Question
The vertices of a triangle are (0,0),(4,0) and \((3,9).\) The area of the circle passing through these three points is
- \(\frac{{14\pi }}{3}\)
- \(\frac{{12\pi }}{5}\)
- \(\frac{{123\pi }}{7}\)
- \(\frac{{205\pi }}{9}\)
Option: 4
Solution:
Solution:
\(a = 4,b = \sqrt {90} ;c = \sqrt {82} \)
Area of the triangle \( = \frac{1}{2} \times 4 \times 9 = 18\)
The circumradius of the triangle \((R) = \frac{{abc}}{{4\Delta }}\)
Area of the circle \( = \pi {R^2} = \pi {\left( {\frac{{abc}}{{4\Delta }}} \right)^2} = \frac{{\pi {{(4 \cdot \sqrt {90} \cdot \sqrt {82} )}^2}}}{{{{(4 \cdot 18)}^2}}}\)
\( = \pi \times \frac{{16 \times 90 \times 82}}{{16 \times 18 \times 18}} = \frac{{205}}{9}\pi \)
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