# ABCD is an isosceles trapezoid in which AD = BC and AB < CD. The distance from A to lines BC, CD, and BD are 15, 18, and 10, respectively. If K is the area of ABCD, find `sqrt(2)* K` .

This value is 549.

## Expert Answers Refer to the attached image. Denote `OG = OF = b , ` `OC = OD = a , ` angle `DCB = ABE = alpha , ` and angles `BDC = ABD = beta .`

Then the area of the trapezoid is `1 / 2 * 18 * ( 2a + 2b ) = ( a + b ) * 18 . ` Let's find `a ` and `b .`

From the right triangles containing alpha we get `tan beta = 18 / ( a + b ) ` and `sin beta = 10 / ( 2b ) , ` `tan alpha = 18 / ( a - b ) , ` and `sin alpha = 15 / ( 2b ) .`

Use the facts that `cos x = sin x / ( cos x ) ` and `sin^2 x + cos^2 x = 1 ; ` transform slightly and obtain a system of equations for a and b:

`1 + ( ( a + b ) / 18 )^2 = ( (2b)/10)^2 , ` `1 + ( ( a - b ) / 18 )^2 = ( (2b)/15)^2 .`

Subtract the second equation from the first, and it gives

`((2b)/10)^2 - ((2b)/15)^2 = (4ab)/18^2 , ` or `b^2 (1/10^2-1/15^2) = ab/18^2,`

or `b*(9-4)/30^2 = a/18^2, ` or `b/5=a/9. ` Substitute `a = 9/5 b ` into the second equation and find `b = 45 / (4 sqrt(2)), ` so `a = 81/(4 sqrt(2)).`

Finally, the area multiplied by `sqrt(2) ` is equal to 549.

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