Applying the idea we just learned, 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
$\Rightarrow S u r f a c e A r e a = l b + 2 b h + 2 h l$

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b=breadth of the cuboid and
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Q. 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

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Applying the idea we just learned, 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 ⇒SurfaceArea=lb+2bh+2hl

Applying the idea we just learned, 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
$\Rightarrow S u r f a c e A r e a = l b + 2 b h + 2 h l$