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by bhumika.k.shah » Sat Feb 27, 2010 12:47 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?


169 * sqrt 3 / 3
B.84.5
C.75 * sq rt 3
D.169 * sq rt 3 / 4
E.225 * sq rt 3 / 4

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by sanju09 » Sat Feb 27, 2010 1:25 am
bhumika.k.shah 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?


169 * sqrt 3 / 3
B.84.5
C.75 * sq rt 3
D.169 * sq rt 3 / 4
E.225 * sq rt 3 / 4

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Area of an equilateral triangle of height h = h^2/√3,

and if h = √ {(-2 + 7) ^2 + (9 + 3) ^2} = 13,

then area of the equilateral triangle XYZ = 13^2/√3 = [spoiler]169 √3/3[/spoiler].

[spoiler]A[/spoiler]
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by kstv » Sat Feb 27, 2010 3:41 am
The height is distance between pts P(-2,9) and Q(-7,-3) =√(-2+7)²+(9+3)² = 13

In the triangle the height of the equilateral triangle is 13.
In 30° 60° 90° triangle the base is 13/√3, which is 1/2 of the base of the equilateral triangle
Area = 13*13/√3
IMO A
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