Question

\(\square\)ABCD is a rectangle. Taking AD as a diameter, a semicircle AXD is drawn which intersects the diagonal BD at X. If AB = 12 cm, AD = 9 cm, then find the values of BD and BX.

Hint:

Recognizing that AX is an altitude to the hypotenuse of \(\triangle DAB\) is the key to solving the second part of the problem. Remember the useful corollaries of similarity in right triangles: \(AB^2 = BX \cdot BD\) and \(AD^2 = DX \cdot BD\).

Answer 1

Step 1: Understanding the Concept: 
We can use the Pythagorean theorem in the right-angled triangle formed by the sides of the rectangle to find the length of the diagonal BD. Then, using the property that the angle in a semicircle is a right angle, we can find BX. 
 

Step 2: Key Formula or Approach: 
1. Pythagorean Theorem: In a right-angled triangle, \((\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Height})^2\). 

2. Angle in a Semicircle Theorem: The angle subtended by a diameter at any point on the circumference is a right angle (90\(^\circ\)). 

3. Property of Right-Angled Triangles: In a right-angled triangle, if an altitude is drawn to the hypotenuse, then \((\text{Leg})^2 = (\text{Adjacent segment of hypotenuse}) \times (\text{Whole hypotenuse})\). 
 

Step 3: Detailed Explanation: 
Given: \[\begin{array}{rl} \bullet & \text{ABCD is a rectangle, so \(\angle DAB = 90^\circ\).} \\ \bullet & \text{AB = 12 cm, AD = 9 cm.} \\ \end{array}\] 1. Find the value of BD: In right-angled \(\triangle\)DAB, BD is the hypotenuse. By Pythagorean theorem: \[ BD^2 = AB^2 + AD^2 \] \[ BD^2 = 12^2 + 9^2 = 144 + 81 = 225 \] \[ BD = \sqrt{225} = 15 \text{ cm} \] 2. Find the value of BX: A semicircle is drawn with AD as the diameter. The diagonal BD intersects the semicircle at point X. According to the Angle in a Semicircle Theorem, the angle subtended by the diameter AD at point X on the circumference is 90\(^\circ\). \[ \therefore \angle AXD = 90^\circ \] This means that seg AX is perpendicular to the diagonal BD. Now, consider the right-angled \(\triangle\)DAB. seg AX is the altitude from vertex A to the hypotenuse BD. Using the property of right-angled triangles (related to geometric mean): \[ AB^2 = BX \times BD \] We have AB = 12 cm and we found BD = 15 cm. \[ 12^2 = BX \times 15 \] \[ 144 = 15 \times BX \] \[ BX = \frac{144}{15} = \frac{48}{5} = 9.6 \text{ cm} \]

Step 4: Final Answer: 
The length of BD is 15 cm and the length of BX is 9.6 cm.