Question

Two concentric circles are of radii 5 cm and 4 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Hint:

Right triangles in concentric circle problems almost always involve standard Pythagorean triplets like (3, 4, 5).
Once you identify the half-chord is 3 cm, you can immediately double it to get the total chord length of 6 cm.

Answer 1

Step 1: Recall the direct chord-length formula.
For two concentric circles, the length of a chord of the bigger circle that just touches the smaller circle is given by $\text{Chord}=2\sqrt{R^2-d^2}$, where $R$ is the radius of the bigger circle and $d$ is the distance from the centre to the chord.
Step 2: Identify d for this tangent chord.
Because the chord is tangent to the smaller circle, the perpendicular distance from the common centre to the chord equals the smaller circle's radius, so $d=4$ cm, while $R=5$ cm.
Step 3: Substitute and simplify.
\[ \text{Chord} = 2\sqrt{5^2-4^2} = 2\sqrt{25-16}=2\sqrt9=2\times3=6 \text{ cm} \]
\[ \boxed{6 \text{ cm}} \]