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

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Math Expert
Joined: 02 Sep 2009
Posts: 50621
In the figure shown, CE has length 40 feet, EA has length 20 feet, and  [#permalink]

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22 Mar 2018, 22:53
00:00

Difficulty:

25% (medium)

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 [ 5.75 KiB | Viewed 557 times ]

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Joined: 07 Jan 2016
Posts: 818
Location: India
GMAT 1: 710 Q49 V36
Re: In the figure shown, CE has length 40 feet, EA has length 20 feet, and  [#permalink]

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22 Mar 2018, 23:24
1
Bunuel wrote:

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
Intern
Joined: 11 Jan 2018
Posts: 19
Re: In the figure shown, CE has length 40 feet, EA has length 20 feet, and  [#permalink]

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23 Mar 2018, 00:04
Bunuel wrote:

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
Manager
Joined: 28 Jan 2018
Posts: 53
Location: Netherlands
Concentration: Finance
GMAT 1: 710 Q50 V36
GPA: 3
In the figure shown, CE has length 40 feet, EA has length 20 feet, and  [#permalink]

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23 Mar 2018, 09:02
Bunuel wrote:

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

$$\frac{AE}{AC} = \frac{AD}{AB} = \frac{DE}{BC}$$

$$Angle DAE = BAC, AED = ACB, ADE = ABC$$
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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