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In the figure, AB = AE = 8, BC = CD = 13, and DE = 2. What is the area [#permalink]
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08 Nov 2008, 17:58
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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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Last edited by Bunuel on 11 Mar 2015, 03:04, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.



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Re: In the figure, AB = AE = 8, BC = CD = 13, and DE = 2. What is the area [#permalink]
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08 Nov 2008, 18:33
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gorden wrote: In the figure ABCDE, angle A = angle E =90 degree, 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 D. 100 Draw a propendicular from D to AB and name it O. DO = 8 and AO= 2 1. area of rectangle AODE = 8x2 = 16 2. area of triangle BOD = 1/2 (8x6) = 24 Also draw a propendicular from C to BD and name it M. BD = sqrt(BO^2 + DO^2) = sqrt (8^2+ 6^2) = 10 BM = DM = 5 and CM = sqrt(BC^2  BO^2) = 12 3. area of triangle BCD = 1/2 (12810) = 60 so total = 16 + 24 + 60 = 100
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Re: In the figure, AB = AE = 8, BC = CD = 13, and DE = 2. What is the area [#permalink]
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10 Mar 2015, 11:37
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Re: In the figure, AB = AE = 8, BC = CD = 13, and DE = 2. What is the area [#permalink]
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23 Mar 2016, 11:49
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Re: In the figure, AB = AE = 8, BC = CD = 13, and DE = 2. What is the area [#permalink]
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04 Apr 2016, 14:36
It took me 6 minutes to solve this one => here is my approach => i got the diagonal as 10 so the area of bigger triangle using herons formula => 60 and the area of the trapezium =40 => total area => 100 hence D
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Re: In the figure, AB = AE = 8, BC = CD = 13, and DE = 2. What is the area [#permalink]
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25 Aug 2017, 06:39
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Re: In the figure, AB = AE = 8, BC = CD = 13, and DE = 2. What is the area [#permalink]
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30 Aug 2017, 08:55
gorden wrote: Attachment: The attachment 20120325234018989_3525.jpg is no longer available 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 Attachment:
areaofabcde.png [ 38.37 KiB  Viewed 563 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 = \(\frac{8 + 2}{2} * 8\) = 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 345 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 51213 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
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Re: In the figure, AB = AE = 8, BC = CD = 13, and DE = 2. What is the area
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