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555-605 (Medium)|   Geometry|      
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Bunuel
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In the figure above, ED = 1, CD = 2, and AE =6*3^(1/2). What is the 27 Dec 2015, 10:26
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Note: Figure not drawn to scale.

In the figure above, ED = 1, CD = 2, and \(AE = 6\sqrt{3}\). What is the perimeter of ΔABC ?

A. \(3\sqrt{3}\)
B. \(10+5\sqrt{3}\)
C. \(10\sqrt{3}\)
D. \(15+5\sqrt{3}\)
E. \(25\sqrt{3}\)

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2015-12-27_2122.png [ 15.1 KiB | Viewed 7528 times ]
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D
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peachfuzz
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In the figure above, ED = 1, CD = 2, and AE =6*3^(1/2). What is the 27 Dec 2015, 22:33
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Bunuel

Note: Figure not drawn to scale.

In the figure above, ED = 1, CD = 2, and AE =6\sqrt{3}. What is the perimeter of ΔABC ?

A. \(3\sqrt{3}\)
B. \(10+5\sqrt{3}\)
C. \(10\sqrt{3}\)
D. \(15+5\sqrt{3}\)
E. \(25\sqrt{3}\)

Show Spoiler
Attachment:
2015-12-27_2122.png
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line segment AE = \(6\sqrt{3}\)
AC+\(\sqrt{3}\) = \(6\sqrt{3}\)
AC=\(5\sqrt{3}\)
use property of a 30-60-90 triangle

5+\(5\sqrt{3}\)+10


D. \(15+5\sqrt{3}\)
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gvij2017
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In the figure above, ED = 1, CD = 2, and AE =6*3^(1/2). What is the 14 Nov 2019, 00:15
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AE =6\sqrt{3}

I think above expression is 6/(3)^(1/2)= 2*Sqrt (3)
If it is so, there is no option for perimeter of ABC.
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In the figure above, ED = 1, CD = 2, and AE =6*3^(1/2). What is the 14 Nov 2019, 00:19
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HOW AE =SQUAREROOT 3+AC?
CAN YOU PLZ EXPLAIN ME THIS?
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gvij2017
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In the figure above, ED = 1, CD = 2, and AE =6*3^(1/2). What is the 14 Nov 2019, 01:49
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CDE is right angle triangle.
ED = 1
CD= 2
Then CE = sqrt(3) . This is followed by properties of right angle triangle.
(1, sqrt3, 2) indicates angle 30°-60°-90°

Hope it clarifies your doubt.
Pooja132
HOW AE =SQUAREROOT 3+AC?
CAN YOU PLZ EXPLAIN ME THIS?
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ThatDudeKnows
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In the figure above, ED = 1, CD = 2, and AE =6*3^(1/2). What is the 03 Jul 2022, 08:55
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Bunuel

Note: Figure not drawn to scale.

In the figure above, ED = 1, CD = 2, and \(AE = 6\sqrt{3}\). What is the perimeter of ΔABC ?

A. \(3\sqrt{3}\)
B. \(10+5\sqrt{3}\)
C. \(10\sqrt{3}\)
D. \(15+5\sqrt{3}\)
E. \(25\sqrt{3}\)

Show Spoiler
Attachment:
2015-12-27_2122.png
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Let's look at CDE first.
\(1^2+CE^2 = 2^2\)

\(1+CE^2 = 4\)

\(CE^2 = 3\)

\(CE = \sqrt{3}\)

Perimeter of CDE = \(1+2+\sqrt{3}\) = \(3+\sqrt{3}\)

AE = \(6\sqrt{3}\) and CE = \(\sqrt{3}\), so AC = \(5\sqrt{3}\).

ABC and CDE are similar triangles. The longer leg of ABC is 5x the longer leg of CDE, so the perimeter is 5x.

\(5(3+\sqrt{3}) = 15+5\sqrt{3}\)

Answer choice D.
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