How would I go about this Geometry problem (details below)?
913445
if it is equilateral i, it is also equiangular so all angels equal 60 degrees...\nso solve 19x + 3=60sub that in to ...23y-3x = 60 for x and solve for y and sub y into 5y + z = 60 and solve for z
328129
just draw a triangle has right angle which dimensions 1,sqrt(3),and hypothenus 2. if you want it, i can send to you
716347
Of three points, A,B,C\n Check the slope of A-B,  B-C, A-C.  If any of the slopes are the same, you have parallel lines and thus no triangle.  So, to do (2), do (4) first.\n  If you know the distance between A-B, B-C, A-C, you can calculate the area with the formula, so (5) also depends on (4).\n\n You write your own code.  yA isn't a site to do your homework, since if you don't do it yourself, you don't pass the class!
4519
(4/3)*pi*r^3\n\n[four thirds pi r cubed]\n\ngood luck!
903397
You might try asking this question under "Mathematics."
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.
572318
1) Let x:the length , y:the width of the picture.Note:x,y>0\nThe first equation we have is : x=y+4 (1)\nThe combined area has the length is x+4,the width is y+4       \nso the second equation we have is: (x+4)*(y+4)=140 (2)\nReplace (1) in (2)-->(x+4)x=140--> x^2+4x-140=0 -->x=10 or x=-14 .since x>0 then x=10(cm)-->y=6(cm).\n2)  Let x: the length,y:the width.Note:x,y>0\n    The first equation we have is x=y+3 (1)\n    The second is: the area=x*y=550 (2)\n    Replace (1) in (2)--> (y+3)y=550--> y^2+3Y-550=0 --> y=22 or y=-25.Since y>0 then y=22(cm)-->x=25(cm)\n3)  Let x,y:two positive integers.Supposed that x>y>0 \n    we have two equation:x*y=240(1) and x=y+8(2)\n    Replace (2) in (1)-->y(y+8)=240--> y^2+8y-240=0 -->y=12 or y=-20.Since y>0 then y=12-->x=20.
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.
93089
This seems like a homework problem, so I'll give a few hints.\n\nThe arclength is radius times the angle subtended.\n\nTo find the angle subtended, consider the isosceles triangle created inside the circle where the chord is the hypotenuse and the radius of the circle is the two other legs. \n\nSince the isosceles triangle created is not a right triangle you need to use a specific formula to find the angle "C" subtended by the chord.
87753
Just add up the lengths of the three sides. Works for any triangle - wether it's equilateral, isosceles or obtuse.
1056342
Hint, try to work out the radius!  Do your homework yourself.
975485
H*W= area\nheight times width\n\none side has a height of 8 and its width is 9 thus the area would be 72
89202
Hi again, just graph the three boundary lines and shade the correct sides.  This will give you a shaded triangle with vertices (2,3), (4,5), and (4,0)  [I think; I don't have graph paper handy].  The maximum and minimum values must occur at one of these 3 sets of coordinates.\n\nfor (2,3), 3x + y = 9\nfor (4,5), 3x + y = 17\nfor (4,0), 3x + y = 12\n\nSo 9 is the minimum value and 17 is the maximum.\n\nCheck it out yourself to be sure the graphs really intersected where I thought they did!
970422
The solution is too convoluted to put it here without diagrams and a good equation editor.  If you wish, I can email a Word document detailing the solution.  I figure that the cone is (1 + 2*sqrt(2/3) + sqrt(3))*r (about 4.365 r) high and has a base radius of ((sqrt(1/2) + sqrt(4/3) + sqrt(3/2))*r or about 3.087.
974060
Use "ray tracing" and ratio and proportion. First draw a horizontal line from the point source to the screen that is 2.20 + 1.40 m long. At a distance of 2.20 m from the point source let the horizontal line be a perpendicular bisector to a line that represents the aperture (unknown height as yet). At the screen let the horizontal line be a perpendicular bisector of a line that is 14.0 cm (0.14 m) high. Now draw two lines from the point source to the ends of the line representing the spot of light on the screen. Where these two lines cross the line of unknown length, 2.20 m from the point source, is the aperture.\n\nYou now have two isosceles triangles with a common apex at the point source, the base of the smaller one being the aperture and the base of the larger being the line representing the spot of light on the screen. Since these are similar triangles, you know the ratio of their heights is the same as the ratio of their bases. Plug in the numbers and solve for the aperture width.
543695
volume = (1/3) b h = 1/3 pi r 2 h \nSurface area = pr2 + prs\ns= the height of the outside of the cone, while h = the height of the cone at its center
847510
Several ways. The first that comes to mind is to figure the area with Heron's formula (area = sqrt(s(s - a)(s - b)(s - c)), where s = (a + b + c)/2). The area can also be figured as 1/2 base times altitude, so you know the altitudes of the triangle. The altitudes give you right triangles where you know the hypotenuse (a side of the triangle), and a leg, so you can compute the trig functions of all the angles. Then the angles can be computed as arcsin or arccos of the angle (arcsin(altitude/side))