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# M22-06

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Math Expert
Joined: 02 Sep 2009
Posts: 58404

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16 Sep 2014, 01:16
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Difficulty:

5% (low)

Question Stats:

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

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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. $$4\sqrt{3}$$

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Joined: 02 Sep 2009
Posts: 58404

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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 $$= \frac{BC*AB}{2} = \frac{2*2*\sqrt{3}}{2} = 2\sqrt{3}$$.

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

Sent from my [device_name] using [url]GMAT Club Forum mobile app[/url]
Math Expert
Joined: 02 Sep 2009
Posts: 58404

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01 Jan 2018, 08:12
GmatAssasin24 wrote:
How can we consider it as 30-60-90 triangle since there is no clue to assume it?

Sent from my [device_name] using [url]GMAT Club Forum mobile app[/url]

Let me ask you: what is the ration of sides in 30-60-90 triangle? What ratio is given in the question?
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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 $$= \frac{BC*AB}{2} = \frac{2*2*\sqrt{3}}{2} = 2\sqrt{3}$$.

Is it 3BC^2[/m] or BC^2[/m]
Math Expert
Joined: 02 Sep 2009
Posts: 58404

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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 $$= \frac{BC*AB}{2} = \frac{2*2*\sqrt{3}}{2} = 2\sqrt{3}$$.

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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# M22-06

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