The area of the floor of a cubical room is 192 m 2. The length of the longest rod that can be kept in that room is :
24 m
This problem asks us to find the length of the longest rod that can fit inside a cubical room, given the area of its floor. The key steps involve using the floor area to find the room's dimensions and then calculating the space diagonal.
A cubical room is a three-dimensional shape where all sides (length, width, and height) are equal. Let's denote the length of one side (edge) of the cube as '$a$'.
The area of a square is calculated by squaring its side length. So, for the floor of the cubical room:
Area = $a \times a = a^2$
We are given that the floor area is 192 m². Therefore:
$$a^2 = 192 \text{ m}^2$$
To find the side length '$a$', we need to calculate the square root of 192:
$$a = \sqrt{192} \text{ m}$$
To simplify $\sqrt{192}$, we look for the largest perfect square that divides 192. We find that $192 = 64 \times 3$, and 64 is a perfect square ($8^2 = 64$).
$$a = \sqrt{64 \times 3} = \sqrt{64} \times \sqrt{3}$$
$$a = 8\sqrt{3} \text{ m}$$
So, the side length of the cubical room is $8\sqrt{3}$ meters.
The length of the longest rod that can fit inside a cube is equal to the length of its space diagonal. The formula for the space diagonal ($d$) of a cube with side length '$a$' is:
$$d = a\sqrt{3}$$
Now, we substitute the value of '$a$' we found ($a = 8\sqrt{3}$ m) into this formula:
$$d = (8\sqrt{3}) \times \sqrt{3} \text{ m}$$
To simplify, we multiply the terms:
$$d = 8 \times (\sqrt{3} \times \sqrt{3}) \text{ m}$$
We know that $\sqrt{3} \times \sqrt{3} = 3$. Substituting this value:
$$d = 8 \times 3 \text{ m}$$
$$d = 24 \text{ m}$$
The length of the longest rod that can be kept in the cubical room is 24 meters.
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