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3 inside a cubic box with edge 35 cm a ball has been introduced that is tangent to the faces of the cube as shown in the f

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3. Inside a cubic box with edge 35 cm, a ball has been introduced that is tangent to the faces of the cube as shown in the following figure. Calculate the volume included by the free space in the cube-shaped box.

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Answer to a math question 3. Inside a cubic box with edge 35 cm, a ball has been introduced that is tangent to the faces of the cube as shown in the following figure. Calculate the volume included by the free space in the cube-shaped box.

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111 Answers

Let's start by finding the radius of the ball. Since the ball is tangent to the faces of the cube, the diameter of the ball is equal to the edge length of the cube. Therefore, the radius of the ball is half of this length:

Let

r

be the radius of the ball. We know that the diameter of the ball is equal to the edge length of the cube, which is 35 cm. Therefore,

2r = 35

, and

r = \frac{35}{2} = 17.5

cm.

Next, we need to calculate the volume of the ball:

The volume of a ball is given by the formula

V = \frac{4}{3} \pi r^3

.

Substitute the radius

r = 17.5

cm into the formula:

V=\frac{4}{3}\pi(17.5)^3\approx\frac{4}{3}\times3.1416\times17.5^3\approx4.1889\times17.5^3\approx22449.85

cubic cm.

The volume of the free space in the cube-shaped box is equal to the difference between the volume of the cube and the volume of the ball:

The volume of the cube is given by

V_{\text{cube}} = s^3

, where

s

is the edge length of the cube.

Substitute the edge length

s = 35

cm into the formula:

V_{\text{cube}} = 35^3 = 42875

cubic cm.

Therefore, the volume included by the free space in the cube-shaped box is:

V_{\text{free space}}=V_{\text{cube}}-V_{\text{ball}}=42875-22449.35=20425.65

cubic cm.

\boxed{20425.65\text{ cubic cm}}

is the volume included by the free space in the cube-shaped box.

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