Question

question 22
your answer is correct.
a spherical balloon contains water that is 3 inches deep at the deepest point. if the balloon has a diameter of 8 inches, what is the area of the top surface of the water?
a) $9\pi$ in$^2$
b) $16\pi$ in$^2$
c) $2\pi\sqrt{15}$ in$^2$
d) $\pi$ in$^2$
e) $15\pi$ in$^2$
f) none of the above.

Explanation:

Step1: Find the radius of the sphere

The diameter of the sphere is 8 inches, so the radius \( R=\frac{8}{2} = 4\) inches.

Step2: Analyze the depth of water and the sphere

The depth of water at the deepest point is 3 inches. Let's consider a cross - sectional circle of the sphere (a great circle). Let the center of the sphere be \( O\), and the depth of water be \( h = 3\) inches. The distance from the center of the sphere to the surface of the water (in the cross - sectional view) is \( d=R - h=4 - 3 = 1\) inch.

Step3: Find the radius of the circular surface of the water

Let the radius of the circular surface of the water be \( r\). Using the Pythagorean theorem in the right - triangle formed by the radius of the sphere \( R\), the distance from the center to the water surface \( d\), and the radius of the water surface \( r\). We have \(r^{2}=R^{2}-d^{2}\). Substituting \( R = 4\) and \( d = 1\), we get \(r^{2}=4^{2}-1^{2}=16 - 1=15\).

Step4: Calculate the area of the circular surface of the water

The area of a circle is given by the formula \( A=\pi r^{2}\). Since \( r^{2}=15\), the area of the top surface of the water (which is a circle) is \( A = 15\pi\) square inches.

Answer:

c) \(15\pi\space\text{in}^2\)