Points and Height- From MGMT question banks

[javaido]

Hi,

Need help with the following problems from the Geometry question bank.

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/ Sqrt. 3
B. 84.5
C. 75 sqrt. 3
D. 169 sqrt 3 /4
E. 225 sqrt 3/ 4

Thanks!

[ryan.m.doyle]

Hi,

Need help with the following problems from the Geometry question bank.

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/ Sqrt. 3
B. 84.5
C. 75 sqrt. 3
D. 169 sqrt 3 /4
E. 225 sqrt 3/ 4

Thanks!

Distance for a line = sqrt((x1-x2)^2 + (y1 - y2)^2) - if you have trouble with this formula, draw it out and think p theorem a^2 + b^2 = c^2
sqrt ((-2--7)^2 + (9--3)^2)
sqrt(5^2 + 12^2)
sqrt(25+144)
sqrt(169)
=13

Once you know the height of the equilateral, it helps to know the shortcut that the height of an equilateral = (x*sqrt(3)/2) where x is a side - this is easy to remember if you draw a line down the middle of an equilateral. It forms and 30, 60, 90 triangle with sides, 1:sqrt(3):2

So you know 13 = (x*sqrt(3)/2)
solve for x = (26*sqrt(3)/3)

So area of a triangle is 1/2BH, where height is 13 and base is x or (26*sqrt(3)/3)
1/2*13 *26 *sqrt(3) / 3
13 * 13 *sqrt(3) / 3
169 * sqrt (3) / 3
So a?

Is that the OA?

[akhp77]

Height of triangle = h = PQ = 13 as already calculated above

Base of the triangle = 2x

tan60 = h/x

x = 13/sqrt(3)

Area = xh = 13*13 / sqrt(3)

Ans: A

[brian_crusader]

Hi,
PQ = 13 as mentioned.

For any equilateral triangle,
Area = (height)^2 / sqrt(3) OR sqrt(3)*(side)^2 / 4

This is derived from area = (0.5) * base * height.

And for any euilateral triangle
height = ( sqrt(3) / 2 ) * (side)

Property of 30-60-90 triangles, side opposite 60 = ( sqrt(3) / 2 ) * hypotenuse.

Hope it helps..

Cheers,
Brian.

[StaceyKoprince]

Please be sure to read (and follow!) the forum guidelines when posting. This is the GMATPrep folder. Questions from the MGMAT Question Banks belong in the MGMAT Strategy Guide folder.

It looks like your fellow students have possibly already answered your question; if not, and if you'd like an expert to respond, please post your question in the appropriate folder. Thanks!

[sugarray2000]

i have a question about this problem. i understand that at the end, the height of the equilateral triangle is 13. based on this, if the sides of the triangle are 2x, the sides of the triangle are 13, x, and 2x. this is due to the line going straight down the middle of the triangle making the 2x line equal to x. solving this using the pythagorean theorem, i get a completely different answer than is listed. can someone tell me why this does not work?

[ChrisB]

Hi,

I think the error you've made is in how you've thought about the height of the equilateral triangle and how it fits into the sides of a 30 60 90 triangle. To help yourself see what's happened draw an equilateral triangle, then draw a line from the top point of the triangle to the base. The line you'll have drawn is the height of the equilateral triangle as well as the 2nd longest side of the 30 60 90 triangle that is formed by cutting the equilateral triangle in two. This height is 13 (as has been shown before) and we'll call it h.

The sides of the 30 60 90 triangle are in the proportion of 1:3^(1/2):2. To find the length of the base of the 30 60 90 triangle you must divide 13 by 3^(1/2). That is because the side of length 13 is opposite the 60 degree angle and is thus the second longest side. This yields (13/3) * 3^(1/2). We'll call this b from now on.

Recall though that the area of the 30 60 90 triangle is (b*h) / 2, and the area of the equilateral triangle is 2 times this amount or b*h. Because we're looking for the area of the equilateral triangle the final answer is (169/3)*3^(1/2).

Thanks!
Chris