Exercise 48 Page 565

Let's begin with recalling the $45^{\circ }\text{- }45^{\circ }\text{- }90^{\circ }$ Triangle Theorem. This theorem tells us that in an isosceles right triangle the legs $\ell$ are congruent and the length of the hypotenuse is $\sqrt{2}$ times the length of a leg.

Let's look at the given picture and name the missing vertices and sides. Notice that $FB=EC$ and $FE=BC$ since $BCEF$ is a rectangle. Additionally, since $\overline{AB}$ and $\overline{CD}$ are congruent, the trapezoid $ABCD$ is isosceles. Using this information, we can conclude that $△ABF\cong △DCE.$

Since we are given that $m\angle ABC=135^{\circ }$ and we know that $m\angle FBC$ is a right angle, the measure of $\angle ABF$ is $45^{\circ }.$ This means that both $△ABF$ and $△DCE$ are $45^{\circ }\text{- }45^{\circ }\text{- }90^{\circ }$ triangles. Therefore, $AF$ and $BF$ are congruent and $x=7.$

Next, we recall that in a $45^{\circ }\text{- }45^{\circ }\text{- }90^{\circ }$ triangle the hypotenuse is $\sqrt{2}$ times the length of the leg. This means that $AB$ and $DC$ are $7\sqrt{2}.$

Finally, we are given that $AD=27.$ Using the Segment Addition Postulate, we can write that the length of $\overline{AD}$ is equal to the sum of the lengths of $\overline{AF},\overline{FE}$ and $\overline{ED}.$ After that, we will substitute appropriate values and solve for $y.$ Let's solve the above equation. The value of $y$ is $13.$ Let's add this information to our picture. Now, as we have found all the side lengths of $ABCD,$ we can evaluate its perimeter. The perimeter of trapezoid $ABCD$ is $40+14\sqrt{2}.$