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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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New post 22 Mar 2018, 22:53
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Question Stats:

85% (02:20) correct 15% (02:44) wrong based on 38 sessions

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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


Attachment:
2018-03-23_1049.png
2018-03-23_1049.png [ 5.75 KiB | Viewed 557 times ]

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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 22 Mar 2018, 23:24
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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


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, 00:04
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


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, 09:02
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


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

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

\(Angle DAE = BAC, AED = ACB, ADE = ABC\)
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In the figure shown, CE has length 40 feet, EA has length 20 feet, and &nbs [#permalink] 23 Mar 2018, 09:02
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In the figure shown, CE has length 40 feet, EA has length 20 feet, and

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