TS EAMCET · Maths · Properties of Triangles
If \(a=3, b=5, c=7\) are the sides of a triangle ABC, then its circumradius is
- A \(\frac{7}{\sqrt{3}}\)
- B \(\frac{15}{2}\)
- C \(\frac{15 \sqrt{3}}{4}\)
- D \(\frac{\sqrt{3}}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{7}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
\(s = \frac{a+b+c}{2}\) \(s = \frac{3+5+7}{2} = \frac{15}{2}\) \(K = \sqrt{s(s-a)(s-b)(s-c)}\) \(K = \sqrt{\frac{15}{2}(\frac{15}{2}-3)(\frac{15}{2}-5)(\frac{15}{2}-7)} = \sqrt{\frac{15}{2} \cdot \frac{9}{2} \cdot \frac{5}{2} \cdot \frac{1}{2}} = \frac{15\sqrt{3}}{4}\)…
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