In right trapezoid ABCD, segment AB is parallel to segment CD, \angle DAB = 90^\circ, AD =… — GRE General Test

In right trapezoid $ABCD$ , segment $AB$ is parallel to segment $CD$ , $\angle DAB = 90^\circ$ , $AD = 12$ , $CD = 15$ , and $BC = 13$ . Point $E$ lies on segment $CD$ such that quadrilateral $ABCE$ is a parallelogram. What is the perimeter of triangle $ADE$ ?

Answer: 30

Answer

Decomposing right trapezoid

ABCD

by dropping a perpendicular from

B

CD

forms a right triangle with height

12

and hypotenuse

13

. The Pythagorean theorem gives the base of this right triangle as

\sqrt{13^2 - 12^2} = 5

. Subtracting this from

CD = 15

yields

AB = 10

. Because

ABCE

is a parallelogram,

CE = AB = 10

, which leaves

DE = CD - CE = 15 - 10 = 5

. Triangle

ADE

is a right triangle with legs

AD = 12

and

DE = 5

, giving hypotenuse

AE = \sqrt{12^2 + 5^2} = 13

. The perimeter of triangle

ADE

12 + 5 + 13 = 30

Step-by-Step Solution

Calculate the horizontal projection of segment

BC

onto base

CD

The length of the horizontal projection is

\sqrt{13^2 - 12^2} = 5

Segment

AD = 12

defines the perpendicular distance between parallel lines

AB

and

CD

Determine the length of parallel base

AB

AB = 15 - 5 = 10

The total length of base

CD = 15

is the sum of

AB

and the horizontal projection of slant side

BC

Calculate the length of segment

DE

DE = 15 - 10 = 5

Quadrilateral

ABCE

is a parallelogram, which implies

CE = AB = 10

Compute the hypotenuse

AE

and the total perimeter of triangle

ADE

AE = \sqrt{12^2 + 5^2} = 13

, so

\text{Perimeter} = 12 + 5 + 13 = 30

Triangle

ADE

is a right-angled triangle with right angle at vertex

D

Key Concept

Trapezoid height decomposition, parallelogram side properties, and Pythagorean theorem application

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