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

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Joined: 08 Nov 2008
Posts: 38
In the figure, AB = AE = 8, BC = CD = 13, and DE = 2. What is the area  [#permalink]

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Updated on: 11 Mar 2015, 03:04
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20120325234018989_3525.jpg [ 5.13 KiB | Viewed 4282 times ]
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

Originally posted by gorden on 08 Nov 2008, 17:58.
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.
SVP
Joined: 29 Aug 2007
Posts: 2349
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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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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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
2
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 582 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 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

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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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19 Oct 2018, 01:41
gorden wrote:
Attachment:
20120325234018989_3525.jpg
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

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

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Karishma
Veritas Prep GMAT Instructor

Re: In the figure, AB = AE = 8, BC = CD = 13, and DE = 2. What is the area   [#permalink] 19 Oct 2018, 01:41
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