Question

⊙n and ⊙o are congruent. (overline{pq}) is a chord of both circles.

1. if (no = 12) in. and (overline{pq}=8) in., how long is the radius to the nearest tenth of an inch?
2. if (no = 30) mm and radius (=16) mm, how long is (overline{pq}) to the nearest tenth of a millimeter?
3. if radius (=12) m and (overline{pq}=9) m, how long is (overline{no}) to the nearest tenth?

Explanation:

Step1: Use the property of congruent circles and perpendicular chords

Since \( \odot N\) and \( \odot O\) are congruent and \( NO\) bisects \( \overline{PQ}\) perpendicularly. Let the radius be \( r\), half - chord length \( a=\frac{PQ}{2}\), and let \( x\) be the distance from the center of one circle to the intersection point of \( NO\) and \( PQ\). If \( NO = d\), then \( x=\frac{d}{2}\) (because the circles are congruent). By the Pythagorean theorem \( r^{2}=x^{2}+a^{2}\)

For problem 14:

Given \( NO = 12\) in, so \( x = 6\) in and \( PQ=8\) in, so \( a = 4\) in.

$$r^{2}=6^{2}+4^{2}=36 + 16=52$$

$$r=\sqrt{52}\approx7.2$$

For problem 15:

Given \( NO = 30\) mm, so \( x = 15\) mm and \( r = 16\) mm. Using \( r^{2}=x^{2}+a^{2}\), we can solve for \( a\)

$$a=\sqrt{r^{2}-x^{2}}=\sqrt{16^{2}-15^{2}}=\sqrt{(16 + 15)(16-15)}=\sqrt{31}\approx5.6$$

\( PQ = 2a\approx11.2\)

For problem 16:

Given \( r = 12\) m and \( PQ=9\) m, so \( a = 4.5\) m. Using \( r^{2}=x^{2}+a^{2}\), we solve for \( x\)

$$x=\sqrt{r^{2}-a^{2}}=\sqrt{12^{2}-4.5^{2}}=\sqrt{144 - 20.25}=\sqrt{123.75}\approx11.1$$

Since \( NO = 2x\), \( NO\approx22.2\)

Answer:

1. \(7.2\) inches
2. \(11.2\) millimeters
3. \(22.2\) meters