Question

In the given figure, PA is the tangent to the circle with centre O such that OA = 10 cm, AB = 8 cm and AB \(\perp\) OP. Find the length of PB.

Hint:

An important property of right-angled triangles to remember:
The square of the altitude to the hypotenuse equals the product of the two segments of the hypotenuse:
\[ AB^2 = OB \times PB \]
Substituting the values immediately gives \( 8^2 = 6 \times PB \implies 64 = 6 \times PB \implies PB = \frac{32}{3}\text{ cm} \).
This direct geometric relation avoids setting up full triangle similarity statements.

Answer 1

Step 1: Understanding the Question:
The topic of this question is Geometry of Circles and Triangles (Properties of Right-angled triangles and similarity).
We are given a circle with center \( O \) and radius \( OA = 10\text{ cm} \).
\( PA \) is the tangent to the circle at point \( A \).
A perpendicular line segment \( AB \) is drawn from \( A \) to the line segment \( OP \) such that \( AB \perp OP \). The length of \( AB \) is \( 8\text{ cm} \).
We need to calculate the length of the segment \( PB \).

Step 2: Key Formula or Approach:
- Since \( PA \) is a tangent, it is perpendicular to the radius \( OA \), making \( \angle OAP = 90^{\circ} \).
- Since \( AB \perp OP \), \( \angle OBA = \angle PBA = 90^{\circ} \).
- We can use Pythagoras theorem in the right-angled triangle \( \Delta OBA \) to find the length of \( OB \).
- Use similarity of triangles \( \Delta OBA \sim \Delta ABP \) to solve for \( PB \).

Step 3: Detailed Explanation:
1. In right-angled triangle \( \Delta OBA \) (since \( \angle OBA = 90^{\circ} \)):
Using Pythagoras theorem:
\[ OA^2 = OB^2 + AB^2 \]
Substitute \( OA = 10\text{ cm} \) and \( AB = 8\text{ cm} \):
\[ 10^2 = OB^2 + 8^2 \]
\[ 100 = OB^2 + 64 \]
\[ OB^2 = 100 - 64 = 36 \]
\[ OB = 6\text{ cm} \]
2. In the main right-angled triangle \( \Delta OAP \) with \( \angle OAP = 90^{\circ} \):
\( AB \) is the altitude drawn to the hypotenuse \( OP \).
In a right-angled triangle, the altitude to the hypotenuse divides the triangle into two similar triangles which are also similar to the original triangle.
Therefore:
\[ \Delta OBA \sim \Delta ABP \]
3. Write down the ratio of corresponding sides for these similar triangles:
\[ \frac{PB}{AB} = \frac{AB}{OB} \]
4. Substitute the known values \( AB = 8\text{ cm} \) and \( OB = 6\text{ cm} \):
\[ \frac{PB}{8} = \frac{8}{6} \]
5. Solve for \( PB \):
\[ PB = \frac{8 \times 8}{6} = \frac{64}{6} \]
Reduce the fraction to its lowest terms:
\[ PB = \frac{32}{3}\text{ cm} \approx 10.67\text{ cm} \]

Step 4: Final Answer:
The length of segment PB is \(\frac{32}{3}\text{ cm}\) (or approximately \(10.67\text{ cm}\)).