A chord 21 cm long is drawn in a circle of diameter 25 cm. The perpendicular distance of the chord from the centre is:

Question

Prepp Team · Accepted Answer

Understanding the Circle and Chord Problem

This problem asks us to find the perpendicular distance from the center of a circle to a chord. We are given the length of the chord and the diameter of the circle. This involves applying basic geometric principles related to circles, specifically the properties of a chord and the radius.

Given Information:

Length of the chord = 21 cm
Diameter of the circle = 25 cm

Step-by-Step Solution for Perpendicular Distance

First, let's find the radius of the circle. The radius is half the diameter.

Radius \(r = \frac{\text{Diameter}}{2} = \frac{25 \text{ cm}}{2} = 12.5 \text{ cm}\)

Next, recall a key property of circles: a perpendicular drawn from the center of a circle to a chord bisects the chord. This means it divides the chord into two equal parts.

Let the chord be AB, and the center of the circle be O. Let D be the point where the perpendicular from O meets the chord AB. Then OD is the perpendicular distance we want to find, and D is the midpoint of AB.

Length of each half of the chord = \(\frac{\text{Length of chord}}{2} = \frac{21 \text{ cm}}{2} = 10.5 \text{ cm}\)

So, AD = DB = 10.5 cm.

Now, consider the triangle ODA. This is a right-angled triangle because OD is perpendicular to AB. The sides of this triangle are:

OA = radius (r) = 12.5 cm (This is the hypotenuse)
AD = half of the chord length = 10.5 cm (This is one leg)
OD = perpendicular distance from the center = ? (This is the other leg)

We can use the Pythagorean theorem in the right-angled triangle ODA. The theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

\((\text{OA})^2 = (\text{AD})^2 + (\text{OD})^2\)

\(r^2 = (\frac{\text{chord length}}{2})^2 + (\text{perpendicular distance})^2\)

Let the perpendicular distance be \(d\).

\((12.5)^2 = (10.5)^2 + d^2\)

Calculate the squares:

\(12.5^2 = 12.5 \times 12.5 = 156.25\)

\(10.5^2 = 10.5 \times 10.5 = 110.25\)

Substitute these values back into the equation:

\(156.25 = 110.25 + d^2\)

Now, isolate \(d^2\):

\(d^2 = 156.25 - 110.25\)

\(d^2 = 46\)

To find \(d\), take the square root of both sides:

\(d = \sqrt{46}\)

The perpendicular distance of the chord from the center is \(\sqrt{46}\) cm.

Checking the Options

Let's look at the given options:

√41
√23
√56
√46

Our calculated distance is \(\sqrt{46}\) cm, which matches option 4.

Geometric Element	Value
Chord Length	21 cm
Diameter	25 cm
Radius	12.5 cm
Half Chord Length	10.5 cm
Perpendicular Distance	\(\sqrt{46}\) cm

Conclusion

By using the property that the perpendicular from the center bisects the chord and applying the Pythagorean theorem to the right-angled triangle formed by the radius, half the chord, and the perpendicular distance, we found the distance to be \(\sqrt{46}\) cm.

Revision Table: Circle and Chord Properties

Concept	Description / Formula
Radius (r)	\(\frac{\text{Diameter}}{2}\)
Perpendicular from Center	Bisects the chord.
Pythagorean Theorem	In a right triangle with legs a, b and hypotenuse c: \(a^2 + b^2 = c^2\). Used here as \(r^2 = (\frac{\text{chord}}{2})^2 + d^2\).

Additional Information on Circle Geometry

Understanding chords and their relationship with the circle's center is fundamental in geometry. A chord is simply a line segment connecting two points on the circle's circumference. The longest chord in a circle is the diameter.

The property used in this problem - that a perpendicular from the center to a chord bisects the chord - is derived from the congruence of the two right-angled triangles formed (ODA and ODB in our example, where OA=OB=radius, OD is common, and angles at D are 90°). Conversely, the line joining the center to the midpoint of a chord is perpendicular to the chord.

These properties are useful for solving various problems involving distances, lengths, and angles within circles.