rectangular beams/pythagorean theorem

Question

The strongest rectangular beam that can be cut from a circular log is one having a cross section in which the diagonal joining two vertices is trisected by perpendicular segments dropped from the other vertices. If AB = a, BC = b, CE = x, and DE = y show that b/a = square root of 2 over 1 PLEASE HELP!!! This is problem #23 on p. 417 of the Geometry for Enjoyment and Challenge by McDougal Littell (1991)

Answer 1

If AB = a,

BC = b,

CE = x,

and DE = y

show that b/a = square root of 2.

 Using pythagprean thorem

 Draw the figure : relationships are

 From traingle BCD

BD² = BC² + CD²

(3y)² = b² + a²

Simplify

9y² = b² + a²---------------> (1)

From traingle CED

CD² = CE² + DE²

a² = x² + y² ------------------> (2)

From traingle CEB

BC² = CE² + EB²

b² = x² + (2y)² ----------------> (3)

b² = x² + y² + 3y²

Note : x² + y² = a²

b² = a² + 3y²

Subtract a² from each side.

b² - a² = 3y² - a²

Simplify

b² - a² = 3y²

substitute 3y² =b² - a² in equation (1)

3(3y²) = b² + a²

3 (b² - a²)= b² + a²

Simplify

3b² - 3a² = b² + a²

Subtract b² from each side.

3b² - b²- 3a² = b² + a² - b²

Simplify

2b² - 3a² =  a²

Add 3a² to each side.

2b² - 3a² + 3a² = a² + 3a²

Simplify

2b² = 4a²

Divide each side by 2a².

2b² / 2a² = 4a² / 2a²

Simplify

b² / a² = 4 /2 = 2/1

Take square root both sides

b/a = √2/1

There fore

b / a = √2

Answer 2

Draw the figure

Using Pythagorean theorem

From triangle BCD

BD² = BC² + CD²

(3y)² = b² + a²

9y² = b² + a²---------------> (1)

From traingle CED

CD² = CE² + DE²

a² = x² + y² ------------------> (2)

From triangle CEB

BC² = CE² + EB² 

b² = x² + (2y)² ----------------> (3)

b² = x² + y² + 3y²

b² = a² + 3y²

b² - a² = 3y²

substitute 3y² =b² - a² in equation 1 

3 (b² - a²)= b² + a²

3b² - 3a² = b² + a²

2b² = 4a²

b² / a²= 4 /2 = 2/1

Take square root both sides

b/a = √2/1