We are given a circle with a radius of 5 cm and a chord of length 8 cm. We need to find the distance from the circle's center to this chord.
The line segment drawn from the center of a circle perpendicular to a chord bisects the chord. This creates a right-angled triangle.
Let:
The half-length of the chord is $\frac{c}{2} = \frac{8 \text{ cm}}{2} = 4 \text{ cm}$.
According to the Pythagorean theorem, in the right-angled triangle formed:
$r^2 = (\frac{c}{2})^2 + d^2$
$5^2 = 4^2 + d^2$
$25 = 16 + d^2$
$d^2 = 25 - 16$
$d^2 = 9$
$d = \sqrt{9}$
$d = 3 \text{ cm}$
The distance from the center of the circle to the chord is 3 cm.
In a circle, a ten cm long chord is at a distance of 12 cm from the centre of the circle. The length of the diameter of the circle (in cm) is:
Chord AB of a circle of radius 10 cm is at a distance 8 cm from the centre O. If tangents drawn at A and B intersect at P., then the length of the tangent AP (in cm) is:
A chord 21 cm long is drawn in a circle of diameter 25 cm. The perpendicular distance of the chord from the centre is:
In a circle with radius 5 cm, a chord is at a distance of 3 cm from the centre. The length of the chord is:
O is the centre of this circle. Tangent drawn from a point P, touches the circle at Q. If PQ = 24 cm and OQ = 10 cm, then what is the value of OP?