GeminiHeat wrote:
Attachment:
mgmat.PNG
In the figure above, ABCD is a square, and the two diagonal lines divide it into three regions of equal area. If AB = 3, what is the length of w, the perpendicular distance between the two diagonal lines?
A. 3√2 – 2√3
B. 3√2 – √6
C. √2
D. 3√2 / 2
E. 2√3 – √6
Let's first calculate the length of diagonal
BDSince BC = DC = 3, we can use the Pythagorean theorem to determine that
BD = 3√2We're told that "the two diagonal lines divide it into three regions of equal area."
The area of square ABCD = (3)(3) = 9
9/3 = 3, which means each region has area 3
So, if we let x = the length of each leg in the
blue RIGHT triangle below...
...then it must be true that: (base)(height)/2 = 3
In other words, (
x)(
x)/2 = 3
Simplify:
x² = 6
Solve:
x = √6We get the following:
Now let's examine the green RIGHT triangle below...
Since this is an isosceles triangle (with two 45 degree angles), we can let each leg have length y
Upon using the Pythagorean theorem to determine the value of y, we get:
y = √3Using the exact same logic, we can conclude that this other green triangle has the following lengths:
Since we already concluded that
BD = 3√2, we can write the following equation:
√3 +
w +
√3 =
3√2Solve for
w to get:
w =
3√2 -
2√3Answer: A