Quiz #2 Geometry
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Test 2, Geometry
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Question 1 of 10
1. Question
What is the value of the short side of a rectangular building lot if one side is 6 m greater than the other, and the area is 135 m^{2}?
Please, give an answer as a number without measurement units.
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Solution: Let us say that one side of the equals X, then the other side is X +6. Solve the equation X × (X+ 6)=135 or X^{2}+6X135=0. X_{1}=9, X_{2}=15.
Answer: 9 m.

Question 2 of 10
2. Question
Find the difference between the long and short length of a rectangular piece of land if its perimeter is 360 meters, and its area is equal to 8,000 m^{2}.
Please present the answer as a number without measurement unit.
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Solution: Let us say that X is one side, while the other is Y. Then, X+Y=180, X×Y=8,000. According to the math problem, X and Y are the root of the following quadratic equation Z^{2}180Z+8,000=0. Solve Z_{1}=X=100, Z_{2}=Y=80.
Answer: 20 m.

Question 3 of 10
3. Question
The attached figure shows two triangular property lots ABC and BDC. Lot ABC has the form of an equilateral triangle, while lot BDC is an isosceles triangle with equal sides BD=BC. Determine the ratio of these triangles’ areas.
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Solution: Let us say that BE is a height lowered to the side AC in an equilateral triangle, and BF is a height lowered to the side of the CD in isosceles triangle CBD. Let us assume that the side of an equilateral triangle is equal to L. Then, AE=L/2 and the area of the triangle is equal to . In an equilateral triangle CBD, angle BCD is 30^{0}, so the height BF=L/ 2, the side is , and the area of the isosceles triangle is equal to. Thus, the areas of these two triangles is equal, and their ratio is equal to 1.
Answer: 1.

Question 4 of 10
4. Question
A lot in the shape of a right triangle is divided into two parts by an altitude drawn from the vertex with the right angle to the hypotenuse. Find the ratio of large part of area to the small part if the altitude value is 12 cm, and the perimeter of the original right triangle is equal to 60 cm. Please, express ratio as a fraction, for example 2:1.
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 Solution: Assume that one leg of the right triangle is equal to a, the other is b, and c=the triangle Thus, we have a system of three equations with three unknowns.
Solution of the system: a=15, b=20, c=25. Then the area of one triangle is 120 cm^{2}, and the area of the second one is 90 cm^{2}.
Answer: 4:3.

Question 5 of 10
5. Question
A triangular property lot is bounded by the following lines: y=x/2, y=x2, and the axis Y. Define the area of the resulting triangular lot.
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 Solution: The triangular lot is shown in the attached figure. Points A, B, and C are defined by the coordinates A (0, 2), B (2, 0), and C (4, 2). The area of the triangle OBA is equal to 2, and the area of triangle OCB is equal to the area of the triangle OCD minus area of the triangle BCD or 42=2. The total area is 4.
Answer: 4

Question 6 of 10
6. Question
A property lot is the sum of five identical squares, as shown in the attached figure. How can this be divided into two parts of equal area using one straight line?
Answer options
 AB (KB=BD)
 EF (GF=FH)
 CD (AC=CM)
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Solution: The solution is shown below, where AC = CB.

Question 7 of 10
7. Question
It is necessary to make 6 identical timbers with equal crosssection squares from one original round log. The length of each timber has to be the same as the length of the original round log. Suggest a solution and calculate the ratio of one cross section square length to the radius of the round log. Please, round the answer to the nearest integer.
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Question 8 of 10
8. Question
It is necessary to make 6 identical timbers with equal crosssection squares from one original round log. The length of each timber has to be the same as the length of the original round log. Suggest a solution and calculate the area of one crosssection area if the radius of the round log is equal to 15 cm. Please, present an answer as a number without measurement units.
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Solution: Solution of this problem is identical to the solution of the previous problem.
Answer: ≈68 cm^{2}.

Question 9 of 10
9. Question
ABCD is a parallelogram such that AB is parallel to DC and DA parallel to CB. The length of side AB is 20 cm. E is a point between A and B such that the length of AE is 3 cm. F is a point between points D and C. Find the length of DF such that the segment EF divide the parallelogram in two regions with equal areas. Give an answer as a number without measurement units.
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Question 10 of 10
10. Question
How many boxes with a rectangular parallelepiped shape with dimensions 20×40×80 (cm^{3}) will fit on a truck with dimensions 2.4×8×3 (m^{3})?
Please, give an answer without measurement units.
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Solution: The number of boxes that fit on the truck is equal to 2.4×8×3×100×100×100/20/40/80=900.
Answer: 900 boxes.