Question

1. construction an arch over an apartment entrance is 3 feet high and 9 feet wide. find the radius of the circle containing the arc of the arch.

Explanation:

Step1: Set up the equation using the Pythagorean theorem

Let the radius of the circle be \(r\). The distance from the center of the circle to the mid - point of the chord (half of the width of the arch) is \(r - 3\), and half of the width of the arch \(a=\frac{9}{2}=4.5\). According to the Pythagorean theorem in the right - triangle formed by the radius, half of the chord, and the line from the center of the circle to the chord: \(a^{2}+(r - 3)^{2}=r^{2}\).
Substitute \(a = 4.5\) into the equation: \(4.5^{2}+(r - 3)^{2}=r^{2}\).
Expand \((r - 3)^{2}\) using the formula \((A - B)^{2}=A^{2}-2AB + B^{2}\), where \(A = r\) and \(B = 3\). So \(4.5^{2}+r^{2}-6r + 9=r^{2}\).

Step2: Simplify the equation

First, calculate \(4.5^{2}=20.25\). The equation becomes \(20.25+r^{2}-6r + 9=r^{2}\).
Subtract \(r^{2}\) from both sides of the equation. We get \(20.25+9-6r=0\).
Combine like terms: \(29.25-6r = 0\).
Then, solve for \(r\): \(6r=29.25\).

Step3: Calculate the value of \(r\)

Divide both sides of the equation \(6r=29.25\) by 6. \(r=\frac{29.25}{6}=4.875\)

Answer:

The radius of the circle is \(4.875\) feet.