Question

What is the outer surface area of a box (open from top) in the shape of cuboid having length 6 cm, breadth 5cm and height 4cm?

Answer 1

Answer:

Area = 148$cm^{2}$

Step-by-step explanation:

According to the information provided in the question it is given as

Outer surface area of a box  in cuboid shape

Length = l = 6 cm

Breadth = b = 5 cm

Height =h= 4 cm

We need to find the outer surface area of cuboid

$A =2(lb+ bh+hl)$

Putting the value we get Area

$A = 2(6\times 5+5\times 4+ 4\times 6\\A = 2(30+20+24)\\A=2(74)\\A= 148 cm^{2}$

Hence the outer area of box is =148$cm^{2}$

Answer 2

Answer:

Given :-

• A cuboid having length 6 cm, breadth is 5 cm and height 4 cm.

To Find :-

• What is the outer surface of a box i.e, open from top in the shape of cuboid.

Formula Used :-

$\clubsuit$ Total Surface Area or T.S.A Of Cuboid Formula :

$\footnotesize \bigstar \: \: \sf\boxed{\bold{\pink{T.S.A_{(Cuboid)} =\: 2(LB + BH + HL)}}}\: \: \: \bigstar$

where,

• L = Length
• B = Breadth
• H = Height

Solution :-

Given :

• Length = 6 cm
• Breadth = 5 cm
• Height = 4 cm

According to the question by using the formula we get,

$\footnotesize \dashrightarrow \sf\bold{\purple{T.S.A_{(Cuboid)} =\: 2(LB + BH + HL)}}$

$\small \dashrightarrow \sf T.S.A_{(Cuboid)} =\: 2\bigg\{(6 \times 5) + (5 \times 4) + (4 \times 6)\bigg\}$

$\small \dashrightarrow \sf T.S.A_{(Cuboid)} =\: 2\bigg\{(30) + (20) + (24)\bigg\}$

$\small \dashrightarrow \sf T.S.A_{(Cuboid)} =\: 2\bigg\{30 + 20 + 24\bigg\}$

$\small \dashrightarrow \sf T.S.A_{(Cuboid)} =\: 2\bigg\{50 + 24\bigg\}$

$\small \dashrightarrow \sf T.S.A_{(Cuboid)} =\: 2\bigg\{74\bigg\}$

$\small \dashrightarrow \sf T.S.A_{(Cuboid)} =\: 2 \times 74$

$\small \dashrightarrow \sf\bold{\red{T.S.A_{(Cuboid)} =\: 148\: cm^2}}$

$\therefore$ The outer surface area of a box i.e, open from top in the shape of cuboid or the total surface area of cuboid is 148 cm² .

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