a cuboid of length 20m , breadth 15m and height 12m is lying on a table. the cuboid is cut into two halves by a plane which is perpendicular to the top surface and passes through a pair diagonally opposite points of that surface . then the second cut is made by a plane which is parallel to the surface of table again dividing it into two parts. of four equal pieces into which the cuboid is divided, one piece is now removed from its place. find the total surface area of remaining portion.

• 1380
  3/4 of 1440=1080
  25*6=150
  10*15=150
  ans 1080+150+150=1380
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• 2580
  
  20*15+25*6+6*15+20*16
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• ans is 1320
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• pls explain it properly
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• 1440
  2(300+240+180)
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• types of areas that we will get:-
  A1=15*12=180
  A2=20*12=240
  A3=6*20=120
  A4=15*6=90
  A5=0.5*15*20=150
  A6=25*6=150
  SO the total surface area is=180+240+120+90+150+150=930 sqmts
  
• 10 years agoHelpfull: Yes(0) No(3)
• area=2(300+180+240)=1440
  area left after parallel cut=2(300+90+120)=2*510
  area after cut along diagonal=1/2(2*510)=510
  total are left=1440-510=930
  
  
• 10 years agoHelpfull: Yes(0) No(1)
• AFTER CUTTING
  L=20
  B=7.5
  H=6
  FOUR IDENTICAL CUBOID ARE FORMED.
  SURFACE AREA OF ONE CUBOID=2(LB+BH+HL)
  =2(20*7.5+7.5*6+6*20)
  =630
  NOW CUBOID CUT 4 EQUAL PIECES 4*630=2520
  REMAINING SURFACE AREA=2520-630=1892 SQ. M
  
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• TOTAL SA is 2*(lb+bh+lh)=1440
  surface removed l now remain 20
  b now half 7.5 and h also half 6
  so area of portion removed is (l*b+2*b*h+l*h)=360
  1440-360=1080 ANS....
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