Problem:
he area of △EBD is one third of the area of 3−4−5△ABC. Segment DE is perpendicular to segment AB. What is BD?
Answer Choices:
A. 34
B. 5
C. 49
D. 343
E. 25
Solution:
The area of △ABC is 21⋅3⋅4=6, so the area of △EBD is 31⋅6=2. Note that △ABC and △EBD are right triangles with an angle in common, so they are similar. Therefore BD and DE are in the ratio 4 to 3. Let BD=x and DE=43x. Then the area of △EBD can be expressed as 21⋅x⋅43x=83x2. Because △EBD has area 2, solving yields BD=(D)343.
OR
Because △EBD and △ABC are similar triangles, their areas are in the ratio of the squares of their corresponding linear parts. Therefore (4BD)2=31 and BD=(D)343.
