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M22-06

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M22-06  [#permalink]

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New post 16 Sep 2014, 01:16
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A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

86% (01:26) correct 14% (01:38) wrong based on 74 sessions

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Re M22-06  [#permalink]

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New post 16 Sep 2014, 01:16
Official Solution:

In the right triangle \(ABC\) (\(BC \lt AB \lt AC\)) side \(AC\) is twice as long as side \(B\)C. If the length of side \(AB\) is \(2\sqrt{3}\), what is the area of the triangle?

A. \(2\sqrt{2}\)
B. \(2\sqrt{3}\)
C. 4
D. 6
E. \(4\sqrt{3}\)

We know that \((2\sqrt{3})^2 + BC^2 = (2BC)^2\) or \(12 = 3BC^2\) from where \(BC = 2\). The area of the triangle \(= \frac{BC*AB}{2} = \frac{2*2*\sqrt{3}}{2} = 2\sqrt{3}\).

Answer: B
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Re: M22-06  [#permalink]

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New post 01 Jan 2018, 06:06
The right triangle is a 30-60-90 triangle. Side AB is 2*sqrt(3). Since AB is >BC , A is 60. AC is 4 and BC is 2. The area of the triangle is (1/2) * (2) *(2*sqrt(3)). This comes to 2*sqrt(3) , i.e. option B.
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Re: M22-06  [#permalink]

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New post 01 Jan 2018, 07:44
How can we consider it as 30-60-90 triangle since there is no clue to assume it?

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Re: M22-06  [#permalink]

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New post 01 Jan 2018, 08:12
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Re: M22-06  [#permalink]

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New post 11 Sep 2019, 01:15
Bunuel wrote:
Official Solution:

In the right triangle \(ABC\) (\(BC \lt AB \lt AC\)) side \(AC\) is twice as long as side \(B\)C. If the length of side \(AB\) is \(2\sqrt{3}\), what is the area of the triangle?

A. \(2\sqrt{2}\)
B. \(2\sqrt{3}\)
C. 4
D. 6
E. \(4\sqrt{3}\)

We know that \((2\sqrt{3})^2 + BC^2 = (2BC)^2\) or \(12 = 3BC^2\) from where \(BC = 2\). The area of the triangle \(= \frac{BC*AB}{2} = \frac{2*2*\sqrt{3}}{2} = 2\sqrt{3}\).

Answer: B



Is it 3BC^2[/m] or BC^2[/m]
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Re: M22-06  [#permalink]

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New post 11 Sep 2019, 01:30
mangamma wrote:
Bunuel wrote:
Official Solution:

In the right triangle \(ABC\) (\(BC \lt AB \lt AC\)) side \(AC\) is twice as long as side \(B\)C. If the length of side \(AB\) is \(2\sqrt{3}\), what is the area of the triangle?

A. \(2\sqrt{2}\)
B. \(2\sqrt{3}\)
C. 4
D. 6
E. \(4\sqrt{3}\)

We know that \((2\sqrt{3})^2 + BC^2 = (2BC)^2\) or \(12 = 3BC^2\) from where \(BC = 2\). The area of the triangle \(= \frac{BC*AB}{2} = \frac{2*2*\sqrt{3}}{2} = 2\sqrt{3}\).

Answer: B



Is it 3BC^2[/m] or BC^2[/m]


The point is that on the right hand side we have (2BC)^2, which is 4BC^2.

\((2\sqrt{3})^2 + BC^2 = (2BC)^2\);

\(12 + BC^2 = 4BC^2\);

\(12 = 3BC^2\).

Hope it's clear.
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Re: M22-06   [#permalink] 11 Sep 2019, 01:30
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