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Geometry - Triangle Area

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Geometry - Triangle Area

by lola27 » Thu Jan 05, 2012 2:50 am
The (x, y) coordinates of points P and Q are (-2, 9) and (-7, -3), respectively. The height of equilateral triangle XYZ is the same as the length of line segment PQ. What is the area of triangle XYZ?

A. 169*sqrt3 /3
B. 84.5
C. 75*sqrt3
D. 169*sqrt3 /4
E. 225*sqrt3 /4

Can anyone explain how to solve this? Thanks

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by neelgandham » Thu Jan 05, 2012 3:16 am
If
A = Length of the side of the equilateral triangle
H = Height of the equilateral triangle


H = length of line segment PQ = Square root((-2+7)^2 + (9+3)^2) = 13
Area of equilateral triangle = H^2/√3 = 169/√3 = 169*√3/3 Option A

note 1: If (x1,y1) and (x2,y2) are two points in the coordinate system, then the distance between two points would be d = sqrt((x1 - x2)^2 + (y1 - y2)^2).
note 2: H = (√3/2)*A => A = (2/√3)*H. Area of triangle = (1/2)*A*H = (1/2)*(2/√3)*H*H = H^2/√3
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by lola27 » Thu Jan 05, 2012 3:44 am
Thanks neelgandham. i guess i need to learn those too rules. i didn't know the distance between two points formula
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by neelgandham » Thu Jan 05, 2012 4:41 am
Here are some formulae, you might want to learn.

1) A Circle in the coordinate system with center (a,b) and radius r can be represented by the equation
(x-a)^2 + (y-b)^2 = (r^2)

2) Coordinates of the midpoint (X,Y) of the line segment PQ, where point P = (x1,y1) and point Q =(x2,y2), are X =(x1+x2)/2 and Y =(y1+y2)/2.

3) Slope m, is a measure of the steepness of the line, is denoted by (y2-y1)/(x2-x1), where (x1,y1) and (x2,y2) are points on the line.

4) y = mx+c is the point intercept form of the line equation, Where m is the slope of the line; c is the y-intercept of the line

5) Equation of a vertical line, x=a (where a is a constant)

6) Equation of a vertical line, y=a (where a is a constant)
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by GMATGuruNY » Thu Jan 05, 2012 6:03 am
lola27 wrote:The (x, y) coordinates of points P and Q are (-2, 9) and (-7, -3), respectively. The height of equilateral triangle XYZ is the same as the length of line segment PQ. What is the area of triangle XYZ?

A. 169*sqrt3 /3
B. 84.5
C. 75*sqrt3
D. 169*sqrt3 /4
E. 225*sqrt3 /4

Can anyone explain how to solve this? Thanks

[size=Normal][/size]
To determine the distance between 2 points, make the distance the hypotenuse of a right triangle.
Then apply the pythagorean formula or -- better yet -- look for a special triangle:
Image
Since PQ is the hypotenuse of a 5-12-13 triangle, PQ=13.

The height of an equilateral triangle creates a 30-60-90 triangle:
Image
The sides of a 30-60-90 triangle are proportioned s : s√3 : 2s.
In ∆WXY, s√3 = 13.
Thus, s = 13/√3 and 2s = 26/√3.
Area = (1/2)bh = (1/2)(26/√3)(13) = 169/√3 = (169√3) / (√3*√3) = (169√3)/3.

The correct answer is A.

Note that the solution above does not require knowledge of any special formulas.
While knowing formulas can be be helpful, very few are needed to solve most GMAT problems.
Two important take-aways:

1. DRAW your own figures.
2. LOOK for special triangles.

These two strategies -- all by themselves -- are sufficient to solve many GMAT problems.
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