sides of a triangle are 10 ,17, and 21. there is a square in the triangle. It touches two of the traingle edges and its base rest on the longest side of the triangle. what is the side of the square.

• first off all take out the area of triangle of sides 10 17 and 21. s=1/2(10+17+21) = 24. then area of triangle :- A=84 by hero's formula. now in the triangle draw a perpendicular to base 21 and name it as h and the sides of square as d.
  again area of triangle = 1/2 *b*h =84. h=(84*2)/21.now imagine a smaller triangle as i cant show u here just above the square from the vertex from where the perpendicular is drawn.both the triangle the smaller and the actual one are similar now both being similar . perpendicular/base of bigger=perpendicular of smaller/base of smalle. 8/21=(8-d)/d = solving we get d=168/29.
• 7 years agoHelpfull: Yes(16) No(1)
• let the side of the square be x
  see,if ABC is the triangle taken,n be DEFG the square resting on BC.then triangle ADE becomes similar to ABC( since DE||GF,as it is a square).
  now DE/BC=height1/height2--------1
  let height2 be the height of ABC=h
  so height1 becomes h-x--------2 (how?see height of ADE which will be total height -niche ka portion ,which is nothing but side of the square )
  put eq 2 in 1
  DE/BC=(h-x)/h
  x/21=(h-x)/h---------3
  now we need to find h to find x
  so,
  equating area of ABC
  1/2*b*h=under root(s(s-10)(s-17)(s-21); (s=(10+17+21)/2
  =>s=24)
  => 1/2*21*h=under root 24*14*7*3
  => h=8---------------4
  put eq 4 in 3
  => x/21=(8-x)/8
  => x=168/29
  this is 100% correct
• 7 years agoHelpfull: Yes(6) No(0)
• Check this out:
  http://www.qbyte.org/puzzles/p076s.html
• 7 years agoHelpfull: Yes(2) No(0)
• As this is a good question, I am posting the solution below:
  
  By Heron's Formula, the area, A, of a triangle with sides a, b, c is given by A = square root[s(s − a)(s − b)(s − c)], where s = ½(a + b + c) is the semi-perimeter of the triangle.
  
  Then s = ½(10 + 17 + 21) = 24, and A = 84.
  
  View the diagram here: https://ibb.co/chNhVk
  
  Now drop a perpendicular of length h onto the side of length 21.
  We also have A = ½ × base × perpendicular height.
  Hence A = 21h/2 = 84, from which h = 8.
  
  Notice that the triangle above the square is similar to the whole triangle. (This follows because its base, the top of the square, is parallel to the base of the whole triangle.)
  
  Let the square have side of length d.
  
  Considering the ratio of altitude to base in each triangle, we have 8/21 = (8 − d)/d = 8/d − 1.
• 5 years agoHelpfull: Yes(1) No(0)