Question

find the exact area and perimeter of each of the shaded regions below. assume things that look like squares are squares, etc. 1. a = 100-(πr²) p = 20 + 20π = 100 - 25π meter = 20+20π 2. area = ______ perimeter = ______ 4.

Explanation:

Step1: Analyze problem 2

The side - length of the square inscribed in the circle with radius \(r = 7\) m. The diameter of the circle \(d=14\) m is the length of the square's diagonal. Using the Pythagorean theorem for the square (\(a^{2}+a^{2}=d^{2}\), where \(a\) is the side - length of the square), we have \(2a^{2}=14^{2}=196\), so \(a^{2}=98\) and \(a = 7\sqrt{2}\) m. The area of the circle is \(A_{circle}=\pi r^{2}=\pi\times7^{2}=49\pi\) \(m^{2}\), and the area of the square is \(A_{square}=98\) \(m^{2}\). The area of the shaded region \(A = 49\pi-98\) \(m^{2}\). The perimeter of the shaded region is the circumference of the circle \(P = 2\pi r=14\pi\) m.

Step2: Analyze problem 3

The outer rectangle has dimensions \(l = 10\) ft and \(w = 8\) ft, so its area \(A_{outer}=10\times8 = 80\) \(ft^{2}\). The inner rectangle has dimensions \(l = 8\) ft and \(w = 5\) ft, so its area \(A_{inner}=8\times5 = 40\) \(ft^{2}\). The area of the shaded region \(A=80 - 40=40\) \(ft^{2}\). The perimeter of the shaded region is \(P=(10 + 8)\times2+(8 + 5)\times2=36 + 26 = 62\) ft.

Step3: Analyze problem 4

The area of the outer - circle with radius \(R=5 + 3=8\) cm is \(A_{outer}=\pi R^{2}=\pi\times8^{2}=64\pi\) \(cm^{2}\). The area of the inner - circle with radius \(r = 5\) cm is \(A_{inner}=\pi r^{2}=\pi\times5^{2}=25\pi\) \(cm^{2}\). The area of the shaded region \(A = 64\pi-25\pi=39\pi\) \(cm^{2}\). The perimeter of the shaded region is \(P = 2\pi R+2\pi r=2\pi\times8+2\pi\times5=16\pi + 10\pi=26\pi\) cm.

Answer:

Problem 2:
Area \(=49\pi - 98\) \(m^{2}\)
Perimeter \(=14\pi\) m
Problem 3:
Area \(=40\) \(ft^{2}\)
Perimeter \(=62\) ft
Problem 4:
Area \(=39\pi\) \(cm^{2}\)
Perimeter \(=26\pi\) cm