I tried to visit your site, https://www.mathbeyondtherealm.com/, but the link didn't work for me...

Jason

Kevin Fite
www.mathbeyondtherealm.com
Detroit,Mi

The trapezoid can be broken into a rectangle of width 3 and height a, a triangle of hypotenuse 2, height a, and width b, and a triangle of hypotenuse 3, height a, and width c.

a^2 + b^2 = 2^2

a^2 + c^2 = 3^2

So, c^2 - b^2 = 3^2 - 2^2 = 5

Also, b + c + 3 = 6, so b + c = 3

c^2 - b^2 = (c - b) * (c + b) = (c - b) * 3

So, 3 * (c - b) = 5, c - b = 5/3

c + b = 3

2c = 3 + 5/3

c = 7/3

b = 2/3

a^2 + (2/3)^2 = 4

a^2 = 32/9, a = 4/3 * 2^.5

The area of the rectangle = 3 * a = 4 * 2^.5

The area of One triangle = .5 * a * b = 4/9 * 2^.5

The area of the other triangle = .5 * a * c = 14/3 * 2^.5

So, the area of the entire trapezoid = (4 + 4/9 + 14/9) * 2^.5 = 6 * 2^.5

Thank you, I am glad we are in agreement. (You have a typo in the last but one line, which you can always fix by editing your message, but the last line is correct). A sketch always helps:

Of course, you can calculate the area of a trapezoid with the sides of length 4, 3, 3, 6, sides of length 3 and 6 being parallel, in exactly the same way, it is not harder at all - I was wrong in this detail. But there is a little catch if the sides are of length 2, 3, 3, 5, sides of length 3 and 5 being parallel.

Possibly the simplest construction of the trapezoid ABCD (AB = 2, BC = 6, CD = 3, AD = 3, BC || AD) is as follows:

1. Draw a line segment BC of the length 6 units.

2. Draw 2 circles b and g centered at the points B and C and with radii 2 and 3 units, respectively.

3. Translate the circle b by AD = 3 units right (AD being the shorter parallel side of the trapezoid) into the circle b' centered at the point B' (such that BB' = AD = 3 and B' is on the line BC). The translated circle b' intersects the circle g at the point D.

4. Draw a parallel to the line BC through the point D. The parallel intersects the circle b at the point A.

Regards, Vladimir

area of rectangle = base x height
= 2 x root 5.

area of triangle = 1/2 x base x height
= 1/2 x 2 x root 5
= root 5

total area = triangle area + other triangle's area + rectangle's area
= root 5 + root 5 + 2 x root 5
= 2x root 5 + 2x root 5
= 4 x root 5.
:)

The problem in San Francisco Bay Area Math Bowl competition is stated as follows:

Find the area of a trapezoid with the sides of length 2, 3, 3, and 6, where the sides of length 3 and 6 are parallel.

The given (and incorrect) solution is 4Ö5. The correct solution is 6Ö2, as Brian calculated above. You are both solving a slightly different problem:

Find the area of a trapezoid with the sides of length 3, 2, 3, and 6, where the sides of length 2 and 6 are parallel.

This makes the trapezoid isosceles, which simplifies the solution. The height of the trapezoid is now

h = AG = Ö(AB² - BG²)

or

h = DH = Ö(CD² - CH²)

Since the trapezoid is now isosceles,

BG = CH = (BC - AD)/2 = (6 - 2)/2 = 2

and the height is

h = Ö(3² - 2²) = Ö5

The area then really is

A = (BC + AD)/2 · h = (6 + 2)/2 · Ö5 = 4Ö5

but it is solution of a different problem. Indeed, this probably identifies the typos in the original problem as given in the San Francisco Bay Area Math Bowl competition.
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Now, suppose that we have to find the area of a trapezoid with the sides of length 2, 3, 3, and 5, where the sides of length 3 and 5 are parallel. The little catch is that now the trapezoid has an obtuse angle at its longer base:

but the solution presented by Brian still works. Height of the trapezoid is

h = AG = Ö(AB² - BG²)

and also

h = DH = Ö(CD² - CH²)

Hence

AB² - BG² = CD² - CH²
CH² - BG² = CD² - AB² = 3² - 2² = 9 - 4 = 5

On the other hand,

CH - BG = BC - AD = 5 - 3 = 2

We have

CH² - BG² = (CH + BG)·(CH - BG) = 5
CH - BG = 2

Therefore

CH + BG = 5/2
CH - BG = 2

Solving for CH and BG by adding and subtracting the 2 equations:

2·CH = 5/2 + 2 = 9/2
2·BG = 5/2 - 2 = 1/2

CH = 9/4
BG = 1/4

Calculating the trapezoid height:

h = Ö(AB² - BG²) = Ö(2² - (1/4)²) = Ö(4 - 1/16) = Ö{(64 - 1)/16} = Ö(63/16) = 3/4·Ö7

Check by calculating the trapezoid height from the other equation for h:

h = Ö(CD² - CH²) = Ö(3² - (9/4)²) = Ö(9 - 81/16) = Ö{(144 - 81)/16} = Ö(63/16) = 3/4·Ö7

as required. The area of the trapezoid is then

A = (BC + AD)/2 · h = (5 + 3)/2 · 3/4·Ö7 = 3Ö7
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Thank you all.

Regards, Vladimir