Answer: [tex]81 ( \pi - 2)\text { square cm}[/tex]
Step-by-step explanation:
Let O be the radius of the circle and AB be the chord of length 18√2 cm.
Since, the radius of the circle is 18 cm,
Therefore, OA = OB = 18 cm,
Let, N be the mid point of the chord then,
AN = NB = 9√2 cm
In triangle AON,
Since, [tex]\angle ANO = 90^{\circ}[/tex] ( by the property of circle)
[tex]\angle AON = sin^{-1} (\frac{9\sqrt{2} }{18} )[/tex]
[tex]\angle AON = sin^{-1} (\frac{\sqrt{2} }{2} )[/tex]
[tex]\angle AON = sin^{-1} (\frac{1}{\sqrt{2} } )[/tex]
[tex]\angle AON = 45^{\circ}[/tex]
Similarly, in triangle BON,
[tex]\angle BON = 45^{\circ}[/tex]
⇒ [tex]\angle AOB = \angle AON +\angle BON = 45^{\circ}+45^{\circ} = 90^{\circ}[/tex]
Therefore, the area of sector AOB = [tex]\frac{90^{\circ}}{360^{\circ}} \pi (18)^2[/tex]
= [tex]\frac{1}{4} \pi (18)^2[/tex]
= [tex]81\pi \text{ square cm}[/tex]
Area of triangle AOB = [tex]\frac{1}{2} \times ON\times AB[/tex]
= [tex]\frac{1}{2} \times \sqrt{OA^2-AN^2}\times AB[/tex]
= [tex]\frac{1}{2} \times \sqrt{324-162}\times 18\sqrt{2}[/tex]
= [tex]\frac{1}{2} \times \sqrt{162}\times 18\sqrt{2}[/tex]
= [tex]\frac{1}{2} \times \sqrt{2}\times \sqrt{81}\times 18\sqrt{2}[/tex]
= [tex]162\text{ square cm}[/tex]
Therefore, the area of smaller region by the chord AB = Area of sector AOB - Area of triangle AOB = [tex]81\pi - 162[/tex]
= [tex]81 ( \pi - 2)\text { square cm}[/tex]
