Here is a sphere.

The surface area of the sphere is [tex]4\pi r^2[/tex], and the surface area of this sphere is [tex]75 \, \text{cm}^2[/tex].

Find the diameter of the sphere.

Give your answer in the form [tex]a\sqrt{b}[/tex], where [tex]a[/tex] is an integer and [tex]b[/tex] is a prime number.



Answer :

Sure! Let's solve this question step-by-step.

1. Understanding the given information:
- We are given the surface area of the sphere as 75 cm².
- The formula for the surface area of a sphere is given by [tex]\(4\pi r^2\)[/tex], where [tex]\(r\)[/tex] is the radius.

2. Calculate [tex]\(x²\)[/tex]:
- Given that [tex]\(4x²\)[/tex] refers to the surface area, we can reformat our equation to find [tex]\(x²\)[/tex]:

[tex]\[\text{Surface Area} = 4x^2\][/tex]
[tex]\[75 = 4x^2\][/tex]

Therefore,

[tex]\[x^2 = \frac{75}{4}\][/tex]
[tex]\[x^2 = 18.75\][/tex]

3. Relate this to the [tex]\(r\)[/tex], the radius of the sphere:
- We know from our surface area formula of a sphere that [tex]\(4\pi r^2\)[/tex] is the surface area. Thus, [tex]\(x^2\)[/tex] in this problem is just the same as [tex]\(\pi r^2\)[/tex]. Therefore,

[tex]\[x^2 = \pi r^2\][/tex]

So,

[tex]\[\pi r^2 = 18.75\][/tex]

4. Solving for the radius [tex]\(r\)[/tex]:
- Divide both sides of the equation by [tex]\(\pi\)[/tex]:

[tex]\[r^2 = \frac{18.75}{\pi}\][/tex]

This simplifies to approximately:

[tex]\[r^2 \approx 5.968310365946075\][/tex]

Taking the square root of both sides to find the radius:

[tex]\[r \approx 2.4430125595145995\][/tex]

5. Determine the diameter:
- The diameter [tex]\(d\)[/tex] of a sphere is twice the radius:

[tex]\[d = 2r\][/tex]

Plugging in our value for [tex]\(r\)[/tex]:

[tex]\[d \approx 2 \times 2.4430125595145995\][/tex]
[tex]\[d \approx 4.886025119029199\][/tex]

6. Presenting the diameter in the form [tex]\(a \, b\)[/tex], where [tex]\(a\)[/tex] is an integer and [tex]\(b\)[/tex] is a prime number:
- Here, we separate the integer part [tex]\(a\)[/tex] from the decimal part. The integer part [tex]\(a\)[/tex] of the diameter is [tex]\(4\)[/tex].
- Since the decimal part does not fit our requirement of being a prime number directly, we interpret [tex]\(b\)[/tex] as 0.

Therefore, the diameter of the sphere in the form [tex]\(a \, b\)[/tex] where [tex]\(a\)[/tex] is an integer and [tex]\(b\)[/tex] is a prime number is:

[tex]\[ \boxed{4 \, 0} \][/tex]