Question

ABC is a triangle. AB = 5 cm, AC = √41 cm and BC = 8 cm. AD is perpendicular to BC. What is the area (in cm2) of triangle ABD?

A) 12 B) 6 C) 10 D) 20

Answer 1

Answer:

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.