The height of triangle is increased by 40%.What can be the maximum percentage increase in length of the base so that the increase in area is restricted to a maximum of 60%?
1)50%
2) 20%
3)14.28%
4) 25%

• area of triangle is given by a= 0.5*b*h
  increase area by 60% means updated area will be, a+0.6a=1.6a
  also increase height by 40% updated height will be 1.4h
  so putting these updated values in area equation will give b=1.1428b
  therefore increase in base will be 14.28%
• 10 years agoHelpfull: Yes(25) No(0)
• suppose a1=100 & h1=10 den l1=20
  now a2=160 & h2=140 so, l2=160/7
  So %increase= (l2-l1)100/l1
  hence answer is 14.28
• 10 years agoHelpfull: Yes(9) No(0)
• answer is option 3) 14.28,area is 1/2*h*b,so according to question 1/2*(40/100h+h)*b1=1/2*b*h+60/100*1/2*b*h,so b1=14.28
• 10 years agoHelpfull: Yes(3) No(0)
• Let the height be h and base be b.
  Height increases 40%,So new height=7/5*h
  Area increases by 60%. so, new area =8/5*1/2*b*h
  let new base be x, then new area=1/2*7/5*x=8/5*1/2*b*h
  which will result into x=8/7*b
  so, base will increase by 1/7th of b=14.28%
• 10 years agoHelpfull: Yes(3) No(0)
• a=0.5bh ------(1)
  (a+0.6a)=0.5(b+x)(h+0.4h)
  1.6a=0.5(b+x)1.4h------------(2)
  solving 1 and 2 we get
  x=0.1428b
  x=14.28
• 9 years agoHelpfull: Yes(3) No(0)
• Any change in 2d can be cal from =(x%+y%+xy%/100)
• 10 years agoHelpfull: Yes(2) No(1)
• 40+b+(40*b)/100 =60
  =>7*b/5 = 20
  =>b= 100/7 =14.28%
• 10 years agoHelpfull: Yes(1) No(0)
• Area of the Triangle (A) = h×b/2
  New height h'= (h+0.4h)=1.4h
  New Area of the triangle A'=(A+0.6A) = 1.6A=1.6(h×b/2)-------(1)
  But New Area A' = h'×b'/2 --------(2)
  -------------
  On solving Eq.(1) and (2), we get h'=1.1428h
  Thus Maximum % increase =(h'-h)×100 = 14.28
  -----
  Answer is 14.28%
• 10 years agoHelpfull: Yes(0) No(0)
• (160/100)*(1/2)*b*h=(1/2)*(140/1000*h*(100+x)/100
  x=200/14
  x=14.28%
  so increase in length of base=14.28%
• 10 years agoHelpfull: Yes(0) No(0)
• answer is 50%
  
• 10 years agoHelpfull: Yes(0) No(5)
• new ht
  h1=140%h=1.4h
  new base
  b1=(100+x)%b=(1+x)b
  new Area
  A1=(160%)A=1.6A
  .5*h1*b1=1.6*(.5*b*h)
  .5*1.4h*(1+x)b=1.6*.5*bh
  we get x=.1428=14.28%
• 9 years agoHelpfull: Yes(0) No(1)
• Triangle ACD and triangle BCD are similar triangles so
  AC/DC=DC/BC
  2/6=6/BC
  BC=18
  Diameter D=AC+BC=2+18=20; r=10
  Area of semicircle= pi* 10^2/2=50*pi
  
• 9 years agoHelpfull: Yes(0) No(1)
• 14.28 is the answer
• 7 years agoHelpfull: Yes(0) No(0)
• area of triangle a =0.5 *h*b-----eq1.
  new area of triangle=a+a*(60/100)=1.6a.
  new height =h+h*(40/100)=1.4h.
  so new eq will be 1.6a=0.5*1.4h*(b+x)------eq2 [let assume b increases (b+x)]
  a=0.5*h*b ------*1.6
  1.6a=0.5*1.4h*(b+x)---------*1
  ..................................................
  1.6a=0.8*h*b
  1.6a=0.7*h*b+0.7*h*x
  - - -
  .......................................
  0=0.1h*b-0.7*h*x
  0.1*h*b=0.7*h*x
  x=(0.1/0.7)*b
  x=0.1428b
  x=14.28%------ans
• 5 years agoHelpfull: Yes(0) No(0)
• et the height be h and base be b.
  
  Height increases 40%,So new height = h + 40% of h =7/5*h
  
  Area of triangle = (½ * h * b)
  
  Area increases by 60%. so, new area,
  
  = (½ * h * b) + 60% of (½ * h * b)
  
  =8/5*1/2*b*h
  
  Let new base be x, then
  
  New area,=1/2*7/5*x=8/5*1/2*b*h
  
  which will result into x=8/7*b
  
  So, base will increase by 1/7th of b=14.28%.
• 3 years agoHelpfull: Yes(0) No(1)