Quiz #4 Geometry
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Test 4, Geometry
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Question 1 of 10
1. Question
The roof of a hangar is part of a cylinder, as shown in the attached figure (left). Find the crosssectional area ABC of the hangar, shown as a shaded in the attached figure (right), with an angle AOC=90^{0} and radius OC=6 m. Give an answer as a number without measurement units.
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Solution: The area of sector ABCD is equal to π × R²/4, since the angle AOC is 900. The area of triangle AOC is R²/2, because triangle AOC is a right triangle and AO=OC=R. Therefore, the crosssectional area of the hangar (segment ABC) is π×R²/4R²/2=R²(π/21)/2≈10.3 m².

Question 2 of 10
2. Question
The crosssection of the garage hangar is part of a parabola. The hangar height h=20 m, while the width of the hangar L=10 m. Define constants A and B for the parabolic (quadratic) function:
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Solution: According to the problem the crosssection of the garage hangar is part of a parabola. Obviously, for X=0, value of Y must be equal to h = 20 m. Therefore, B = 20 m. For X=5 m value of Y must be equal to 0. Therefore, A=4/5. Value A has a unit 1/m. Thus, function Y has the following form:
Answer: A=4/5; B=20.

Question 3 of 10
3. Question
The crosssection of the garage hangar is part of a parabola:
Define the width L and the height h of the hangar
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Solution: The crosssection of the garage hangar is part of a parabola If X=0, Y=h=20 (height of the hangar); if Y=0, X=L/2=5 m (half of the hangar width). Therefore, the height is equal to 20 m and the width is 10 m.
Answer: Height=20 m; width=10 m.

Question 4 of 10
4. Question
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Question 5 of 10
5. Question
From point A, mountain peak C is visible at an angle of 30^{0}. From point B, the mountain peak becomes visible at an angle of 45^{0}. The distance between points A and B is equal to 1,000 m. Find the approximate height of the mountain. Result round to the nearest integer number in meters.
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Solution: The right triangle BCD is isosceles, that is, BD = CD. On the other hand in a right triangle ACD length CD=1/2×AC. Thus, if CD=X, AC=2X, AD=1,000+X or. From this quadratic equation, it can be found that X ≈ 1366 m.
Answer: ≈ 1366 m.

Question 6 of 10
6. Question
From point A, mountain peak C is visible at an angle of 30^{0}. From point B, the mountain peak becomes visible at an angle of 60^{0}. The distance between points A and B is equal to 1,000 m. Find the approximate height of the mountain in meters.
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Solution: Comparing the triangles ACD and BCD and using the Pythagorean Theorem gives
Answer: ≈850 m.

Question 7 of 10
7. Question
From the top of the first mountain AB, the peak of the second mountain DC is seen at an angle of 30^{0}; meanwhile, from the foot of the first mountain, the top of the second mountain is visible at an angle of 60^{0}. Determine the height DC of the second mountain if the height AB is equal to 100 m.
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Solution: Assume that the length EC=X and AD = Y. It can be seen from the figure that BC=2X, AC=2Y, (1); (2). Solving this system of two equations with two unknowns, we obtain X = 50 and Y=. Thus, the height of the second mountain is 150 m and the distance between the first and the second mountain is equal to 85 m.

Question 8 of 10
8. Question
The level of the liquid in the cylindrical container is equal to h = R, where R is the radius of a cylinder’s base circle. The level of the same amount of liquid in the cubeshaped container with a side 2R is equal to X. What is the ratio R/X?
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Solution: The volume of liquid in the cylinder is . The same volume of liquid in a cubic container is , where X is the height of the liquid in the cube. Since V_{1}=V_{2}, we have R/X=4/π.
Answer: R/X = 4/π.

Question 9 of 10
9. Question
The attached figure (left) shows a conical container with water in it. A base radius of the container is R and a height is equal to L. At what height H relative to L will the water level be if all of the water from the conical container is poured into a cylindrical tank (right figure) with a radius of the base equal to R? The answer must be expressed as a ratio H/L.
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Question 10 of 10
10. Question
It is necessary to build a cubeshaped tank with a capacity for 64 liters of water. How many square meters of the material would be required to make such a cube?
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 Solution: One liter of water occupies a volume of 10 cm × 10 cm × 10 cm. Thus, the 64 l correspond to a cube with dimensions of 40 cm × 40 cm × 40 cm. The area of one face of the cube is equal to 1,600 cm^{2}. The total area of the six faces of the cube is equal to 1,600 × 6 = 9,600 cm^{2 }
Answer: 9,600 cm².