Q7:

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.

Given:

Two concentric circles with a common center $O$.

Radius of the larger circle, $R = 5$ cm.

Radius of the smaller circle, $r = 3$ cm.

A chord $AB$ of the larger circle touches the smaller circle at point $P$.

To Find:

The length of the chord $AB$.

Step 1: Establishing Geometric Relationships

Let $O$ be the center of the concentric circles. Let $AB$ be the chord of the larger circle that is tangent to the smaller circle at point $P$.

By the property of tangents: A tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore, $OP \perp AB$. [Theorem: The tangent at any point of a circle is perpendicular to the radius through the point of contact.]

Step 2: Applying the Perpendicular Bisector Theorem

In the larger circle, $OP$ is a line segment from the center $O$ perpendicular to the chord $AB$.

According to the theorem: A perpendicular drawn from the center of a circle to a chord bisects the chord. [Theorem: The perpendicular from the center of a circle to a chord bisects the chord.]

Therefore, $AP = PB$.

Step 3: Calculating the Length of $AP$ using the Pythagorean Theorem

Consider the right-angled triangle $\triangle OPA$, where $\angle OPA = 90^\circ$.

Using the Pythagorean Theorem: $OA^2 = OP^2 + AP^2$.

Given $OA = R = 5$ cm and $OP = r = 3$ cm.

$5^2 = 3^2 + AP^2$

$25 = 9 + AP^2$

$AP^2 = 25 - 9$

$AP^2 = 16$

$AP = \sqrt{16} = 4$ cm.

Step 4: Determining the Total Length of the Chord $AB$

Since $AP = PB$ and $AP = 4$ cm, then $PB = 4$ cm.

The total length of the chord $AB = AP + PB$.

$AB = 4 \text{ cm} + 4 \text{ cm} = 8 \text{ cm}$.

Final Answer: The length of the chord of the larger circle is 8 cm.