Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If the two regions, an equilateral triangle with side t and [#permalink]
09 Apr 2008, 22:44

If the two regions, an equilateral triangle with side t and square with side s, have the same area, what is the ratio t:s? A. 2:3 B. 16:3 C. 4:sqrt(3) D. 2:sqrt(3)^4 E. 4:sqrt(3)^4

This the way I solved the question:

The area of the triangle can be calculated as height*base*1/2 (30-60-90 >> X : X*sqrt(3) : 2X >> t/2 : t*sqrt(3)/2 : t ) Height = sqrt(3)t/2 Base = t Area = sqrt(3)*t^2/4

The area of square is s^2 sqrt(3)*t^2/4 = s^2 >> s = 3*t/2

The ratio: t/s = t/(3*t/2)

Completely blind here: what did I do wrong?? Please, point my mistake... Thank you

If the two regions, an equilateral triangle with side t and square with side s, have the same area, what is the ratio t:s? A. 2:3 B. 16:3 C. 4:sqrt(3) D. 2:sqrt(3)^4 E. 4:sqrt(3)^4

This the way I solved the question:

The area of the triangle can be calculated as height*base*1/2 (30-60-90 >> X : X*sqrt(3) : 2X >> t/2 : t*sqrt(3)/2 : t ) Height = sqrt(3)t/2 Base = t Area = sqrt(3)*t^2/4

If the two regions, an equilateral triangle with side t and square with side s, have the same area, what is the ratio t:s? A. 2:3 B. 16:3 C. 4:sqrt(3) D. 2:sqrt(3)^4 E. 4:sqrt(3)^4

This the way I solved the question:

The area of the triangle can be calculated as height*base*1/2 (30-60-90 >> X : X*sqrt(3) : 2X >> t/2 : t*sqrt(3)/2 : t ) Height = sqrt(3)t/2 Base = t Area = sqrt(3)*t^2/4

Well, we are given equilateral triangle 60 - 60 - 60 A height bisects the angle of the equilateral triangle. Therefore, when we draw a height we will get 2 similar triangles: 30-60-90 and 30-60-90.

If the two regions, an equilateral triangle with side t and square with side s, have the same area, what is the ratio t:s? A. 2:3 B. 16:3 C. 4:sqrt(3) D. 2:sqrt(3)^4 E. 4:sqrt(3)^4

This the way I solved the question:

The area of the triangle can be calculated as height*base*1/2 (30-60-90 >> X : X*sqrt(3) : 2X >> t/2 : t*sqrt(3)/2 : t ) Height = sqrt(3)t/2 Base = t Area = sqrt(3)*t^2/4

The area of square is s^2 sqrt(3)*t^2/4 = s^2 >> s = 3*t/2

The ratio: t/s = t/(3*t/2)

Completely blind here: what did I do wrong?? Please, point my mistake... Thank you