Explanation:

ABC is a triangle with AB = 5 cm, AC = √41 (=6.403124237) cm and BC = 8 cm. AD is perpendicular to BC.
Let BD = x cm, and DC = (8-x) cm.
From triangle ABD: AD^2 = AB^2=BD^2 = 5^2-x^2 = 25-x^2 ,,,(1)
From triangle ACD: AD^2 = AC^2=CD^2 = 41-(8-x)^2 = 41–64+16x-x^2 …(2)
Equate (1) and (2)
25-x^2 = -23+16x-x^2, or
48 = 16x, or
x = 3 cm = BD
Area of triangle ABC can be got from Heron’s relation,
2s = 5+8+√41 = 19.40312424 or s = 9.701562119
Area of triangle ABC = [9.701562119(9.701562119–5)(9.701562119–8)(9.701562119-6.403124237)]^0.5
= [9.701562119*4.701562119*1.701562119*3.298437882]^0.5
= 256^0.5 = 16 sq cm
BD:BC = 3:8 so the
Area of ABD =(3/8)*16 = 6 sq cm.