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.
✓
Answer
In rect. ABCD, $\angle BAD = 90^{\circ}$. By Pythagoras: $BD = \sqrt{AB^{2} + AD^{2}} = \sqrt{12^{2} + 9^{2}} = \sqrt{144 + 81} = \sqrt{225} = 15$ cm.
Angle in a semicircle $\angle AXD = 90^{\circ}$. In $\Delta BAD$, $AX \perp BD$.
By area of triangle: $\frac{1}{2} \times AB \times AD = \frac{1}{2} \times BD \times AX \Rightarrow AX = \frac{12 \times 9}{15} = 7.2$ cm.
In $\Delta ABX$, $BX = \sqrt{AB^{2} - AX^{2}} = \sqrt{144 - 51.84} = \sqrt{92.16} = 9.6$ cm.
Angle in a semicircle $\angle AXD = 90^{\circ}$. In $\Delta BAD$, $AX \perp BD$.
By area of triangle: $\frac{1}{2} \times AB \times AD = \frac{1}{2} \times BD \times AX \Rightarrow AX = \frac{12 \times 9}{15} = 7.2$ cm.
In $\Delta ABX$, $BX = \sqrt{AB^{2} - AX^{2}} = \sqrt{144 - 51.84} = \sqrt{92.16} = 9.6$ cm.
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