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Question 1 of 10
1. Question
A television screen is 100 centimeters diagonally, the ratio of the television screen height to the width is ¾. What is the height of the TV screen?
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Question 2 of 10
2. Question
The figure shows a podium to honor athletes. The widths of all three steps have the same length. What is the area of the front side surface? Dimensions are in millimeters.
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Correct answer is 1 m²
Question 3 of 10
3. Question
A building lot has a quadrilateral shape, and its lengths are in the ratio 3:4:5:6. Find the longest side if the perimeter is equal to 540 m.
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Answer: The lengths of the quadrilateral are as follows: 540×3/(3+4+5+6)=90 m; 540×4/18=120 m; 540×5/18=150 m; 540×6/18=180 m.
Question 4 of 10
4. Question
A building lot has a quadrilateral shape, and its angles are in the ratio 3:4:5:6. Find the smallest angle of the quadrilateral.
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The sum of interior angles of a simple (and planar) quadrilateral is equal to 360^{0}. Therefore, the angle values are as follows: 360×3/(3+4+5+6) =60^{0}, 360×4/(3+4+5+6)=80^{0 }, 360×5/(3+4+5+6)=100^{0 }, 360×6/(3+4+5+6) =120^{0}.
Answer: 60°^{.}
Question 5 of 10
5. Question
A right triangle–shaped land ABC is divided by an altitude AD to the hypotenuse BC into two pieces, as shown in the attached figure. Find the perimeters of the triangular lots ABD and ADC if AB=120 m, AC=160 m.
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A hypotenuse BC of the right triangle ABC with legs equal to 120 m and 160 m is equal to 200 m (Pythagorean Theorem). Let us assume that the altitude AD=X and BD=Y. Because triangles ABD and ABC are similar, , or X=AD=96 m, Y=BD=72 m, and DC=200-72=128 m. Thus, the perimeter of triangle ABD is equal to 288 m, and the perimeter of triangle ADC is 384 m.
Answer: The perimeter of the triangle ABD is equal to 288 m and the perimeter of the triangle ADC is 384 m.
Question 6 of 10
6. Question
A right triangle land ABC is divided by altitude AD into two triangles, as shown in the attached figure. Find the ratio of the area S_{1} of triangle ABD to the area S_{2} of the triangle ADC if AB=120 m, AC=160 m.
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Solution: The ratio . According to the results of the previous problem, BD = 72 m, DC = 128 m. Thus, the ratio
S_{1}/S_{2 }= 9/16.
Question 7 of 10
7. Question
The side lengths of a triangle property lot are equal to 20, 21, and 29 m. Find the lot area.
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Solution: Values 20, 21, and 29 are a “Pythagorean Triple” set (a set of positiveintegers a, b, and c that fits the rule a^{2}+b^{2}=c^{2} [20^{2}+21^{2}=29^{2}]). Therefore, the triangular lot is a right triangle and its area is equal to 20×21/2=210 m^{2}.
Answer: 210 m².
Question 8 of 10
8. Question
A building lot has the shape shown in the attached figure. Find the area of the lot if the area of the colored unit is 1 m^{2}.
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Solution: Let us divide the polygon into three right triangles and one rectangle, as shown in the figure below. The area of triangle ABF is 14 m², the area of triangle GCD is 7 m², the area of a triangle HDE is 14 m², and the area of a rectangle FBCG is 56 m². Thus, the total polygon square is 91 m².
Question 9 of 10
9. Question
What is the perimeter of a rectangular building lot with an area of 2,000 m^{2 }if one side is five times longer than the other one? Give the answer in meters.
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Solution: Let us assume that the length of one side of the lot is equal to X. Then, the length of the other side is equal to X / 5. According to the problem, X × X / 5=2,000, or X=100, X/5=20, and the perimeter is 240 m.
Answer: 240 m.
Question 10 of 10
10. Question
What is the short side length of a rectangular lot, if sides are in proportion 3:5, and the area of the rectangular is equal to 135 m^{2}?
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Solution: Let us say one side length is 3X and the other side is equal to 5X. Then, according to the math problem, 3Х×5Х=135 or Х^{2}=9 or Х=3. Thus, 3X=9 and 5X=15.