Understanding the Geometry Problem

This problem involves two circles that share the same center (concentric circles). We are given the radii of both circles. A diameter of the larger circle is involved, and a line segment tangent to the smaller circle originates from a point on the larger circle (specifically, an endpoint of the diameter) and touches the smaller circle at a point D. We need to find the length of a specific segment, AD, connecting one endpoint of the larger circle's diameter to the point of tangency on the smaller circle.

Let the center of the concentric circles be O.

• Radius of the smaller circle, \(r = 12\) cm.
• Radius of the bigger circle, \(R = 13\) cm.
• AB is a diameter of the bigger circle. Thus, O is the midpoint of AB, and \(OA = OB = R = 13\) cm. The length of the diameter \(AB = 2R = 2 \times 13 = 26\) cm.
• BE is a tangent to the smaller circle at point D. A fundamental property of tangents is that the radius drawn to the point of tangency is perpendicular to the tangent line. Therefore, the radius OD of the smaller circle is perpendicular to BE at D. This means \(\angle ODB = 90^\circ\). The length of OD is the radius of the smaller circle, so \(OD = r = 12\) cm.

Step-by-Step Solution to Find AD

Finding the length of BD

Consider the triangle ODB. We know OD = 12 cm (radius of smaller circle) and OB = 13 cm (radius of bigger circle). Since BE is tangent to the smaller circle at D, \(\triangle ODB\) is a right-angled triangle with the right angle at D (\(\angle ODB = 90^\circ\)). OB is the hypotenuse.

We can use the Pythagorean theorem to find the length of BD:

\[OB^2 = OD^2 + BD^2\]

Substitute the known values:

\[13^2 = 12^2 + BD^2\]

\[169 = 144 + BD^2\]

\[BD^2 = 169 - 144\]

\[BD^2 = 25\]

\[BD = \sqrt{25} = 5 \text{ cm}\]

So, the length of BD is 5 cm.

Finding the length of AD using the Law of Cosines

Now consider the triangle ADB. We know the lengths of two sides: AB = 26 cm (diameter of the bigger circle) and BD = 5 cm (calculated above). To find the length of the third side AD, we can use the Law of Cosines if we know one of the angles. The angle \(\angle ABD\) (which is the same as \(\angle OBD\)) can be determined from the right-angled triangle ODB.

In \(\triangle ODB\), we can find the cosine of \(\angle OBD\):

\[\cos(\angle OBD) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BD}{OB}\]

Substitute the values BD = 5 cm and OB = 13 cm:

\[\cos(\angle ABD) = \cos(\angle OBD) = \frac{5}{13}\]

Now apply the Law of Cosines to \(\triangle ADB\) to find AD:

\[AD^2 = AB^2 + BD^2 - 2 \cdot AB \cdot BD \cdot \cos(\angle ABD)\]

Substitute the values AB = 26 cm, BD = 5 cm, and \(\cos(\angle ABD) = \frac{5}{13}\):

\[AD^2 = 26^2 + 5^2 - 2 \cdot 26 \cdot 5 \cdot \left(\frac{5}{13}\right)\]

\[AD^2 = 676 + 25 - 2 \cdot \left(\frac{26}{13}\right) \cdot 5 \cdot 5\]

\[AD^2 = 676 + 25 - 2 \cdot 2 \cdot 25\]

\[AD^2 = 676 + 25 - 100\]

\[AD^2 = 701 - 100\]

\[AD^2 = 601\]

Now, find the square root of 601 to get the length of AD:

\[AD = \sqrt{601}\]

Calculating the Final Answer (to one decimal place)

We need to calculate the value of \(\sqrt{601}\) and round it to one decimal place.

We know that \(24^2 = 576\) and \(25^2 = 625\), so \(\sqrt{601}\) is between 24 and 25.

Let's check values close to the middle:

\[24.5^2 = (24 + 0.5)^2 = 24^2 + 2 \cdot 24 \cdot 0.5 + 0.5^2\]

\[24.5^2 = 576 + 24 + 0.25\]

\[24.5^2 = 600.25\]

Since \(24.5^2 = 600.25\) is very close to 601, \(\sqrt{601}\) is approximately 24.5.

Let's check 24.6 to be sure:

\[24.6^2 = (24.5 + 0.1)^2 = 24.5^2 + 2 \cdot 24.5 \cdot 0.1 + 0.1^2\]

\[24.6^2 = 600.25 + 4.9 + 0.01\]

\[24.6^2 = 605.16\]

Since 601 is closer to 600.25 than to 605.16, \(\sqrt{601}\) is closer to 24.5 than to 24.6.

Therefore, rounding \(\sqrt{601}\) to one decimal place gives 24.5.

The length of AD is approximately 24.5 cm.

Revision Table: Key Geometric Concepts

Additional Information: Alternative Approaches

While the Law of Cosines is a direct approach using calculated lengths and angles, another method could involve using coordinate geometry. By setting the center O at the origin (0,0), A and B can be placed at (-13,0) and (13,0). The point D lies on the smaller circle \(x^2 + y^2 = 12^2\). Since OD is perpendicular to the tangent line passing through D and B(13,0), the slope of OD (\(y_D/x_D\)) and the slope of BD (\(y_D/(x_D-13)\)) are negative reciprocals. Also, D is the foot of the perpendicular from O to the line BE. The line BE passes through B(13,0) and D. The perpendicular from O(0,0) to BE is OD. The coordinates of D can be found by using the property that the foot of the perpendicular from O(0,0) to a line \(ax+by+c=0\) is given by a formula, or by finding the intersection of the circle and the line perpendicular to OD passing through B. This approach is more algebraic but yields the same result for the coordinates of D, and consequently, the length of AD using the distance formula. Both methods rely on the core geometric properties of tangents and circles.