Question

Plz solve the Q11 with steps

Answer 1

a cubical box has all its dimensions equal
i.e. length = breadth = height

Let the length of the cubical box = a

Volume of cubocal box = a^3

Given that volume = 0.001331 m^3

$Volume = 0.001331 \\ \\ {a}^{3} = 0.001331 \\ \\ a = \sqrt[3]{0.001331} \\ \\ a = \sqrt[3]{0.11 \times 0.11 \times 0.11} \\ \\ a = \sqrt[3]{ {0.11}^{3} } \\ \\ a = 0.11 \: m$

So length of the cubical box is 0.11 m

Now, total surface area of cube = 6a^2

TSA = 6 × 0.11^2 = 6 × 0.0121 = 0.0726 m^2

Hence, surface area of the cubical box is 0.0726 m^2.

Answer 2

The cubical box has all sides equal.

We know that,

Volume of cubical box=0.001331m^3
a^3

(side)^3=0.001331

(side)^3=(0.11)^3

side = 0.11

Length of all the sides of a cubic box is 0.11 m

We know that,

Total surface area of cube
= 6 (side)^2

= 6 (0.11)^2

= 6 (0.0121)

= 0.0726 m^2

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