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Mathematical Reasoning Geometry and Measurement Flashcards

6 cards from real GED practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 6 Mathematical Reasoning Geometry and Measurement flashcards as text

  1. A right circular cone has a slant height of 13 cm and a base radius of 5 cm. What is the total surface area of the cone?

    Answer: 90π cm²

    Total surface area of a cone = πr² + πrl, where r is the base radius and l is the slant height. = π(5)² + π(5)(13) = 25π + 65π = 90π cm².

  2. Two similar triangles have areas of 36 cm² and 100 cm². If the perimeter of the smaller triangle is 24 cm, what is the perimeter of the larger triangle?

    Answer: 40 cm

    The ratio of areas of similar figures equals the square of the ratio of corresponding lengths. Area ratio = 36/100 = 9/25, so the side-length ratio = √(9/25) = 3/5. Perimeter of larger triangle = 24 × (5/3) = 40 cm.

  3. A rectangular prism has a length of 8 m, a width of 5 m, and a height of 3 m. If all three dimensions are doubled, by what factor does the volume increase?

    Answer: 8

    Original volume = 8 × 5 × 3 = 120 m³. New volume = 16 × 10 × 6 = 960 m³. Factor = 960 ÷ 120 = 8. When all three linear dimensions are doubled, volume increases by 2³ = 8.

  4. Point A is located at (−3, 4) and point B is located at (5, −2) on a coordinate plane. What is the midpoint of segment AB, and what is the length of AB?

    Answer: Midpoint (1, 1); Length 10

    Midpoint = ((−3+5)/2, (4+(−2))/2) = (2/2, 2/2) = (1, 1). Length = √[(5−(−3))² + (−2−4)²] = √[8² + (−6)²] = √[64 + 36] = √100 = 10.

  5. A circular swimming pool has a diameter of 14 feet. A square deck is built around the outside of the pool, with each side of the square equal to the diameter of the pool. What is the area of the deck only (the part NOT covered by the pool)?

    Answer: 196 − 49π ft²

    Area of square = 14² = 196 ft². Area of circular pool = π × r² = π × 7² = 49π ft². Deck area = 196 − 49π ≈ 196 − 153.94 ≈ 42.06 ft².

  6. An architect scales a building drawing so that 1.5 cm on the drawing represents 9 meters in real life. On the drawing, two windows are 4.5 cm apart. A doorway is shown as 0.5 cm wide. What is the actual width of the doorway in real life?

    Answer: 3 m

    Scale: 1.5 cm = 9 m, so 1 cm = 6 m. Doorway = 0.5 cm × 6 m/cm = 3 m. The distance between windows (4.5 cm × 6 = 27 m) is extra information included as a distractor.