PQ is chord of length 6 cm of a circle of radius 5 cm. Tangents to the circle at P and Q meet at T. Length of TP is

Options

A. 4.75 cm
B. 2.75 cm
C. 3.75 cm
D. 4.25 cm

Answer

To find the length of TP, we will use the properties of circles and right triangles. Here’s a step-by-step solution:

### Step 1: Understanding the Geometry
We have a circle with a radius of 5 cm and a chord PQ of length 6 cm. We need to find the length of the tangent TP from point T to point P on the circle.

### Step 2: Finding the Midpoint of the Chord
Since PQ is a chord, we can find its midpoint K. The length of the chord PQ is 6 cm, so:
\[ PK = KQ = \frac{PQ}{2} = \frac{6}{2} = 3 \text{ cm} \]

### Step 3: Drawing the Perpendicular from the Center to the Chord
Let O be the center of the circle. We draw a perpendicular line from O to the chord PQ, which meets PQ at K. This perpendicular bisects the chord, so PK = KQ = 3 cm.

### Step 4: Applying the Pythagorean Theorem
In the right triangle OPK, we can apply the Pythagorean theorem:
\[ OP^2 = OK^2 + PK^2 \]
Where:
- OP = radius of the circle = 5 cm
- PK = 3 cm (half the chord length)
- OK = length we need to find

Substituting the values:
\[ 5^2 = OK^2 + 3^2 \]
\[ 25 = OK^2 + 9 \]
\[ OK^2 = 25 - 9 = 16 \]
\[ OK = \sqrt{16} = 4 \text{ cm} \]

### Step 5: Using Similar Triangles
Now, we will use the similarity of triangles. In triangles OPT and PKO:
- Angle OPT = Angle PKO = 90° (both are right angles)
- Angle POT = Angle POK (common angle)

Since the triangles are similar, we can set up the ratio of their corresponding sides:
\[ \frac{PT}{PK} = \frac{OP}{OK} \]

### Step 6: Substituting Known Values
We know:
- PK = 3 cm
- OP = 5 cm
- OK = 4 cm

Substituting these values into the ratio:
\[ \frac{PT}{3} = \frac{5}{4} \]

### Step 7: Solving for PT
Cross-multiplying gives us:
\[ PT \cdot 4 = 3 \cdot 5 \]
\[ 4PT = 15 \]
\[ PT = \frac{15}{4} = 3.75 \text{ cm} \]

### Conclusion
The length of TP is 3.75 cm.

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