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Question
The following figure shows a closed victory-stand whose dimensions are given in cm.
Find the volume and the surface area of the victory stand.
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Solution
Consider the box 1
Thus, the dimensions of box 1 are 60 cm, 40 cm, and 30 cm.
Therefore, the volume of box 1 = 60 x 40 x 30 = 72000 cm3
Surface area of box 1 = 2 ( lb + bh + lh )
Since the box is open at the bottom and from the given figure, we have,
Surface area of box 1 = 40 x 40 + 40 x 30 + 40 x 30 + 2 ( 60 x 30 )
= 1600 + 1200 + 1200 + 3600
= 7600 cm2
Consider the box 2 
Thus, the dimensions of box 2 are 40 cm, 30 cm, and 30 cm.
Therefore, the volume of box 2 = 40 x 30 x 30 = 36000 cm3
Surface area of box 2 = 2 ( lb + bh + lh )
Since the box is open at the bottom and from the given figure, we have,
Surface area of box 2 = 40 x 30 + 40 x 30 + 2 ( 30 x 30 )
= 1200 + 1200 + 1800
= 4200 cm2
Consider the box 3
Thus, the dimensions of the box 3 are 40 cm, 30 cm, and 20 cm.
Therefore, the volume of box 3 = 40 x 30 x 20 = 24000 cm3
Surface area of box 3 = 2 ( lb + bh + lh )
Since the box is open at the bottom and from the given figure, we have,
Surface area of box 3 = 40 x 30 + 40 x 20 + 2 ( 30 x 20 )
= 1200 + 800 + 1200
= 3200 cm2
Total volume of the box
= volume of box 1 + volume of box 2 + volume of box 3
= 72000 + 36000 + 24000
= 132000 cm3
Similarly, total surface area of the box
= surface area of box 1 + surface area of box 2 + surface area of box 3
= 7600 + 4200 + 3200
= 15000 cm2
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