in an isosclees triangle ABC(AB=BC) medians AM and CN intersect at point D at a right angle. Find the area of the quadrilateral NBMD if AC=7.

Answer 1

AD = DC = 7/√2; DN = DM = 7/(2√2); ∠A = ∠C = 75°, so, ∠ B = 30°; AN² = 49/2 + 49/8 = 49·5/8;

AN = 7√10/4; BN = BM = 7√10/4; NM = 3.5. BK is 3rd median. BY Pythagorean Th.eorem BK = √(7√10/2)² - -49/4 = 7√10/4 - 1/4 = 21/2; BD = 2/3BK = 7. S, Area NBMD = MN·BD/2 = 7·3.5/2 = 12.25 ,

because MN orthogonal with BD