if a ladder is 100m long, and distance b/w bottom of ladder and wall is 60. top side of bottom and wall is joint.
what is the maximum size of cube that place b/t them.

• The answer is 35. it is a simple pythagorus thm problem.let the edge of cube be x. now make a square between a right angled triangle of base 60 hypotnuse 100 and height is calculated to be 80. now due to square two right triangles are formed. so the hypo of 1st tri. + hypo. of 2nd tri. should be equal to 100.
  therefore (60-x)^2 +x^2 + x^2 + (80-x)^2 = 100.
  
• 12 years agoHelpfull: Yes(37) No(14)
• 35.....
  using trigonometry we have
  (80/60)=tan(y)....1
  (x/60-x)=tan(y)....2
  solve it... u will have 35
• 12 years agoHelpfull: Yes(16) No(1)
• i think 30
• 12 years agoHelpfull: Yes(11) No(8)
• let x be side of cube
  80/x=60/60-x
  ans: x=34.28
  
• 9 years agoHelpfull: Yes(9) No(2)
• how?
  i m not getting que. exactly.
• 12 years agoHelpfull: Yes(6) No(5)
• sorry friends there was a mistake in my previous answer..exact answer is 34
  since line passing through (60,0) and (0,80)is 4x+3y-240=0
  put x=y=35, we get 5>0.
  so point (35,35)lies outside the triangle, but 34 lies inside the triangle
  so size=34
• 12 years agoHelpfull: Yes(6) No(0)
• yes.. i too nt abl to get. . who is givin ans.
  please. .xplain qs. n ans both?
• 12 years agoHelpfull: Yes(5) No(9)
• APPLYING pythagorus theorem-let side of square be x unit ,then the base and height of triangle are devided in 60-x and 80-x .
  (60-x)^2+x^2+(80-x)^2+x^2=100^2=4(x^2)-280*x=0
  4x(x-70)=0
  Xmax=70 but it goes outside the triangle so Xmax=Xmax/2=35
• 12 years agoHelpfull: Yes(5) No(2)
• hypotune=100
  base=60
  so length=80
  max side of cube that can fit into it will be HCF OF 100 60 80
  that is 20
  so diagonal of that cube will be 20*sqrt3=34.65
• 9 years agoHelpfull: Yes(3) No(0)
• Exactly 30
• 12 years agoHelpfull: Yes(2) No(8)
• plz give solution... with details
• 12 years agoHelpfull: Yes(2) No(2)
• it should be square with side 66/2=33
  
  so area of cube is 33^3=35937
• 12 years agoHelpfull: Yes(2) No(6)
• by pythagorus height of triangle formed=80
  area of triangle=1/2* base*height
  =1/2*60*80
  =2400
  to fit the square area must be perfect square
  so 2400-96=2304(which is a perfect square)
  area of sq.=2304
  side=48
  maximum size=4*48=192
• 12 years agoHelpfull: Yes(2) No(11)
• when we are insert a cube into triangle it divides the triangle in to two sub triangles.
  so area old triangle= area of square + area of sub triangle1 + area of sub triangle2.
  1/2(60*80)=1/2((60-x)*x)+1/2((80-x)*x)+x^2..
  if we solve this we will get 34.26
• 9 years agoHelpfull: Yes(2) No(0)
• i m not able to get the answer so plz explain
• 12 years agoHelpfull: Yes(0) No(0)
• dra a right angled triangle and draw a square in it with side x. then at the base u will get two parts one is x and other is 60-x. similarly along height u will get x and 80-x. then c the sollution given by sahil
  
• 12 years agoHelpfull: Yes(0) No(0)
• by solving the equation we get x=70...where is the need to divide 70 by 2...???can anybody explain
• 12 years agoHelpfull: Yes(0) No(1)
• I think ans is 20.
  a/h=8/10;
  (60-x)^2 +x^2 + x^2 + (80-x)^2 = [10x/8]^2+[100-10x/8]^2
  
• 12 years agoHelpfull: Yes(0) No(1)
• i got the question n also solution of ur sahil motwani but got stuck in last concept..
  finally i got the eq. 100=2x*x+(60-x)*(60-x)+(80-x)*(80-x)
  m i going rgt or not???
• 12 years agoHelpfull: Yes(0) No(4)
• y should i divide 70 by 2
  
• 12 years agoHelpfull: Yes(0) No(1)
• it is 20 becz.....draw a square inside the triangle you got and maximise the obtained triangle with sides 80-x,x by differensiation making equals to zero.......u will end up with x=20
• 12 years agoHelpfull: Yes(0) No(2)
• here ladder forms hypotenuse(100m) with the wall(opp or adj side).
  According to Pythagoras theorem we will get
  X^2 + 60^2 = 100^2
  => X^2 = 6400
  => X = 80
  
  Now ABC forms a triangle with side AB=80, BC=100, AC=60.
  consider that a cude DEFG is inscribed in a triangle, such that vertice E lie on the side BC, and lines DG and FG lies on AB and AC respectively.
  
  here size of cube will 30X40 = 120
  
• 11 years agoHelpfull: Yes(0) No(0)
• make triangle abc in which ab is the perpendicular or we can say that is wall 60m and base bc let taken as x and hypotenuse is the ladder that is leaning against the wall so apply pyathagoras theorem
  (10)^2-(60)^2=6400
  so sqrt of 6400 is 80
• 11 years agoHelpfull: Yes(0) No(0)
• well its a piece of cake u all are making it difficult....
  
  we are given a rt triangle with base 60,,hyp=100..so third side is 80..
  
  area=1/2*base*height...(A1) here base =60 height=80
  
  now look at triangle from another veiw....
  area=1/2*base*height
  
  this time take hypotnuse as base of triangle...
  from A1 we can calculate height
  
  this is nothing but the diagnol of the biggest cube
  calculate this...from which we can get the side...
  answer will come out about 33.5
  
  
• 11 years agoHelpfull: Yes(0) No(0)
• if edge of cube is x, so we can say that,
  
  tan(53.13) = x / (60-x)
  
  3187.8 - 53.13x = x
  so,
  x = 58.89 m
  
• 10 years agoHelpfull: Yes(0) No(0)
• As RAVINDRA KUMAR YADAV stated the equation is 4x+3y-240=0. Now we have equal x and y for a square. Lets put z in place of x and y. => 4z + 3z = 240.... => 7z = 240.... => z = 240/7..... which is aprox 34.28
• 9 years agoHelpfull: Yes(0) No(0)
• let size of cube be x.
  / 1
  / 1 80-x
  / 1
  / 1 ------ 1
  1 1 x
  ------------ 1
  (60-x),x
  
  80/60=x/60-x,x=34
• 9 years agoHelpfull: Yes(0) No(0)
• Refer here for proper explanation..!!
  http://www.examveda.com/if-a-ladder-is-100m-long-and-distance-between-bottom-of-a-ladder-and-wall-is-60m-what-is-the-maximum-size-of-the-cube-that-can-be-placed-between-1513/
• 8 years agoHelpfull: Yes(0) No(1)