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Interpreting Geometric Diagrams - Accuplacer Qas

  1. In a diagram, line tt intersects parallel lines ll and mm. An angle formed by lines tt and ll measures (4x−30)°(4x - 30)°. A consecutive interior angle formed by lines tt and mm measures (x+10)°(x + 10)°. What is the measure of the larger of the two angles?
    Answer: 130° When a transversal intersects two parallel lines, consecutive interior angles are supplementary, meaning their sum is 180°. Therefore, (4x−30)+(x+10)=180(4x - 30) + (x + 10) = 180. Combining like terms gives 5x−20=1805x - 20 = 180. Adding 20 to both sides gives 5x=2005x = 200, so x=40x = 40. Substitute x=40x=40 back into the expressions for the angles. The first angle is 4(40)−30=160−30=130°4(40) - 30 = 160 - 30 = 130°. The second angle is 40+10=50°40 + 10 = 50°. The larger of the two angles is 130°.
  2. Two lines in the xy-plane are described. Line A passes through the points (1, 6) and (-2, 0). Line B passes through the points (0, 2) and (4, 0). Which statement best describes the relationship between the two lines?
    Answer: They are perpendicular. First, find the slope of each line using the formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}. The slope of Line A is 0−6−2−1=−6−3=2\frac{0 - 6}{-2 - 1} = \frac{-6}{-3} = 2. The slope of Line B is 0−24−0=−24=−12\frac{0 - 2}{4 - 0} = \frac{-2}{4} = -\frac{1}{2}. Two lines are perpendicular when the product of their slopes equals -1. The product of these slopes is 2×(−12)=−12 \times (-\frac{1}{2}) = -1. Therefore, the lines are perpendicular.
  3. Triangle FGH is similar to triangle JKL. The length of side FG is 9 cm and the length of the corresponding side JK is 12 cm. If the perimeter of triangle FGH is 24 cm, what is the perimeter of triangle JKL?
    Answer: 32 For similar polygons, the ratio of their perimeters is equal to the ratio of their corresponding side lengths. The ratio of the sides is JKFG=129=43\frac{JK}{FG} = \frac{12}{9} = \frac{4}{3}. Let P(FGH) be the perimeter of triangle FGH and P(JKL) be the perimeter of triangle JKL. Then P(JKL)P(FGH)=43\frac{P(JKL)}{P(FGH)} = \frac{4}{3}. Substituting the known perimeter, we get P(JKL)24=43\frac{P(JKL)}{24} = \frac{4}{3}. To solve for P(JKL), multiply both sides by 24: P(JKL)=43×24=4×8=32P(JKL) = \frac{4}{3} \times 24 = 4 \times 8 = 32 cm.
  4. A cylinder has a radius of 4 cm and a height of 10 cm. If the radius is doubled and the height is halved, what is the ratio of the new volume to the original volume?
    Answer: 2:1 The formula for the volume of a cylinder is V=πr2hV = \pi r^2 h. The original volume is Vorig=π(42)(10)=160πV_{orig} = \pi (4^2)(10) = 160\pi. The new radius is 4×2=84 \times 2 = 8 cm, and the new height is 10/2=510 / 2 = 5 cm. The new volume is Vnew=π(82)(5)=π(64)(5)=320πV_{new} = \pi (8^2)(5) = \pi (64)(5) = 320\pi. The ratio of the new volume to the original volume is 320π160π=320160=2\frac{320\pi}{160\pi} = \frac{320}{160} = 2. The ratio is 2 to 1.
  5. A figure consists of a semicircle placed on top of a rectangle. The width of the rectangle is 10 feet and also serves as the diameter of the semicircle. The height of the rectangle is 5 feet. What is the total area of the figure, in square feet?
    Answer: 50+12.5π50 + 12.5\pi The total area is the sum of the area of the rectangle and the area of the semicircle. The area of the rectangle is length ×\times width =10×5=50= 10 \times 5 = 50 square feet. The diameter of the semicircle is 10 feet, so its radius is 10/2=510/2 = 5 feet. The area of a full circle is πr2\pi r^2, so the area of the semicircle is 12πr2=12π(52)=12(25π)=12.5π\frac{1}{2}\pi r^2 = \frac{1}{2}\pi (5^2) = \frac{1}{2}(25\pi) = 12.5\pi square feet. The total area is 50+12.5π50 + 12.5\pi square feet.

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