# rectangular beams/pythagorean theorem

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)
asked Dec 25, 2012 in GEOMETRY

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

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

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