Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
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Step-by-Step Solution
Step 1: Visualize the setup
We begin by visualizing the problem. Imagine two circles sharing the same center, O. The larger circle has a radius of 5 cm, and the smaller one has a radius of 3 cm. A chord of the larger circle, let's call it AB, is tangent to the smaller circle at a point P.
Step 2: Identify perpendicularity
When a chord of a larger circle touches the smaller circle, it means the chord is tangent to the smaller circle. The radius drawn to the point of tangency is always perpendicular to the tangent. Therefore, the line segment OP (radius of the smaller circle) is perpendicular to the chord AB.
Step 3: Apply chord property
A perpendicular from the center of a circle to a chord bisects the chord. Since OP is perpendicular to AB, it bisects AB at point P. This means AP is equal to PB.
Step 4: Form a right-angled triangle
We have established that OP is perpendicular to AB. This forms a right-angled triangle OPA, where OA is the hypotenuse (radius of the larger circle), OP is one leg (radius of the smaller circle), and AP is the other leg (half the length of the chord).
Step 5: Apply Pythagorean Theorem
Using the Pythagorean theorem in the right-angled triangle OPA, we can find the length of AP. We know OA (radius of larger circle) is 5 cm and OP (radius of smaller circle) is 3 cm. Substituting these values, we find AP to be 4 cm.
Step 6: Calculate the length of the chord
Since P is the midpoint of the chord AB, the total length of the chord AB is twice the length of AP. Therefore, the length of the chord is 2×4=8 cm.