Asked in: March 2025 board paper\(\square ABCD\) 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.

Answer: BD = 15 cm, BX = 9.6 cm

[Figure: rectangle ABCD; semicircle on AD drawn inside the rectangle cuts diagonal BD at X; seg AX joined.]

In \(\Delta BAD\), \(\angle BAD = 90^\circ\) (angle of a rectangle). By Pythagoras theorem, \(BD^2 = AB^2 + AD^2 = 144 + 81 = 225\) \(\therefore BD = 15\) cm.

\(\angle AXD = 90^\circ\) (angle inscribed in a semicircle), so seg AX \(\perp\) BD.

In right-angled \(\Delta BAD\), AX is the perpendicular to the hypotenuse BD, so \(\Delta BAD \sim \Delta BXA\) and \(AB^2 = BX \times BD\).

\(\therefore 144 = BX \times 15\) \(\therefore BX = \dfrac{144}{15} = 9.6\) cm.