gorden wrote:

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In the figure, AB = AE = 8, BC = CD = 13, and DE = 2. What is the area of region ABCDE ?

A. 76

B. 84

C. 92

D. 100

E. 108

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areaofabcde.png [ 38.37 KiB | Viewed 360 times ]
Area of figure ABCDE =

area of trapezoid ABDE +

area of ∆ BCD.

This problem contains two special right triangles with Pythagorean triplets, which makes lengths for area of ∆ BCD easy to find.

1.

Connect B to D and find area of trapezoid ABDE

Area ABDE = \(\frac{b_1 + b_2}{2} * h\)

Given: AB = 8, DE = 2, and AE = 8

Area of trapezoid ABDE = \(\frac{8 + 2}{2} * 8\) =

402. Find area of ∆ BCD -- we need the base and height.

3. Base of ∆ BCD

To find base of ∆ BCD,

draw a line from D to AB (in the diagram,

DX) that is parallel to AE in order to create a right triangle BDX where

DX is one leg of the right triangle

BX is the other leg, and

BD, the base we need, is the hypotenuse

DX = 8

BX = 6 (AX = DE; they are parallel sides of the new rectangle created by the pink line. BX = AB - AX.

BX = 8 - 2 =

6)

This is a 3-4-5 right triangle: 6: 8:

10. Base BD =

104. Height of ∆ BCD

Because ∆ BCD is an isosceles triangle, its altitude will create two congruent right triangles (see second small figure in diagram). Draw an

altitude from C to BD

BD is bisected. Each right triangle has

short leg of length

5Hypotenuse CD = 13. This is a 5-12-13 triangle.

Height of ∆ BCD is

125. Area of ∆ BCD =\(\frac{10*12}{2}\)=

60Add the areas of the trapezoid and the triangle: 40 + 60 = 100

Answer D