JEE Main202429 Jan 2024Morning ShiftMathematicsStraight LinesActual
Let 5 , a 4 , be the circumcenter of a triangle with vertices A ( a , - 2 ) , B ( a , 6 ) and C a 4 , - 2 . Let α denote the circumradius, β denote the area and γ denote the perimeter of the triangle. Then α + β + γ is
Options
- A60
- B53
- C62
- D30
Correct answer
B. 53
Step-by-step solution
Given: A a , - 2 , B a , 6 and C a 4 , - 2 ⇒ A B 2 = a - a 2 + 6 + 2 2 ⇒ A B 2 = 64 . . . i ⇒ B C 2 = a - a 4 2 + 6 + 2 2 ⇒ B C 2 = 9 a 2 16 + 64 . . . i i ⇒ A C 2 = a - a 4 2 + - 2 + 2 2 ⇒ A C 2 = 9 a 2 16 . . . i i i Using i , i i and i i i , ⇒ A C 2 + A B 2 = B C 2 So, △ A B C is right angled at A . So, circumcentre will be mid-point of B C . ⇒ S ≡ a + a 4 2 , 6 - 2 2 ⇒ S ≡ 5 a 8 , 2 = 5 , a 4 ⇒ a = 8 ⇒ B C = 9 × 64 16 + 64 ⇒ B C = 100 = 10 ⇒ A B = 8 ⇒ A C = 9 × 64 16 ⇒ A C = 6 Now, circumradius is given by, R =