Calculating Perpendicular Distance from Circle Center to Chord

This problem involves finding the perpendicular distance from the center of a circle to a given chord, using the area of the circle and the length of the chord.

Understanding the Problem Setup

We are given:

• Area of the circle = 616 cm²
• Length of the chord XY = 10 cm

We need to find the distance from the center of the circle to the chord XY, along a line perpendicular to the chord.

Step 1: Find the Radius of the Circle

The area of a circle is given by the formula \(A = \pi r^2\), where \(A\) is the area and \(r\) is the radius. We can use the given area to find the radius.

Given Area = 616 cm²

Using \(\pi = \frac{22}{7}\):

\(616 = \frac{22}{7} \times r^2\)

To find \(r^2\), we can rearrange the equation:

\(r^2 = 616 \times \frac{7}{22}\)

Divide 616 by 22:

\(616 \div 22 = 28\)

So,

\(r^2 = 28 \times 7\)

\(r^2 = 196\)

Now, find the radius \(r\) by taking the square root of \(r^2\):

\(r = \sqrt{196}\)

\(r = 14\) cm

The radius of the circle is 14 cm.

Step 2: Relate Perpendicular Distance, Radius, and Chord Length

A fundamental property of circles states that the perpendicular drawn from the center of a circle to a chord bisects the chord. This means if we draw a line segment from the center (let's call it O) perpendicular to the chord XY at point M, then M is the midpoint of XY.

The length of the chord XY is 10 cm. Since M is the midpoint, XM = MY = \(\frac{1}{2} \times \text{XY}\).

\(\text{XM} = \frac{1}{2} \times 10\) cm

\(\text{XM} = 5\) cm

Now, consider the triangle OMX. O is the center, M is the midpoint of the chord where the perpendicular meets, and X is an endpoint of the chord. OX is the radius of the circle (r = 14 cm).

The line segment OM is the perpendicular distance from the center to the chord, which we need to find. Triangle OMX is a right-angled triangle with the right angle at M.

Step 3: Use the Pythagorean Theorem

In the right-angled triangle OMX, the Pythagorean theorem states that the square of the hypotenuse (the side opposite the right angle, which is OX, the radius) is equal to the sum of the squares of the other two sides (OM, the perpendicular distance, and XM, half the chord length).

Let the perpendicular distance OM be \(d\).

According to the Pythagorean theorem:

\(\text{OX}^2 = \text{OM}^2 + \text{XM}^2\)

Substitute the values we know:

\(14^2 = d^2 + 5^2\)

\(196 = d^2 + 25\)

Now, solve for \(d^2\):

\(d^2 = 196 - 25\)

\(d^2 = 171\)

Finally, find the distance \(d\) by taking the square root:

\(d = \sqrt{171}\)

The perpendicular distance from the center of the circle to the chord XY is \(\sqrt{171}\) cm.

Summary of Calculation

We found the radius from the area, determined the half-chord length, and used the Pythagorean theorem in the right triangle formed by the radius, half-chord, and perpendicular distance.

• Radius \(r = 14\) cm
• Half-chord length = 5 cm
• Perpendicular distance \(d = \sqrt{r^2 - (\text{half-chord})^2} = \sqrt{14^2 - 5^2} = \sqrt{196 - 25} = \sqrt{171}\) cm

Revision Table: Circle Geometry Concepts

Additional Information: Properties of Chords

Understanding chords and their properties is crucial in circle geometry. Here are a few key points:

• A chord is a line segment connecting two points on the circumference of a circle.
• The longest chord in a circle is the diameter.
• Equal chords are equidistant from the center. Conversely, chords equidistant from the center are equal in length.
• The perpendicular bisector of a chord passes through the center of the circle. This is related to the property used in the problem (perpendicular from center bisects chord).
• Angles subtended by a chord at the center and at the circumference have a specific relationship.

These properties are often used in solving various problems related to circles, chords, and their distances from the center.