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

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 = \(\frac{8 + 2}{2} * 8\) = 40

2. 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 = 10

4. 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 5

Hypotenuse CD = 13. This is a 5-12-13 triangle. Height of ∆ BCD is 12

5. Area of ∆ BCD =\(\frac{10*12}{2}\)= 60

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

Answer D

Area of trapezoid ABDE = (1/2) * Sum of parallel sides * Altitude = (1/2)*(8 + 2)*8 = 40

To get the length of BD, drop an altitude (say DP) from D to AB. DP = AE = 8, PB = 8-2 = 6. So the triangle DPB is right angled with sides 6-8-10 (multiple of 3-4-5 pythagorean triplet)
Hence BD = 10

Area of triangle BCD = (1/2)*altitude*BD
Since BCD is an isosceles triangle, when we drop the altitude say CM on BD, the base BD will be divided into two equal halves of 5 each and we will get two right triangles with sides in the ratio 5-12-13. Altitude of triangle BCD = 12
Area of triangle BCD = (1/2)*12*10 = 60

Area of ABCDE = Area of ABDE + Area of BCD = 40 + 60 = 100

Answer (D)

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