This problem requires finding the radius of a circle using the properties of tangents and the Pythagorean theorem.
Understanding the Geometry
Let 'O' be the center of the circle, 'Q' be the external point, and 'P' be the point where the tangent from Q touches the circle.
- The length of the tangent PQ = 21 cm.
- The distance from the center O to the point Q is OQ = 29 cm.
- The radius OP is perpendicular to the tangent PQ at the point of tangency P. This forms a right-angled triangle ΔOPQ.
- We need to find the radius, which is the length of OP.
Applying the Pythagorean Theorem
In the right-angled triangle ΔOPQ:
- The hypotenuse is OQ.
- The other two sides are OP (radius, 'r') and PQ (tangent length).
According to the Pythagorean theorem:
$ \OP^2 + PQ^2 = OQ^2 $
Calculation Steps
- Substitute the known values into the equation: $ r^2 + (21 \text{ cm})^2 = (29 \text{ cm})^2 $
- Calculate the squares: $ r^2 + 441 \text{ cm}^2 = 841 \text{ cm}^2 $
- Isolate $r^2$: $ r^2 = 841 \text{ cm}^2 - 441 \text{ cm}^2 $ $ r^2 = 400 \text{ cm}^2 $
- Find the radius 'r' by taking the square root: $ r = \sqrt{400 \text{ cm}^2} $ $ r = 20 \text{ cm} $
The radius of the circle is 20 cm.
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