Question 3

We will find the distance of each chord from the center. Then we will add these distances.

Step 1 — Distance of first chord

Let the first chord be $AB$. Its length is $\mathbf{6 \text{ cm}}$. The radius of the circle is $\mathbf{5 \text{ cm}}$. The perpendicular from the center bisects the chord. So, half the chord length is $\mathbf{3 \text{ cm}}$. Let $d_1$ be the distance from the center to chord $AB$. We use the Pythagorean theorem.

$d_1^2 + 3^2 = 5^2$

$d_1^2 + 9 = 25$

$d_1^2 = 25 - 9$

$d_1^2 = 16$

$\boxed{d_1 = 4 \text{ cm}}$

Step 2 — Distance of second chord

Let the second chord be $CD$. Its length is $\mathbf{8 \text{ cm}}$. The radius of the circle is $\mathbf{5 \text{ cm}}$. The perpendicular from the center bisects the chord. So, half the chord length is $\mathbf{4 \text{ cm}}$. Let $d_2$ be the distance from the center to chord $CD$. We use the Pythagorean theorem.

$d_2^2 + 4^2 = 5^2$

$d_2^2 + 16 = 25$

$d_2^2 = 25 - 16$

$d_2^2 = 9$

$\boxed{d_2 = 3 \text{ cm}}$

Step 3 — Total distance between midpoints

The chords are on opposite sides of the center. The distance between their midpoints is the sum of $d_1$ and $d_2$.

$\text{Distance} = d_1 + d_2$

$= 4 + 3$

$\boxed{7 \text{ cm}}$

Answer

The distance between the midpoints of the two chords is $\mathbf{7 \text{ cm}}$.