Proposed Problem

Click the figure below to see the complete problem 322.

See more:

Complete Problem 322

Level: High School, SAT Prep, College geometry

## Monday, July 13, 2009

### Problem 322: Square, Inscribed circle, Tangent, Triangle area

Subscribe to:
Post Comments (Atom)

Let DF = p and BE = q and let the side of the square = 2a. From rt. angled tr. AEF we've, (2a-p)^2+(2a-q)^2=(p-a+q-a)^2 and this yields pq = 2a^2 ------(1) while Tr. CEF= 4a^2 -Tr.AEF -Tr.FDC -Tr.CBE = 4a^2-(2a-p)(2a-q)/2 -ap -aq = 2a^2 - pq/2 = 2a^2 - a^2 by eqn. (1). In other words, Tr. CEF = a^2 =(1/4)* (Sq. ABCD). QED

ReplyDeletevihaanup@gmail.com