Take point H on AD such that HD=6 and AH=1. Extend EG and intersect AH at point P. Then PE ∥ HC ∥ DF. So HP : AH = 2 : (2+9), and thus HP = 2/11. Also, HC = 2DF = 2a, where a is the side length of the square. EP : HC = 9 : (2+9) implies GP = 7a/11. In Rt△DGP, Pythagoras theorem gives (7a/11)² + a² = (6+2/11)², i.e., a² = 68²/170 ≈ 27.2
Пікірлер: 27