Problem:
Triangle ABC has a right angle at B. Point D is the foot of the altitude from B,AD=3, and DC=4. What is the area of â–³ABC?
Answer Choices:
A. 43​
B. 73​
C. 21
D. 143​
E. 42
Solution:
By the Pythagorean Theorem, AB2=BD2+9,BC2=BD2+16, and AB2+BC2=49. Adding the first two equations and substituting gives 2⋅BD2+25=49. Then BD=23​, and the area of △ABC is 21​⋅7⋅23​=(B)73​​.
OR
Because △ADB and △BDC are similar, 3BD​=BD4​, from which BD=23​. Therefore the area of △ABC is 21​⋅7⋅23​=(B)73​​.
