Problem:
In △ABC,AB=BC=29, and AC=42. What is the area of △ABC?
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.