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GRE Prep Club Team Member
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In the figure above, ABCD is a square, and the two diagonal lines divi
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30 Mar 2021, 06:00
30 Mar 2021, 06:00
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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
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Re: In the figure above, ABCD is a square, and the two diagonal lines divi
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28 Apr 2021, 06:14
28 Apr 2021, 06:14
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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 BD
Since BC = DC = 3, we can use the Pythagorean theorem to determine that BD = 3√2
We'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 = √6
We 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 = √3
Using 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√2
Solve for w to get: w = 3√2 - 2√3
Answer: A
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In the figure above, ABCD is a square, and the two diagonal lines divi
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30 Mar 2021, 09:50
30 Mar 2021, 09:50
3
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
Awesome Question, A devilish one for sure!
Let us name the regions (Left to Right) as \(A_1\), \(A_2\), and \(A_3\)
Area of Square \(= 3^2 = 9\)
So, \(A_1\) = \(A_2\) = \(A_3 = 3\)
Let, sides of triangle be \(x\)
Area of smaller triangles (\(A_1\) and \(A_3\)) = \(\frac{1}{2}x^2 = 3\)
\(x = \sqrt{6}\)
Now, In smaller triangles sides would be \(\sqrt{6} : \sqrt{6} : \sqrt{12}\)
Drop a perpendicular from \(D\) to the hypotenuse of the smaller triangle and name it as \(H\)
\(\frac{1}{2}(\sqrt{12})(H) = 3\)
\(H = \sqrt{3}\)
Finally, \(\frac{w}{2} + H = \frac{1}{2}\) (Diagonal of the square)
\(\frac{w}{2} + \sqrt{3} = \frac{3\sqrt{2}}{2}\)
\(w + 2\sqrt{3} = 3\sqrt{2}\)
\(w = 3\sqrt{2} - 2\sqrt{3}\)
Hence, option A
