 7\n\nLast 7 * 5.\n\nThen add them all together.\n\n22 NOV 06, 0420 hrs, GMT.
787829
SOHCAHTOA\nsinθ = O/H = a/c\ncosθ = A/H = b/c\ntanθ = O/A = a/b
63969
A hexagon has 6 sides.
891617
The triangle area formula is\n\nA = 1/2 b h\n\na = 1/2(5)(2)\n\nA = 5/2(2)\n\nA = 10/2\n\nA = 5\n\nThe area of the triangle is 5 feet\n\nThe answer is " a "\n\n- - - - - - - - - - - -s-
238182
p1    p2\n--- = ---\nq1    q2\n\n 5     2\n--- = ---\n 2    q2\n\nCross multiply,\n\n4/5
692268
1. c\n2. c\n3. c\n4. c\n5. c
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width = 5 \nlength = 12