A triangle of side 10,17,21.a square in the triangle whose two corner touch tow side of triangle & one side of square on biggest side of triangle side. find out side of square??
(a)5.4 (b)5.8 (c)6.7 (d)7.2

• first find the area using the heros formula
  A=sqrt[(s)(s-a)(s-b)(s-c)]
  A=sqrt[24*14*7*3]
  A=sqrt[7056]
  A=84
  
  now if you name the triangle PQR with PQ=10 QR=21 PR=17
  and then draw a perpendicular from P to the side QR and name it PM
  
  now you can find the height of the triangle PQR by
  1/2 * base * height= Area(by heroes formula)
  1/2 * 21 * h =84
  h= 8
  
  if u imagine the two edges of square touching the sides of triangle let that be A on side PQ and B on side PR
  and the perpendicular PM cutting AB @ C
  now let the side of square be = x cm total height of PQR = 8
  then height of triangle PAB is PC=8-x
  
  
  the the triangle PAB is similar to triangle PQR
  
  then AB/QR=PC/PM
  x/21=8-x/8
  8x=168-21x
  29x=168
  x=168/29
  x=5.79 ~ 5.8 cm
  
  Option B ok
  any problem contact me @ iqbalj6303@yahoo.co.in
  
• 10 years agoHelpfull: Yes(36) No(1)
• if any problem in the solution i explained plz refer to this link
  
  http://www.qbyte.org/puzzles/p076s.html
• 10 years agoHelpfull: Yes(12) No(0)
• a=10,b=21,c=17
  s=(a+b+c)/2
  s= 48
  area (heroins formula) = 3 * 8 * 7 *2*3
  area = b*h/2
  h= area * 2 /b
  = 8
  
  side of square = (b * h)/ (b+h)
  = 21 * 8 /(21 +8)
  = 5.79 ~ 5.8
• 7 years agoHelpfull: Yes(2) No(0)