JEE Advanced Model Paper 17 with solutions for online practice

PA = PB = 1
QA = QC = 2
RB = RC = 3
⇒ PQ = 3 ; QR = 5 ; PR = 4

⇒ ΔPQR will be a right angled triangle with ∠P = 90°
∠Q = sin−1

4

5

and ∠R = sin−1

3

5

Area of ΔABC = A (ΔPQR) - A (ΔPAB) - A (ΔQAC) - A (ΔRBC)
= 6 - (

1

2

) -

8

5

-

27

10

=

60−5−16−27

10

=

6

5

Circum - radius of ΔABC = 1