Find the length of the hypothenus of a right triangle with sides of 20 cm and 15 cm.?
965204
Pythagorean Theorem\n\nc²  = √a ² = b²\n\nc²  = √(20)² + (15)²\n\nc² = √400 + 225\n\nc² = √625\n\nc = 25\n\nThe answer is c = 25\n\nThe length os the hypotenus is 25 cm\n\n- - - - - -s-
634321
Every time the ray of light goes from one mirror to the other, it goes through a certain distance d along the mirrors axis (in the 20m direction).   The law of reflection says that the angle of incidence is the same angle as the one the ray will leave the mirror surface, so, we can build a right triangle, with the reflected ray of light as the hypotenuse.  So, we have:\n\nr*Sin( theta ) = d    and   r*Cos( theta ) = a\n\nDividing the first over the second:\n\nSin( theta ) / Cos ( theta ) = d / a\n\nOr:    Tan( theta ) = d / a\n\n            --->   d = a*Tan( theta )\n\ntheta being 30 degrees and a being 10cm = 0.1m\n\nNow, just divide 20m over d and you're good to go.\n\nHope I helped.
252781
Well, the circumference (C) is 2 * pi * radius, so just divide 15 by (2*pi).\n\nANSWER : 15 / (2 * pi) = about 2.39 cm
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....
498476
Unfortunately, this is wrong.  (And I do believe that I am the first one here with the correct answer...hint, hint)\n\nHere is how it should be solved algebraically:\n\nThe areas of the sides are:\nL*W = 80 \nL*H = 20 \nW*H = 16 \n\nTo find the lengths of the sides use the equations for the areas:\n\nH = 16/W\n\nsubstitute for H\nL*16/W = 20\n\nW = 80/L\nsubstitute for W\n\nL*16/(80/L) = 20\nL^2 = 1600/16 \nL = sqrt(100) = 10\n\nW= 80/10 = 8 \n\nH = 160/ 80 = 2\n\nV = L*W*H = 160 [in^3...don't forget the units!]\n\nTo check:\n\nMultiply:\n\nL*H = 10 * 2 = 20\n\nL*W = 10 * 8 = 80\n\nW*H = 8 * 2 = 16\n\n\nYou could have also solved it geometrically be realizing that:\n\nV = L*W*H \n\nV^2 = L*L*W*W*H*H = L*H*L*W*W*H = 20*80*16 = 25600\n\nV = sqrt(25600) = 160 in^3\n\nEither way you get the same answer for volume.  The algebraic way, however, gives you more information about the solid (ie, the lengths of the sides) and could be useful for any follow-up questions one might expect.
640092
450 cm\n1/2 base x height = area of triangle\n\n15 x 30=450 cm
882787
let \nc=hypotenuse\na,b = legs\n\ntherefore:\nc=25\na=2b-10\n\nform pythagorean theorem;\n\na^2+b^2 = c^2\n(2b - 10)^2 + b^2 = 25^2\n4b^2 - 40b + 100 + b^2 = 625\n5b^2 - 40b - 525 = 0\n\ndividing both sides by 5\nb^2 - 8b - 105 = 0\n\nfactoring the left side\n\n(b - 15)(b + 7) = 0\n\nb + 7 = 0\nb = -7 absurd (no side can be negative)\n\nb - 15 = 0\nb = 15\n\nsolve for a:\n\na = 2b - 10\na = 2*15 - 10\na = 20\n\ntherefor the two legs of the right triangle are\n20 cm and 15 cm\n\nchecking:\nc^2 = 20^2 + 15^2\nc^2 = 400 + 225\nc^2 = 625\nc = 25 cm.\n\n\nthat's it
195018
Draw all this out so you can see it - it won't be to scale, but don't let that fool you.....\n\n\ndiameter = 50cm (radius = 25cm)\nchord = 20cm from diameter (thus it's 20cm from the center)\n\nOK, you'll need to assume that the line segment forming the diameter touches the chord at point B on the circle - i.e. we'll say the diameter is line segment AB and the chord is line segment DB. \n\nC is the center of the circle, so we have radius CB = 25cm. \n\nAlso, we'll have line segment CE drawn perpendicular from the center (C) to point E on the chord, forming a right angle and thus *bisecting* chord DB.\n\nYou have the hypotenuse of triangle ECB (radius CB) as 25cm, and side EC as 20cm (distance from the center to the chord - always perpendicular).\n\nLine segment EB will be the second side of a right triangle, so your formula:\n\nEC^2 + EB^2 = CB^2\n20^2 + EB^2 = 25^2\nEB^2 = 25^2 - 20^2\nEB^2 = 625 - 400 = 225\nEB = 15cm\n\nSince EB = half the length of the chord DB, DB = 2*15cm = 30cm.
717001
Since the triangle is Isoceles, if you split the triangle down the middle you can turn in into a right triangle. You can use the Pythagorean theorem. (a^2 + b^2 = c^2.)\n\n13 will be the hypotenuse and 5 will be the base. To find the altitude you just fill everything in. \n\n5^2 + b^2= 13^2\n25 + b^2 = 169\nb^2 = 144\nb = 12
960920
0.07896 km\n153 mm\n0.578  km
534524
A nonagon can be broken into 9 isoceles triangles.\nThe vertex angles will be 360/9 = 40 degrees.\nThe two sides adjacent to the vertex have length 1 meter\nEach of the 9 triangles can be split into 2 right triangels with angles of 20 and 70 degrees.\nThe side opposite the 20 degree angle is sin(20) meters.\nThe side adjacent the 20 degree angle is cos(20) meters.\n\nEach of those triangles will have area (1/2)sin(20)cos(20 m^2\n\nThe total area of the 18 triagles combined is 9sin(20)cos(20).
80712
If you mean 1 cm and 5 cm, then the answer would be\n\n3 cm\n4 cm\nand a 5 cm hypotenuse.
903414
Because the diagonal creates two right triangles you can use the Pythagorean theorem to find the length of the other side:\n\n5^2 +x^2 = 9^2\n25 + x^2 = 81\nx^2 = 56\nsqrt of 56 = 7.48\n\nNow that you have the length of the other side you can use the formula    area = length x width:\n\n5 x 7.48 = 37.4 feet
535960
The speed has no part in the solution, as it is the same for all three paths. So, you draw a triangly with legs 2.25 and 3 in lenght. the solution is the lenght of the hypotenusa formed. the answer is 3.75 min.
616025
Pretend that:\nx= length of base\n\nUsing the information given:\n4x-8= length of one of the legs\n\n\nProcedure:\n2(4x-8) + x = 29 ---> (there are 2 legs- so you x2)\n8x-16 + x= 29\n9x-16= 29\n9x= 29+16\n9x= 45\nx=5\n\nResult:\nSince x=5 and 4x-8 is one of legs of the triangle, by simplying applying the x=5 into 4x-8, you can get your length of one of the equal legs.\n\n4x-8= 4(5)-8= 20-8= 12\n\nANSWER: D) 12 cm
897915
500 miles\n\nthink of it as a triangle with the hypp. being the 1000miles and the distance east being the opp. side.\n\nuse sin.\n\nsin(degrees)=opp./hypp.\nsin(30)=opp./1000\n(1000)(sin(30))=opp.\n500=opp.\n\n500 miles east
1116580
Brian gave you the answer, but he didn't explain it thoroughly.\n\nPlace your vector at the origin of an x and y plane.  The x and y axes are perpendicular, so you can use the Pythagorean formula to get the magnitude which is the length of the hypotenuse of the right triangle formed by the points (0,0), (10,0), and (0,8). sqrt(10^2+8^2)\n\nNow you need direction which is the angle between the x axis and the vector. Since you have solved for all three lengths of the triangles sides, you can use any inverse trig function to find the angle.  Brian chose tangent.\n\nGood work, Brian!