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22 Sep 2020, 06:10
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A tank with an isosceles-triangular cross section with sides 4,4,and 6 meters, has a length of 10 meters, and is currently resting on the 6 meter by 10 meter face. The tank is filled with water to the depth of 1 meter. If the tank is placed on a triangular face, what would be the depth of the water in the tank?
A. \(\frac{5}{7} [\sqrt{7} - 1]\)
B. \(\frac{10}{7} [\sqrt{7} - 1]\)
C. \(\frac{10}{7} [\sqrt{7}]\)
D. \(\frac{10}{7} [2 \sqrt{7} - 1]\)
E. \(\frac{20}{7} [2 \sqrt{7} - 1]\)
Source: Manhattan Review
22 Sep 2020, 10:49
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We can first answer the question: what fraction of the tank is full? To answer that, all we need to work out is what fraction of the triangle at the end is filled in, by the dark grey water, in the diagram (we only need to be concerned about that area because the tank and the water both have the same length of 10 meters).
In the isosceles triangle, taking the side of 6 as the base, the height, using Pythagoras, is √7. Now the horizontal water line in the diagram creates a smaller triangle (roughly the top half of the 4-4-6 isosceles triangle), and that smaller triangle is similar to the larger 4-4-6 isosceles triangle. Again using the horizontal base, the smaller triangle's height is √7 - 1, since the depth of the water is 1.
So, by similarity, the lengths in the smaller triangle are (√7 - 1)/√7 times the lengths in the bigger triangle. If that's the ratio of their lengths, then the square of that, or (√7 - 1)^2 / 7, is the ratio of their areas. So that's the fraction of the area of the 4-4-6 triangle that is unfilled by the water. We'd subtract that from 1 to find what fraction is filled with water. Simplifying,
1 - (√7 - 1)^2/7 = (7/7) - (8 - 2√7)/7 = (2√7 - 1)/7
is the fraction of the isosceles triangle at the end that is filled with water. So if we turn the tank on its end, that's the fraction of the height that will be filled. If the height is actually 10, the fraction of the height that will be filled with water is 10 times the above, so is 10/7 (2√7 - 1), which is answer D.
In the isosceles triangle, taking the side of 6 as the base, the height, using Pythagoras, is √7. Now the horizontal water line in the diagram creates a smaller triangle (roughly the top half of the 4-4-6 isosceles triangle), and that smaller triangle is similar to the larger 4-4-6 isosceles triangle. Again using the horizontal base, the smaller triangle's height is √7 - 1, since the depth of the water is 1.
So, by similarity, the lengths in the smaller triangle are (√7 - 1)/√7 times the lengths in the bigger triangle. If that's the ratio of their lengths, then the square of that, or (√7 - 1)^2 / 7, is the ratio of their areas. So that's the fraction of the area of the 4-4-6 triangle that is unfilled by the water. We'd subtract that from 1 to find what fraction is filled with water. Simplifying,
1 - (√7 - 1)^2/7 = (7/7) - (8 - 2√7)/7 = (2√7 - 1)/7
is the fraction of the isosceles triangle at the end that is filled with water. So if we turn the tank on its end, that's the fraction of the height that will be filled. If the height is actually 10, the fraction of the height that will be filled with water is 10 times the above, so is 10/7 (2√7 - 1), which is answer D.
22 Sep 2020, 13:14
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The Volume of the water filled = Area of the rhombus DECB * Length
Area of the rhombus = \(\frac{1}{2}\) * h * (Sum of Parallel Sides [DE + BC])
\(\triangle\) ADE is \(\sim\) to \(\triangle\) ABC (By AA theorem of similarity)
From This we get \(\frac{DE}{BC} = \frac{AF}{AG}\)
Therefore \(\frac{DE}{6} = \frac{\sqrt{7} - 1}{\sqrt{7}}\)
\({DE} = 6 *( \frac{\sqrt{7} - 1}{\sqrt{7}})\)
Area of the Rhombus = \(\frac{1}{2} * 1 * (6 + 6 *\frac{\sqrt{7} - 1}{\sqrt{7}})\)
Simplyfying we get \(\frac{1}{2} * \frac{(6 \sqrt{7} + 6 \sqrt{7} - 6)}{\sqrt{7}} = \frac{1}{2} * \frac{(2 *6 \sqrt{7}) - 6)}{\sqrt{7}} = \frac{(6 \sqrt{7} - 3)}{\sqrt{7}}\)
Volume of water = Area of Rhombus * Length = \(\frac{(6 \sqrt{7} - 3)}{\sqrt{7}}\) * 10
If the triangle is resting on the base, then the volume of water = Area of triangle * height of water
Area of the triangle = \(\frac{1}{2} * Base * Height\)
Height is found by using Pythagoras Theorem taking Triangle ABG, where \(AG^2 = AB^2 - BG^2 = 4^2 - 3^2 = 16 - 9 = 7\)
AG = \(\sqrt{7}\)
Area of the triangle = \(\frac{1}{2} * 6 * \sqrt{7} = 3 \sqrt{7}\)
Therefore \(3 \sqrt{7} * h = \frac{(6 \sqrt{7} - 3)}{\sqrt{7}} * 10\)
h = \(10 * \frac{(6 \sqrt{7} - 3)}{3 \sqrt{7}*\sqrt{7}}\)
h = \(\frac{10}{7} * (\frac{6\sqrt{7}}{3} - \frac{3}{3})\)
h = \(\frac{10}{7} * (2\sqrt{7} - 1)\)
Option D
Arun Kumar
Area of the rhombus = \(\frac{1}{2}\) * h * (Sum of Parallel Sides [DE + BC])
\(\triangle\) ADE is \(\sim\) to \(\triangle\) ABC (By AA theorem of similarity)
From This we get \(\frac{DE}{BC} = \frac{AF}{AG}\)
Therefore \(\frac{DE}{6} = \frac{\sqrt{7} - 1}{\sqrt{7}}\)
\({DE} = 6 *( \frac{\sqrt{7} - 1}{\sqrt{7}})\)
Area of the Rhombus = \(\frac{1}{2} * 1 * (6 + 6 *\frac{\sqrt{7} - 1}{\sqrt{7}})\)
Simplyfying we get \(\frac{1}{2} * \frac{(6 \sqrt{7} + 6 \sqrt{7} - 6)}{\sqrt{7}} = \frac{1}{2} * \frac{(2 *6 \sqrt{7}) - 6)}{\sqrt{7}} = \frac{(6 \sqrt{7} - 3)}{\sqrt{7}}\)
Volume of water = Area of Rhombus * Length = \(\frac{(6 \sqrt{7} - 3)}{\sqrt{7}}\) * 10
If the triangle is resting on the base, then the volume of water = Area of triangle * height of water
Area of the triangle = \(\frac{1}{2} * Base * Height\)
Height is found by using Pythagoras Theorem taking Triangle ABG, where \(AG^2 = AB^2 - BG^2 = 4^2 - 3^2 = 16 - 9 = 7\)
AG = \(\sqrt{7}\)
Area of the triangle = \(\frac{1}{2} * 6 * \sqrt{7} = 3 \sqrt{7}\)
Therefore \(3 \sqrt{7} * h = \frac{(6 \sqrt{7} - 3)}{\sqrt{7}} * 10\)
h = \(10 * \frac{(6 \sqrt{7} - 3)}{3 \sqrt{7}*\sqrt{7}}\)
h = \(\frac{10}{7} * (\frac{6\sqrt{7}}{3} - \frac{3}{3})\)
h = \(\frac{10}{7} * (2\sqrt{7} - 1)\)
Option D
Arun Kumar
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