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18 Feb 2012, 17:44
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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/√3
B. 84.5
C. 75√3
D. 169√3 /4
E. 225√3 /4
Attachment:
Triangle.PNG [ 2.68 KiB | Viewed 32140 times ]
Attachment:
Triangle.PNG [ 2.68 KiB | Viewed 32140 times ]
Length of PQ = 13 i.e. this is 5:12:13 Right angle triangle.
We are told that this length (13) is equal to the height of the equilateral triangle XYZ. An equilateral triangle can be cut into two 30-60-90 triangles, where the height of the equilateral triangle is equal to the long leg of each 30-60-90 triangle. We know that the height of XYZ is 13 so the long leg of each 30-60-90 triangle is equal to 13. So the sides of the 30:60:90 triangle are x:x\(\sqrt{3}\):2x.
Now here I am stuck. What will be the three sides of a triangle?
18 Feb 2012, 18:08
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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/√3
B. 84.5
C. 75√3
D. 169√3 /4
E. 225√3 /4
The formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).
So the distance between P and Q is \(d=\sqrt{(-2+7)^2+(9+3)^2}=13\). (Else you can find the length of PQ by realizing that PQ is a hypotenuse of 5:12:13 right triangle)
So we know that the height in equilateral triangle XYZ equals to 13: \(height=13\)
Now, since the height of the equilateral triangle divides it into two 30-60-90 right triangles with the ratio of the sides \(1:\sqrt{3}:2\) then height becomes the leg opposite 60 degrees angle and the hypotenuse, which is the side of an equilateral triangle can be found from the ratio: \(\frac{height}{side}=\frac{\sqrt{3}}{2}\) --> \(side=\frac{26}{\sqrt{3}}\).
Next, \(area_{equilateral}=side^2*\frac{\sqrt{3}}{4}=\frac{169}{\sqrt{3}}\). (Else, \(area_{equilateral}=\frac{1}{2}*height*side=\frac{1}{2}*13*\frac{26}{\sqrt{3}}=\frac{169}{\sqrt{3}}\))
Answer: A.
Check this: https://gmatclub.com/forum/math-triangles-87197.html
Triangle.PNG [ 2.68 KiB | Viewed 30170 times ]
A. 169/√3
B. 84.5
C. 75√3
D. 169√3 /4
E. 225√3 /4
The formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).
So the distance between P and Q is \(d=\sqrt{(-2+7)^2+(9+3)^2}=13\). (Else you can find the length of PQ by realizing that PQ is a hypotenuse of 5:12:13 right triangle)
So we know that the height in equilateral triangle XYZ equals to 13: \(height=13\)
Now, since the height of the equilateral triangle divides it into two 30-60-90 right triangles with the ratio of the sides \(1:\sqrt{3}:2\) then height becomes the leg opposite 60 degrees angle and the hypotenuse, which is the side of an equilateral triangle can be found from the ratio: \(\frac{height}{side}=\frac{\sqrt{3}}{2}\) --> \(side=\frac{26}{\sqrt{3}}\).
Next, \(area_{equilateral}=side^2*\frac{\sqrt{3}}{4}=\frac{169}{\sqrt{3}}\). (Else, \(area_{equilateral}=\frac{1}{2}*height*side=\frac{1}{2}*13*\frac{26}{\sqrt{3}}=\frac{169}{\sqrt{3}}\))
Answer: A.
Check this: https://gmatclub.com/forum/math-triangles-87197.html
Attachment:
Triangle.PNG [ 2.68 KiB | Viewed 30170 times ]
General Discussion
18 Feb 2012, 18:19
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This is where I always get confused:
Height becomes the leg opposite 60 degrees angle and the hypotenuse, which is the side of an equilateral triangle can be found from the ratio: \(\frac{height}{side}=\frac{\sqrt{3}}{2}\) --> \(side=\frac{26}{\sqrt{3}}\).
How come we decide that it will be the leg opposite to 60 degrees angle?
I am fine with the rest, but struggles in the above concept.
Answer: A.
Check this: math-triangles-87197.html[/quote]
Height becomes the leg opposite 60 degrees angle and the hypotenuse, which is the side of an equilateral triangle can be found from the ratio: \(\frac{height}{side}=\frac{\sqrt{3}}{2}\) --> \(side=\frac{26}{\sqrt{3}}\).
How come we decide that it will be the leg opposite to 60 degrees angle?
I am fine with the rest, but struggles in the above concept.
Answer: A.
Check this: math-triangles-87197.html[/quote]
