I am terrible at geometry, so can someone explain to me how to solve this problem?
487445
just pick any number and substitute it for x. lets try 10.\n\nw=4(10)-1   w=39\nx=7(10)+3   x=73\ny=3(10)-4   y=26\n\n\nso the answer is   x, w, y\n\nedit::\n\nmaybe next time you could ask the right question? you asked for the lengths of the SIDES, not the ANGLES. >:(
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.
380004
I don't understand.  Can you explain the problem a little better?
151334
there is a website called www.hotmath.com where they have some books that offer help on homework and explains how to do the problem step by step. \n\ni hope that may help you in geometry! \ndoctor math (google it) might help too
978173
You need to start off with a six-pointed star (constructed with two equilateral triangles).\n\ntriangle AEC and triangle FBD\n\nNow connect A to D, then connect B to E, and finally C to F. If you \ndrew everything correctly, those three segments should meet at a point \nin the center of the star.  \n\nThe 3 line segments you just drew plus the six sides of the star make \n9 lines. There are 5 points of intersection on each line, and the \ntotal number of intersections is 19.
901368
You need to take triangles with leg length 2-sqrt(2) from each corner. That is about 0.586".\n\nThe length of each side of the octagon will be 2 * sqrt(2) - 2. This is about 0.828"\n\nThis was calculated by setting the length of each leg of each triangle to x.\n\nhypotenuse = remaining side of square\n\nx * sqrt(2) = 2 - 2x. And then solve for x.
1017997
The equations are wrongor insufficient. Post the correct data if you want me to help/ Alternately, you could also message me.
855129
Word problems like this can be hard.  Everybody has problems with them sometimes.\n\nIt does help to draw it out.  So you draw a little rectangle like this:\n ______________\n|                            |\n|                            |\n|______________|\n\n2 of the sides are 120 m and the other 2 are 150 m.  So you take 2 times each of these numbers and add them together:\n2(120) + 2(150)\n\nThat's your perimeter, in meters.  Now if your poles are 2 m apart, you don't need one for every meter you get in your answer, you only need 1 every 2 meters.  So you divide your answer by 2, and that gets your number of poles you need to put around the field.
824775
The Area of a triangle is half the area of a square. So take one verticy and take a 90degree line from that up to the other corner.\n\nOh cr@p I'm starting to forget it all now. Too long ago.\n\nIt's something like that.
237985
Cut a 2x2 piece off. Cut the remaining 2x3 piece in half lengthwise.  Put the two resulting 1x3 pieces one atop the other and cut the stack diagonally.  You now have 4 triangles with sides = 1, 3 and sqrt(10).  Arrange them with their right angles touching the corners of the 2x2 square piece, 3-meter sides parallel to and touching the sides of the square piece, and all "pointing" in the same direction, clockwise or counterclockwise.  The result is a square that is sqrt(10) on a side.  Area = 2x5 = sqrt(10)^2 = 10 square meters.  Three cuts were used to make five pieces.\nTo solve this I started with the requirement of the large square being sqrt(10) on a side.  Sqrt(anything) suggests the hypotenuse of a right triangle.  A little mental fiddling produced the sides of 1 and 3 meters.  Luckily a 2x2 piece remained.
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
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
883909
draw FBDs of the 2 blocks independently\nbe careful\nfind the accelrations of the  2 blocks using CONSTRAINT relations\nand comes ur answer\n\ndon't wait for the solution\nit lessens ur thinking power n creativity\n\nif it's still problematic, consult ur lecturer\n\nconcentrate on ur studies \nur still a college guy
943665
this is impossible to answer, there are many shapes that could be made this way.  do you have any other information like right angles or angles at corners?
877622
ok...is what id imaigne you want is a ruler and compass construction of a line segment and you want to divide it into 4 equal parts...this isnt so bad\n\nis what you need is the bisector construction...and you do this 3 times...so you have your segment...go get yoru compass!\n\nok, now that you have your compass...put the pointy part of the compass on one of the segment endpoints...open teh compass so its open to a little more than half the length of the segment...draw an arc above and below the segment, then do the same thing with the other end point...\n\nyou now notice that those two arcs you drew intersect at 2 points...and if you connect those two points of intersectoin you would bisect the line....ok, so we have divided the line in half...well we need to do that again...\n\nso now go back to the endpoint you started on, and put the pointy end of the compass on the endpoint and open the compass to slightly more than half the distance from the endpoint to the bisector...arc above and below the line, repeat teh steps at the place where the bisector cuts the line...you would noice that these two arcs will intersect above and below the line...and you connect those points, and you have now split one half of the line into quarters...do the same thing for the other half\n\nhope this helps.\n\nmatttlocke
744397
Hi Karen,\n\nIt's good that you tried to do your problems. Some ppl just post stuff on yahoo! answers, without trying, to get their answer.\n\nAnyway, This is how you would do you problem. Belive me it's that tough.\n\nDefinition of Perimeter=sum of all sides.\n \nX=one side of pentagon\nY= one side of octagon\n\nGiven-\n5X=8Y\nX=3Y-2\n\nNow substitute X=3Y-2 into 5X=8Y to find Y.\n\nSo...5(3Y-2)=8Y\n15Y-10=8Y\n7Y=10\nY=10/7\n\nSubstitute Y=10/7 into X=3Y-2 to find what X is. \nso...X=3(10/7)-2\nX = (30/7)-2\nX=16/7\n\nEnjoy!
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.