6 If a tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O
at the point Q so that OQ = 13 cm, then the length of the tangent is:
Ans 1. 11 cm
2. 12 cm
3. 13 cm
4. 14 cm

To solve this problem, we can use the Pythagorean theorem in the right triangle formed by the radius of the circle, the tangent, and the line from the center of the circle to the point where the tangent meets the extended line through the center.

Given:
- The radius of the circle \( OP = 5 \) cm.
- The distance \( OQ = 13 \) cm.
- We need to find the length of the tangent \( PQ \).

In the right triangle \( OPQ \):
- \( OP \) is the radius and perpendicular to the tangent at point \( P \).
- \( PQ \) is the tangent we need to find.
- \( OQ \) is the hypotenuse.

Using the Pythagorean theorem:
\[ OQ^2 = OP^2 + PQ^2 \]

Substitute the given values:
\[ 13^2 = 5^2 + PQ^2 \]
\[ 169 = 25 + PQ^2 \]

Solve for \( PQ^2 \):
\[ PQ^2 = 169 - 25 \]
\[ PQ^2 = 144 \]

Taking the square root of both sides:
\[ PQ = \sqrt{144} \]
\[ PQ = 12 \]

Thus, the length of the tangent \( PQ \) is:

**2. 12 cm**

Would you like further details or have any questions?

Here are 5 questions you might want to ask:
1. How do you use the Pythagorean theorem in geometry problems?
2. What are the properties of a tangent to a circle?
3. Can you provide more examples involving circles and tangents?
4. How do you solve problems involving right triangles?
5. What are some tips for visualizing geometric problems?

Tip: Drawing a diagram and labeling all known values can greatly help in understanding and solving geometry problems accurately.

Learn how to calculate the length of a tangent PQ at a point P of a circle with radius 5 cm, meeting a line through the centre O at point Q such that OQ = 13 cm. This problem involves applying the Pythagorean theorem in circle geometry to find the tangent length. Suitable for students in Grades 9-10.

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