I'm Up and Down, and Round and Round - Midpoints and Perpendicular Bisectors of Chords

A chord of length $16 \text{ cm}$ is drawn in a circle of radius $10 \text{ cm}$. Find the distance of the chord from the center of the circle.

Given: Radius $r = 10 \text{ cm}$, Length of chord $AB = 16 \text{ cm}$. Let $OM$ be the perpendicular from center $O$ to chord $AB$. By the theorem, $M$ is the midpoint of $AB$, so: $AM = \frac{1}{2} \times 16 = 8 \text{ cm}$. In right-angled $\triangle OMA$, using Pythagoras theorem: $OA^2 = OM^2 + AM^2$ $10^2 = OM^2 + 8^2$ $100 = OM^2 + 64$ $OM^2 = 100 - 64 = 36$ $OM = \sqrt{36} = 6 \text{ cm}$.

The perpendicular from the center bisects the chord into two $8 \text{ cm}$ segments. We then apply the Pythagoras theorem to the triangle formed by the radius, the distance from the center, and half the chord.

In a circle of radius $5 \text{ cm}$, a chord is at a distance of $3 \text{ cm}$ from the center. Find the length of the chord.

Given: Radius $r = 5 \text{ cm}$, Distance $d = 3 \text{ cm}$. Let the chord be $AB$ and the perpendicular from the center be $OM = 3 \text{ cm}$. In right-angled $\triangle OMA$: $OA^2 = OM^2 + AM^2$ $5^2 = 3^2 + AM^2$ $25 = 9 + AM^2$ $AM^2 = 25 - 9 = 16$ $AM = \sqrt{16} = 4 \text{ cm}$. Since the perpendicular from the center bisects the chord: $AB = 2 \times AM = 2 \times 4 = 8 \text{ cm}$.

We first find half the length of the chord using the Pythagoras theorem in the right triangle formed by the radius and the distance from the center, then double it to find the full chord length.

Two chords $AB$ and $CD$ of a circle are parallel and have lengths $6 \text{ cm}$ and $8 \text{ cm}$ respectively. If the radius is $5 \text{ cm}$ and the chords are on the same side of the center, find the distance between them.

Let $O$ be the center and $r = 5 \text{ cm}$. Let $OM \perp AB$ and $ON \perp CD$. Since $AB = 6 \text{ cm}$, $AM = 3 \text{ cm}$. In $\triangle OMA$: $OM = \sqrt{OA^2 - AM^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = 4 \text{ cm}$. Since $CD = 8 \text{ cm}$, $CN = 4 \text{ cm}$. In $\triangle ONC$: $ON = \sqrt{OC^2 - CN^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = 3 \text{ cm}$. The distance between the chords is $MN = OM - ON$ (as they are on the same side): $MN = 4 - 3 = 1 \text{ cm}$.

Calculate the distance of each chord from the center individually using the Pythagoras theorem. Since they are on the same side of the center, subtract the smaller distance from the larger distance.

Let the circle have center $O$ and the chord be $AB$. Let $OM$ be the perpendicular from $O$ to $AB$. Given $OA = 13 \text{ cm}$ (radius) and $OM = 5 \text{ cm}$ (distance). In right-angled $\triangle OMA$: $OA^2 = OM^2 + AM^2$ $13^2 = 5^2 + AM^2$ $169 = 25 + AM^2$ $AM^2 = 169 - 25 = 144$ $AM = \sqrt{144} = 12 \text{ cm}$ Since the perpendicular from the center bisects the chord: $AB = 2 \times AM = 2 \times 12 = 24 \text{ cm}$ Thus, the length of the chord is $24 \text{ cm}$.

We use the Pythagorean theorem in the right triangle formed by the radius, the distance from the center, and half the chord length. Multiplying the result by 2 gives the full chord length.

Let $r$ be the radius and $O$ be the center. Let $OM \perp AB$ and $ON \perp CD$. Let $OM = x$, then $ON = 17 - x$. In $\triangle OMA$, $AM = \frac{10}{2} = 5 \text{ cm}$: $r^2 = x^2 + 5^2 \quad \text{---(1)}$ In $\triangle ONC$, $CN = \frac{24}{2} = 12 \text{ cm}$: $r^2 = (17 - x)^2 + 12^2 \quad \text{---(2)}$ Equating (1) and (2): $x^2 + 25 = 289 - 34x + x^2 + 144$ $25 = 433 - 34x$ $34x = 408$ $x = 12$ Substitute $x = 12$ into (1): $r^2 = 12^2 + 5^2 = 144 + 25 = 169$ $r = \sqrt{169} = 13 \text{ cm}$

By expressing the radius squared in terms of the distance from the center for both chords, we create an equation to solve for the unknown distance $x$, then find the radius.