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In the figure shown, CE has length 40 feet, EA has length 20 feet, and

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In the figure shown, CE has length 40 feet, EA has length 20 feet, and [#permalink]

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In the figure shown, CE has length 40 feet, EA has length 20 feet, and DE is perpendicular to AC and has length 10 feet. What is the area, in feet^2, of ΔABC?


(A) 225
(B) 450
(C) 900
(D) 1,800
(E) 3,600


[Reveal] Spoiler:
Attachment:
2018-03-23_1049.png
2018-03-23_1049.png [ 5.75 KiB | Viewed 400 times ]
[Reveal] Spoiler: OA

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Re: In the figure shown, CE has length 40 feet, EA has length 20 feet, and [#permalink]

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New post 23 Mar 2018, 00:24
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Bunuel wrote:
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In the figure shown, CE has length 40 feet, EA has length 20 feet, and DE is perpendicular to AC and has length 10 feet. What is the area, in feet^2, of ΔABC?


(A) 225
(B) 450
(C) 900
(D) 1,800
(E) 3,600


[Reveal] Spoiler:
Attachment:
2018-03-23_1049.png


not sure but

de=10 ea=20

ac=60

then cb=30

simialr triangles abc and dea

area= 60x30/2 = 900
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Re: In the figure shown, CE has length 40 feet, EA has length 20 feet, and [#permalink]

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New post 23 Mar 2018, 01:04
Bunuel wrote:
Image

In the figure shown, CE has length 40 feet, EA has length 20 feet, and DE is perpendicular to AC and has length 10 feet. What is the area, in feet^2, of ΔABC?


(A) 225
(B) 450
(C) 900
(D) 1,800
(E) 3,600


[Reveal] Spoiler:
Attachment:
2018-03-23_1049.png


EA/CA = DE/ BC

Therefore
Let BC = X
20/ 60 = 10/X
X=30

Area of an triangle = 1/2 * base * height
Area of Triangle ABC = 1/2 * 60 * 30 = 900
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In the figure shown, CE has length 40 feet, EA has length 20 feet, and [#permalink]

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New post 23 Mar 2018, 10:02
Bunuel wrote:
Image

In the figure shown, CE has length 40 feet, EA has length 20 feet, and DE is perpendicular to AC and has length 10 feet. What is the area, in feet^2, of ΔABC?


(A) 225
(B) 450
(C) 900
(D) 1,800
(E) 3,600


[Reveal] Spoiler:
Attachment:
2018-03-23_1049.png


\(\frac{AE}{AC} = \frac{20}{(40+20)} = \frac{1}{3}\)

Triangle ABC and ADE are similar right triangle, Therefore \(\frac{DE}{BC} = \frac{AE}{AC} = \frac{1}{3} => \frac{10}{BC} = \frac{1}{3} => BC = 30\)

Area of Triangle ABC \(= \frac{(AC*BC)}{2} = \frac{(30*60)}{2} = 900 (C)\)

Key takeaway from this problem: If two triangle are similar, the ratio of their corresponding measurements will be identical and all corresponding internal angle will be identical

\(\frac{AE}{AC} = \frac{AD}{AB} = \frac{DE}{BC}\)

\(Angle DAE = BAC, AED = ACB, ADE = ABC\)
In the figure shown, CE has length 40 feet, EA has length 20 feet, and   [#permalink] 23 Mar 2018, 10:02
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In the figure shown, CE has length 40 feet, EA has length 20 feet, and

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