If two regions (one a triangle with equals sides, T, and a square with side,S) have the same area, what is the ratio of T:S?
2:3
16:3
4: sq rt of 3
2: 4th rt of 3
4: 4th rt of 3
Answer is D......
PLEASE TELL ME WHERE I WENT WRONG....
I made both areas equal to 36. Obviously S will be 6 and T works out to be 6 sq rt 2. IS THIS RIGHT? If so, how do I translate this into the the correct answer?
GMAT PREP - RATIO OF T:S
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i think you could solve this problem algebrically than picking the numbers.
the triangle mentioned in the problem is an equvilateral triangle with side T and the area of an eq. triangle is
[(square root of 3)*T^2]/4
and the area of a square with side S = S^2
given that area of eq. triangle = area of square
[(square root of 3)*T^2]/4 = S^2
rearranging the variables,
(T/S)^2 = 4:(square root of 3)
or (T/S) = (2: 4th root of 3)
answer D
the triangle mentioned in the problem is an equvilateral triangle with side T and the area of an eq. triangle is
[(square root of 3)*T^2]/4
and the area of a square with side S = S^2
given that area of eq. triangle = area of square
[(square root of 3)*T^2]/4 = S^2
rearranging the variables,
(T/S)^2 = 4:(square root of 3)
or (T/S) = (2: 4th root of 3)
answer D
The formula for calculating the Area of the equilateral triangle is
root3*side^2/4
sqrt3*(Tri Side^2)/4 = Squ side^2
sqrt3*(Tri Side^2) = 4 * Squ side^2
(Tri Side^2) / Squ side^2 = 4/sqrt3
taking root on both sides
Tri side/Squ side = 2/ 4th rt 3
Hope that helps....
root3*side^2/4
sqrt3*(Tri Side^2)/4 = Squ side^2
sqrt3*(Tri Side^2) = 4 * Squ side^2
(Tri Side^2) / Squ side^2 = 4/sqrt3
taking root on both sides
Tri side/Squ side = 2/ 4th rt 3
Hope that helps....
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It does help. I had no idea there was a standard formula for the area of an equallateral triangle.
I figured it out by working it out. 1/2* T/2 * T sq rt 3. Is that correct?
Thanks guys.
I figured it out by working it out. 1/2* T/2 * T sq rt 3. Is that correct?
Thanks guys.