perplexus.info :: Geometry : Triangle bounded by rectangles

Triangle bounded by rectangles (Posted on 2024-12-11)

There are two rectangles 1 × 2 and 3 × 4 with parallel sides centered at a common point O. Points F and G are selected on the boundary of the inner and outer rectangle, respectively. What is the maximum possible area of triangle FOG?

No Solution Yet Submitted by Danish Ahmed Khan

No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)

Solution - different answer

| Comment 3 of 4 |

Case 1: Lengths 1 and 3 are parallel.

Point O is (0,0), Let point G be (2, 1.5).

Length OG is 5/2 (triangle base)

Let F be (1, -.5)

Altitude of triangle FOG is 1.

Area: (1/2)(5/2)(1) = 5/4 = 1.25

Case 2: Lengths 1 and 4 are parallel.

Point O is (0,0), Let point G be (2, 1.5).

Length OG is 5/2 (triangle base)

Let F be (.5, -1)

Altitude of triangle FOG is 11/10.

Area: (1/2)(5/2)(11/10) = 55/40 = 1.375

So 55/40 = 1.375 is the maximum possible area of triangle FOG.

Posted by Larry on 2024-12-13 10:55:13

Please log in:

Login:
Password:
Remember me:

Sign up! \| Forgot password

Search:

Search body:

Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info