Find the surface area of the cuboidal box which is open at the top- if-l-length of the cuboid- b - breadth of the cuboid andh -height of the cuboidA- Surface Area -2 lb - bh - lh B- Surface Area -2 lb -2 bh - hlC- Surface Area - lb -2 bh -2 hlD- Surface Area -2 lb - bh -2 hl

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Surface area of a solid

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Question

Find the surface area of the cuboidal box which is open at the top, if:
l=length of the cuboid,
b=breadth of the cuboid and
h=height of the cuboid

Surface Area = 2(lb+bh+lh)

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Surface Area = 2lb+2bh+hl

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Surface Area = lb+2bh+2hl

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Surface Area = 2lb+bh+2hl

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Solution

The correct option is C Surface Area = lb+2bh+2hl

When the cuboidal box is cut open, the total surface area is the sum of the areas of 5 rectangles as shown in the figure.
Surface Area = hl+lb+bh+hl+bh
$∴$ Surface Area = lb+2bh+2hl

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What is the surface area of the cuboidal box which is open at the top, if:
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Find the surface area of the cuboidal box which is open at the top, if: l=length of the cuboid, b=breadth of the cuboid and h=height of the cuboid

The correct option is C Surface Area = lb+2bh+2hl When the cuboidal box is cut open, the total surface area is the sum of the areas of 5 rectangles as shown in the figure. Surface Area = hl+lb+bh+hl+bh ∴ Surface Area = lb+2bh+2hl

Find the surface area of the cuboidal box which is open at the top, if:
l=length of the cuboid,
b=breadth of the cuboid and
h=height of the cuboid

The correct option is C Surface Area = lb+2bh+2hl

When the cuboidal box is cut open, the total surface area is the sum of the areas of 5 rectangles as shown in the figure.
Surface Area = hl+lb+bh+hl+bh
$∴$ Surface Area = lb+2bh+2hl