JEE Mains · Maths · STD 11 - rectangular cartensian co-ordinates
Let \(\left(5, \frac{a}{4}\right)\), be the circumcenter of a triangle with vertices \(A(a,-2), B(a, 6)\) and \(C\left(\frac{a}{4},-2\right)\). Let \(\alpha\) denote the circumradius, \(\beta\) denote the area and \(\gamma\) denote the perimeter of the triangle. Then \(\alpha+\beta+\gamma\) is ...........
- A 60
- B 53
- C 62
- D 30
Answer & Solution
Correct Answer
(B) 53
Step-by-step Solution
Detailed explanation
\(A ( a ,-2), B ( a , 6), C \left(\frac{ a }{4},-2\right), O \left(5, \frac{ a }{4}\right)\) \(AO = BO\) \((a-5)^2+\left(\frac{a}{4}+2\right)^2=(a-5)^2+\left(\frac{a}{4}-6\right)^2\) \(a=8\) \(AB =8, AC =6, BC =10\) \(\alpha=5, \beta=24, \gamma=24\)
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