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In the figure above, Rand Q are points on the x-axis. What is the area

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Re: In the figure above, Rand Q are points on the x-axis. What is the area [#permalink]
1) The coordinates of point P (6, 2 √3) .

as we know the height of the equilateral triangle: 2 √3, we can calculate the hypotenuse (PR), which enable to calculate the area of the equilateral triangle (4/√3 x (PR)^2). Sufficient

2)The coordinates of point Q are (8, 0)

Not able to calculate any of its length of the triangle. Not sufficient.

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Re: In the figure above, Rand Q are points on the x-axis. What is the area [#permalink]
IMO may also be C.

Because, RQ= 4 & PS= 2√3

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Re: In the figure above, Rand Q are points on the x-axis. What is the area [#permalink]
I did it this way. Height of equilateral triangle (h) = (A√3)/2. (A is the side of an triangle)

Plug 2sqrt/3 into H, we can solve for A.

We got the Base. We have both BxH divide by 2.
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Re: In the figure above, Rand Q are points on the x-axis. What is the area [#permalink]
GMATBusters wrote:
In the figure above, R and Q are points on the x-axis. What is the area of equilateral triangle PQR ?

(1) The coordinates of point P are (6, 2 $$\sqrt{3}$$)

Drop a median from P.
Now it become 30-60-90 Triangle and one side opposite to 60 degrees angle is known.
$$1:\sqrt{3}:2$$

Hence, sufficient

(2) The coordinates of point Q are (8, 0)

Definitely not sufficient.

Hence, Option (A)
Re: In the figure above, Rand Q are points on the x-axis. What is the area [#permalink]
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