Problem Solving

Rastis wrote:A piece of wire is bent so as to form the boundary of a square with area A. If the wire is then bent into the shape of an equilateral triangle, what will be the area of the triangle thus bounded in terms of A?

a) (√3/64)A^2

b) (√3/4)A

c) (9/16)A

d) (3/4)A

e) (4√3/9)A

SQUARE
Area of square = (side length)²
So we know that A = (side length)²
Solve to get: side length = √A
So, perimeter = 4√A
IMPORTANT: So, the length of the wire is 4√A

EQUILATERAL TRIANGLE
The wire has length 4√A
So, EACH of the 3 sides of the triangle will have length (4√A)/3

USEFUL FORMULA: Area of equilateral triangle = (√3/4)(side)²
So, area = (√3/4)(4√A/3)²
= (√3/4)(16A/9)
= (4A√3)/9
= E

Cheers,
Brent

Hi Rastis,

In your example, you mention using the number 36. This number is a fine choice, but are you using it for the perimeter of the square OR the area of the square?

If you use it as the AREA of the square, then

Area = 36
Each side = 6
Perimeter = 24

So the Equilateral triangle has a perimeter of 24
Each side = 8

The Equilateral triangle can be split into two 30/60/90 right triangles with sides of 4/4(sqrt3)/8 respectively.

The 4(sqrt3) is the height.

If the explanation references ((sqrt3)/2)8 as the height, then that's technically correct here since the "8" divided by "2" is "4"... but I will admit that that's a strange way to present that information.

GMAT assassins aren't born, they're made,
Rich

Hi Jesse,

In the 30/60/90 triangle, since the "8" is across from the 90 degree angle, 8 = 2X (thus X = 4).

GMAT assassins aren't born, they're made,
Rich