Let P Q R be a triangle of area with a=2, b= 7 2 and c= 5 2 ; where a, b , and c are the lengths of the sides of the triangle opposite to the angles at P, Q and R respectively. Then 2 P- 2 P 2 P+ 2 P equals.

Step-by-step solution

2 P- 2 P 2 P+ 2 P = 2 P-2 P P 2 P+2 P P = 1- P 1- P = 2 ² P 2 2 ² P 2 = ² P 2 = (s-b)(s-c) s(s-a) where s= a+b+c 2 = (s-b)²(s-c)² s(s-a)(s-b)(s-c) = (a+c-b)²(a+b-c)² 16 . ² = (2+ 5 2 - 7 2 )² (2+ 7 2 - 5 2 )² 16 ² = 1 9 16 ² = ( 3 4 )²