A chord 21 cm long is drawn in a circle of diameter 25 cm. The perpendicular distance of the chord from the centre is:
√46
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.
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:
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.
Let's look at the given options:
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 |
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.
| 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 \(r^2 = (\frac{\text{chord}}{2})^2 + d^2\). |
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.
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