Question

name date additional practice problem set unit 3 lesson 9 additional practice problems 1. the diameter of a record is 12 inches. the diameter of the yellow label is 4 inches. what is the area of the outer, playable surface of the record? express your answer in terms of π. explain your reasoning.

Explanation:

Step1: Find the radius of the record and the label

The radius of a circle is half of its diameter.
For the record: \(r_{record}=\frac{12}{2} = 6\) inches.
For the label: \(r_{label}=\frac{4}{2}=2\) inches.

Step2: Use the formula for the area of a circle \(A = \pi r^{2}\) and find the area of the playable surface

The area of the playable surface \(A\) is the area of the record minus the area of the label.
The area of the record \(A_{record}=\pi r_{record}^{2}=\pi\times6^{2}=36\pi\)
The area of the label \(A_{label}=\pi r_{label}^{2}=\pi\times2^{2} = 4\pi\)
Then \(A=A_{record}-A_{label}\)
Substitute the values: \(A = 36\pi-4\pi=(36 - 4)\pi\)

Answer:

\(32\pi\) square inches.