Problem:
In â–³ABC,AB=BC=29, and AC=42. What is the area of â–³ABC?
Answer Choices:
A. 100
B. 420
C. 500
D. 609
E. 701
Solution:
Let D be the midpoint of side AC. Then BD is the altitude to AC and △BDC is a right triangle with BC=29 and DC=21. So BD=292−212​=400​=20. The area of △ABC=21​⋅20⋅42=420.
OR
Heron's formula gives the area of a triangle in terms of the lengths of its sides. If the side lengths are a,b and c, then let s=2a+b+c​. The area is then s(s−a)(s−b)(s−c)​. In this problem, s=229+29+42​=50, and the area is 50⋅21⋅21⋅8​=21400​=21⋅20=420.