EF is a chord of a circle whose length is 10 cm. If radius of the circle is 13 cm, then the length of perpendicular from the centre to the chord is

Options

A. 12 cm
B. 8 cm
C. 7 cm
D. 5 cm

Answer

To solve the problem step by step, we will use the properties of circles and right triangles.

### Step 1: Understand the problem
We have a circle with a radius of 13 cm and a chord EF of length 10 cm. We need to find the length of the perpendicular from the center of the circle (point O) to the chord (point G).

**Hint:** Remember that the perpendicular from the center of a circle to a chord bisects the chord.

### Step 2: Draw the circle and label the points
Draw a circle and label the center as O. Mark the chord EF such that its length is 10 cm. Let G be the point where the perpendicular from O meets the chord EF.

**Hint:** Visualize the circle and the chord to understand the geometric relationships.

### Step 3: Bisect the chord
Since OG is perpendicular to EF, it bisects the chord. Therefore, the lengths of EG and GF are equal. Since EF = 10 cm, we have:
\[ EG = GF = \frac{10}{2} = 5 \text{ cm} \]

**Hint:** Use the property of bisecting chords to find the lengths of the segments.

### Step 4: Apply the Pythagorean theorem
In the right triangle OGF, we can apply the Pythagorean theorem:
\[ OG^2 + GF^2 = OF^2 \]
Where:
- \( OG \) is the length we want to find,
- \( GF = 5 \text{ cm} \),
- \( OF = 13 \text{ cm} \) (the radius).

**Hint:** Identify the sides of the right triangle correctly to apply the theorem.

### Step 5: Substitute the known values
Substituting the known values into the equation:
\[ OG^2 + 5^2 = 13^2 \]
\[ OG^2 + 25 = 169 \]

**Hint:** Make sure to square the lengths correctly.

### Step 6: Solve for OG
Now, isolate \( OG^2 \):
\[ OG^2 = 169 - 25 \]
\[ OG^2 = 144 \]

Taking the square root of both sides gives:
\[ OG = \sqrt{144} = 12 \text{ cm} \]

**Hint:** Remember to take the positive root since lengths cannot be negative.

### Conclusion
The length of the perpendicular from the center of the circle to the chord EF is 12 cm.

**Final Answer:** 12 cm