how many toothpicks does it take to create each side of the right triangle?
472364
if the toothpicks are of the same length, you'l need a side of 3 t'picks, anther of 4 and the last one of 5.\n\nyou can prove it using the pitagorean theorem\na^2 + b^2 = c^2\n3^2 + 4^2 = 5^2\n9 + 16 = 25\n25 = 25
953424
Yes. Right triangles have them all the time. it's called the orthocenter i think....it's been a while since Geometry.
442290
1. pythagoros theorem applies to any right angled triangle.\n\n2. The geometric mean of a set of positive data is defined as the nth root of the product of all the members of the set, where n is the number of members.\n\n3. in an isosceles triangle 2 sides and 2 angles are equal\n\n4.hypotenuse is opp. to the right angle.it is the largest side.\n\n5.true
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.
610782
A circle has 360 degrees. Now imagine that from the icosagon you can make 20 equally-sized triangles. The angle made by each triangle is 360/20=18 degrees. The other two angles of each triangle are both 81 degrees.\n\nThe basic formula for the area of a triangle is 1/2 base x height. How do we find these? Well, imagine that from each of the 20 triangles we make two additional right triangles. In doing this we know several things:\n\n-The hypotenuse is equal to 6.5 cm (the radius).\n-The angles of the smaller triangle are now 9 degrees, 81 degrees, and 90 degrees.\n-The sine of 9 degrees is equal to the opposite side divided by the hypotenuse. If we multiply this value by 6.6 we get the length of the opposite side ("1/2 base").\n-The sine of 81 degrees multiplied by 6.5 gives us the "height".\n-Once you compute each value (1/2 base and height) you multiply them by each other and you get the area of the triangle.\n-Multiply the area of that triangle by 20, since you have 20 of them. Voila.
892974
a right triangle is called a right triangle because there is a 90 degree angle in it.
722996
Drop a perpendicular to side of length 10 from the vertex where the side of length 8 ends.\n\nA right angle triangle is formed,\n\nthe perpendicular is 8*sin 60 = 4sqrt(3)\n\nArea of ||gm is 10*4*sqrt(3) = 69.282 approx.
600610
In a single hand, there are 5 phalanges, and 5 meta phalanges. \n\nI hope this helps!
6544
By using previously existing toothpicks
882048
Here's the answer you want:  With a compass set to the radius of the circle, start marking off radii around the circle, dividing it into 6 equal parts.  Connect any 3 alternate points and you'll have your triangle.\n\nNO MATH REQUIRED!
63969
A hexagon has 6 sides.
904256
there are 6 vertices and you form 4 triangles from the first 2, then 3, then 2, then 1 for a total of 14.\n\nNow we form other triangles that overlap and it gets really complex...\n\nI count 18 smaller triangles, for a total of 32 so far...\n\nNow middle triangles...\n\n36, for a running total of 68
779652
Do you mean a right triangle, if so the answer is 13. 5^2 + 12^2 = 13^2. If not, it depent on the angles.
352939
it should have a right angle.\n\nThus, it's a right triangle
736929
I think I'd start with 9 isosceles triangles.\n\nEach angle at the apex is 360 divided by 9 degrees\n\n=40degrees.\n\nThe other two angles are equal, so subtract 40 degs from 180 degs and halve what's left.\n\nEach base angle of the triangle = 70 degs.\n\nYou already have 40degs and the side opposite it (0.80m)\n\nSo you can use a over sinA = b over sinB to find the other two (equal) sides of the isosceles triangle.\n\nNow drop a perpendicular from the apex of the isosceles triangle to the (middle of) its base, forming two right angled triangles.\n\nEach of these triangles is half of the isosceles triangle.\n(remember the base of the right angled triangle is half the 0.80m)\n\nNow you can use \n\nsin = opp/hyp,  cos = adj/hyp,  tan = opp/adj to find the length of the perpendicular height.\n\nNearly there!\n\nArea of a triangle = half the base times perpendicular height \n\nWhen you've worked this out for one of the isosceles triangles, you just have to remember back in line one, that you are finding the area of 9 identical triangles, so you multiply the area of one by 9 to get the area of the nonagon.\n\nThis uses lots of fairly sinple steps, which you should be able to remember in an exam.  PS don't try to remember the order of the steps we used here, just have confidence that when you start, the solution will unfold for you.  If you wanted to practice, try it in a week with a different polygon!
881588
First, the isosceles triangle has sides of 10, 12, 12.  You can find the angle by law of cosines or, as I prefer, just divide the triangle into two right triangles.  The short side is 5 and the hypotnuese is 12 so the angle at G is arcsin(5/12) and the total angle FGA is twice that (since we split the original triangle in two) <FGA = 2*arcsin(5/12) = 49.249 deg.\n\nIf you make a pyramid, The radius of the hexagon (10 cm), the 12 cm side and the height form a right triangle so the height is:\n\nh = sqrt(12^2 - 10^2) = sqrt(44) = 6.633\n\nFinally, for the area, we need the height of the isosceles triangle which is the tjhird side of the right triangle in the first part. This height is:\n\nc = sqrt(12^2 - 5^2) = sqrt(119) = 10.909\n\nSo the area is: A = 3*10*sqrt(119) = 327.261
39543
There are seven different (even if rotated, they're still different) ways to join the points together. Multiplied by 5, this gives 35 triangles total.