Solution of the Week #574 - Area

I’ve denoted the vertices with letters and also added an extra line between C and D.

As a reminder, we are looking for the area of the quadrilateral CFDE.

If we call CE ‘x’, and note that triangles BEC and BDF and similar, then DF = 17x/32. Because of the right angles we can say that the area is 15(x+(17x/32))/2, which is 735x/64. All we need to do now is to find the value of x.

Since AB and CD are parallel, and AD and CE are parallel, triangles ADB and CED are similar. If CE is x, AD = 17x/15.

Since triangles ADB and BCA are congruent, BC is also 17x/15.

So now we have the right triangle BEC whose legs are 32 and x and whose hypotenuse is 17x/15. From this we can work out that x=60.

Plugging this into to the expression for the area we found earlier, the area is 735 x 60 / 64 = 11025/16, or 689 1/16.