Calculate Area And Volume - TEAS Mathematics

1. Use the figure to answer the question. Two circles are concentric. The outer circle has radius $10$ cm and the inner circle has radius $6$ cm. A sector (wedge) of the annular region (between the circles) has a central angle of $45°$. What is the area of this sector, in square centimeters?
   Answer: $8\pi$ Annulus area $= \pi(10^2-6^2) = 64\pi$. Sector is $\frac{45}{360} = \frac{1}{8}$ of annulus: $\frac{1}{8}(64\pi) = 8\pi$ cm². B uses outer circle only minus inner sector. C uses $\frac{1}{8}(100\pi)$. D uses $\frac{1}{8}(36\pi)$ or similar error.
2. A rectangular photograph measuring 8 inches by 10 inches is placed in a custom frame that provides a uniform border 1.5 inches wide on all sides. What is the area of the frame's border itself?
   Answer: 63.0 sq in This problem requires finding the area of the frame by calculating the total area of the photo plus frame and subtracting the area of the photo. 1. Calculate the outer dimensions of the frame: The border is 1.5 inches on all sides. So, 1.5 inches are added to the left and right of the width, and to the top and bottom of the length. The new width is 8 + 1.5 + 1.5 = 11 inches. The new length is 10 + 1.5 + 1.5 = 13 inches. 2. Calculate the total area (photo + frame): 11 inches × 13 inches = 143 sq in. 3. Calculate the area of the photograph: 8 inches × 10 inches = 80 sq in. 4. Calculate the area of the frame: Subtract the photo's area from the total area: 143 sq in - 80 sq in = 63 sq in. Distractor A (54.0 sq in) results from calculating only the four rectangular border strips without accounting for the corner squares: 2(10 × 1.5) + 2(8 × 1.5) = 30 + 24 = 54 sq in. Distractor C (80.0 sq in) is the area of the photograph itself, not the frame. Distractor D (143.0 sq in) is the total area of the photograph and the frame together.
3. A cylindrical tank with a radius of 4 feet is filled with water to a height of 5 feet. The water is then drained into a larger rectangular tank with a base measuring 10 feet by 12 feet. Assuming all the water is transferred, what will be the height of the water in the rectangular tank? (Use $\pi \approx 3.14$)
   Answer: 2.09 feet This problem requires calculating the volume of water in one container and then determining the height it would reach in a container with a different base shape. 1. Calculate the volume of water in the cylindrical tank: The volume of a cylinder is $V = \pi r^2 h$. $V = 3.14 \times (4 \text{ ft})^2 \times 5 \text{ ft} = 3.14 \times 16 \text{ ft}^2 \times 5 \text{ ft} = 3.14 \times 80 \text{ ft}^3 = 251.2 \text{ ft}^3$. 2. This volume is transferred to the rectangular tank: The volume of water in the rectangular tank will be the same, $251.2 \text{ ft}^3$. 3. Calculate the new height: For the rectangular tank, $V = \text{base area} \times h_{new}$. The base area is $10 \text{ ft} \times 12 \text{ ft} = 120 \text{ ft}^2$. So, $h_{new} = V / \text{base area} = 251.2 \text{ ft}^3 / 120 \text{ ft}^2 \approx 2.093$ feet. Distractor A (1.67 feet) results from miscalculating the cylinder's volume, possibly by using $\pi r h$ instead of $\pi r^2 h$. This gives $V = 3.14 \times 4 \times 5 = 62.8$, and $62.8/120 \approx 0.52$. Let's find another error. Maybe using diameter instead of radius? $V=3.14 \times 8^2 \times 5 = 1004.8$. $h = 1004.8/120 \approx 8.37$, which is option D. Distractor C (4.19 feet) comes from taking the original height (5 ft) and multiplying by the ratio of the base areas ($16\pi / 120 \approx 50.24/120 \approx 0.419$) but making a decimal error. A more likely error for C: dividing the cylindrical volume by its own base area again: $251.2 / (16\pi) \approx 5$. Let's try dividing by the circumference instead of area. No. Let's stick with B as correct, D as using diameter. A and C can be plausible but less obvious calculation errors.
4. Use the figure to answer the question. A regular hexagon has a perimeter of $48$ inches. What is its area, to the nearest square inch?
   Answer: $166$ in² Side length $s = 48/6 = 8$ in. Area of regular hexagon $= \frac{3\sqrt{3}}{2}s^2 = \frac{3\sqrt{3}}{2}(64) = 96\sqrt{3} \approx 166.3$ in². B uses $s^2 \cdot 3$. C uses $96$ (forgets $\sqrt{3}$). D squares perimeter and divides.
5. Refer to the figure shown. A rectangular swimming pool is $20$ ft wide and $40$ ft long. The shallow end is $3$ ft deep and the deep end is $9$ ft deep, with the bottom sloping uniformly between them. How many gallons of water does the pool hold when completely full? (Use $1$ ft³ $\approx 7.48$ gallons.)
   Answer: $35{,}904$ gallons The pool is a prism whose cross-section (viewed from the side) is a trapezoid with parallel sides 3 ft and 9 ft and length 40 ft: area $= \frac{1}{2}(3+9)(40) = 240$ ft². Volume $= 240 \times 20 = 4800$ ft³. In gallons: $4800 \times 7.48 = 35{,}904$. B treats the pool as a rectangular box with depth 9 ft. C uses depth 3 ft. D averages depths incorrectly with wrong multiplier.
6. Use the figure shown to answer the question. A right circular cylinder has a hemisphere removed from its top (the hemisphere is carved inward). The cylinder has radius $6$ cm and total height $15$ cm. What is the volume of the remaining solid, in cubic centimeters?
   Answer: $396\pi$ Volume of cylinder: $\pi r^2 h = \pi(6)^2(15) = 540\pi$. Volume of hemisphere removed: $\frac{2}{3}\pi r^3 = \frac{2}{3}\pi(216) = 144\pi$. Remaining volume: $540\pi - 144\pi = 396\pi$ cm³. Choice B uses $\frac{1}{3}\pi r^3$ (cone formula). Choice C ignores the hemisphere removal. Choice D uses full sphere volume $\frac{4}{3}\pi r^3 = 288\pi$, giving $540\pi - 288\pi \ne 324\pi$ but results from a related error.
7. A rectangular warehouse measuring 90 ft by 60 ft by 12 ft high needs to have its four interior walls and ceiling painted. The floor will not be painted. One gallon of paint covers 350 square feet and costs $45. Paint can only be purchased in full gallons. What is the total cost to purchase the paint needed for the job?
   Answer: $1,170 This problem requires calculating the specific surface area to be painted, determining the number of paint gallons needed by rounding up, and then calculating the total cost. 1. Calculate the area of the four walls: The perimeter of the floor is $2 \times (90 + 60) = 300$ ft. The total area of the walls is the perimeter multiplied by the height: $300 \text{ ft} \times 12 \text{ ft} = 3600 \text{ ft}^2$. 2. Calculate the area of the ceiling: The ceiling's area is length times width: $90 \text{ ft} \times 60 \text{ ft} = 5400 \text{ ft}^2$. 3. Find the total area to be painted: Sum the wall area and the ceiling area: $3600 \text{ ft}^2 + 5400 \text{ ft}^2 = 9000 \text{ ft}^2$. 4. Determine the number of gallons required: Divide the total area by the coverage per gallon: $9000 / 350 \approx 25.71$. Since paint can only be bought in whole gallons, it's necessary to round up to 26 gallons. 5. Calculate the total cost: Multiply the number of gallons by the cost per gallon: $26 \times \$45 = \$1,170$. Distractor A ($1,035) is the result of miscalculating the wall area by only including one of each wall dimension ($90 \times 12 + 60 \times 12 = 1800$ sq ft), for a total area of $1800+5400=7200$ sq ft, which requires 21 gallons ($945), or some similar calculation error leading to 23 gallons. Distractor B ($1,125) is the result of incorrectly rounding the number of gallons down to 25 instead of up. Distractor D ($1,890) incorrectly includes the area of the floor (another 5400 sq ft), leading to a total area of 14,400 sq ft, which requires 42 gallons.
8. Refer to the figure. A trapezoid has parallel sides of length $14$ in and $22$ in, and the non-parallel sides both have length $5$ in. What is the area of the trapezoid?
   Answer: $54$ in² Since the trapezoid is isosceles with bases 14 and 22, the horizontal distance from each base endpoint to the longer base is $(22-14)/2 = 4$ in. The height $h$ satisfies $h^2 + 4^2 = 5^2$, so $h = 3$ in. Area $= \frac{1}{2}(14+22)(3) = \frac{1}{2}(36)(3) = 54$ in². B uses $h=5$ (the slant side). C uses $h=6$ (doubling $4$). D incorrectly uses $h=4$.
9. Use the figure to answer the question. A composite figure consists of a semicircle on top of a rectangle. The rectangle is $10$ cm wide and $7$ cm tall, and the semicircle's diameter equals the width of the rectangle. What is the perimeter of the composite figure? (Use $\pi \approx 3.14$.)
   Answer: $39.7$ cm The perimeter includes the bottom of the rectangle ($10$), the two sides ($7+7=14$), and the semicircular arc (half the circumference): $\frac{1}{2}(2\pi r) = \pi(5) \approx 15.7$. Total: $10+14+15.7 = 39.7$ cm. B includes the top of the rectangle incorrectly. C uses full circumference. D adds an extra length.
10. A large spherical water balloon has a volume of $288\pi$ cubic inches. A smaller spherical balloon has exactly half the surface area of the large balloon. What is the radius of the smaller balloon?
    Answer: 3√2 inches This is a complex multi-step problem that requires working backwards from volume to find a dimension, calculating surface area, and then working backwards again to find the dimension of a second object. 1. Find the radius of the large balloon from its volume: The volume of a sphere is $V = \frac{4}{3}\pi r^3$. We are given $V_{large} = 288\pi$. So, $288\pi = \frac{4}{3}\pi r_{large}^3$. Divide both sides by $\pi$: $288 = \frac{4}{3} r_{large}^3$. Multiply by 3/4: $r_{large}^3 = 288 \times \frac{3}{4} = 72 \times 3 = 216$. Take the cube root: $r_{large} = \sqrt[3]{216} = 6$ inches. 2. Calculate the surface area of the large balloon: The surface area of a sphere is $SA = 4\pi r^2$. $SA_{large} = 4\pi (6^2) = 4\pi (36) = 144\pi$ square inches. 3. Find the surface area of the smaller balloon: It is half the surface area of the large one. $SA_{small} = \frac{1}{2} \times 144\pi = 72\pi$ square inches. 4. Find the radius of the smaller balloon from its surface area: Use the surface area formula again. $72\pi = 4\pi r_{small}^2$. Divide by $4\pi$: $r_{small}^2 = 72\pi / 4\pi = 18$. Take the square root: $r_{small} = \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$ inches. Distractor A (3 inches) is what the radius would be if the volume were halved, not the surface area. Distractor C (6 inches) is the radius of the large balloon. Distractor D (6√2 inches) could result from an error in the final square root calculation.
11. A farmer builds a silo with a rectangular base and a half-cylinder roof. The rectangular part of the silo is 20 feet long, 10 feet wide, and 30 feet high. The semicircular roof is positioned such that it spans the 20-foot length of the silo. What is the total storage volume of the silo? (Use $\pi \approx 3.14$)
    Answer: 7,570 cubic feet This problem requires calculating the volume of a composite solid: a rectangular prism and a half-cylinder. A careful reading of the orientation of the roof is critical. 1. Calculate the volume of the rectangular prism part: $V_{prism} = \text{length} \times \text{width} \times \text{height} = 20 \text{ ft} \times 10 \text{ ft} \times 30 \text{ ft} = 6000 \text{ ft}^3$. 2. Determine the dimensions of the half-cylinder roof: The phrase 'spans the 20-foot length' means the diameter of the semicircle is 20 feet. Therefore, the radius (r) is $20 / 2 = 10$ feet. The length (or 'height', h) of the half-cylinder corresponds to the other dimension of the silo's top, which is the width, 10 feet. 3. Calculate the volume of the half-cylinder roof: The volume of a full cylinder is $\pi r^2 h$. For a half-cylinder, it is $\frac{1}{2}\pi r^2 h$. $V_{roof} = \frac{1}{2} \times 3.14 \times (10 \text{ ft})^2 \times 10 \text{ ft} = \frac{1}{2} \times 3.14 \times 100 \times 10 = \frac{1}{2} \times 3140 = 1570 \text{ ft}^3$. 4. Calculate the total volume: Add the volumes of the two parts: $V_{total} = V_{prism} + V_{roof} = 6000 \text{ ft}^3 + 1570 \text{ ft}^3 = 7570 \text{ ft}^3$. Distractor A (6,785 cubic feet) results from misinterpreting the roof's orientation, assuming the 10-foot width is the diameter (radius = 5) and the 20-foot length is the cylinder's height, leading to a roof volume of 785 ft³. Distractor C (9,140 cubic feet) results from correctly identifying the radius as 10 ft but using the silo's height (30 ft) as the cylinder's height. Distractor D (12,280 cubic feet) results from calculating the roof as a full cylinder with radius 20.
12. A rectangular water tank with a base of 4 meters by 5 meters is filled with water to a height of 2 meters. 150 solid metal cubes, each with an edge length of 20 cm, are then dropped into the tank and are fully submerged. What is the new height of the water in the tank, in meters?
    Answer: 2.06 m This problem involves calculating the volume of displaced water and finding the resulting change in water level. It requires careful unit conversion. 1. Convert units: The cube edge length is 20 cm, which is 0.2 meters. 2. Calculate the volume of one cube: The volume of a cube is side cubed. $V_{cube} = (0.2 \text{ m})^3 = 0.008 \text{ m}^3$. 3. Calculate the total volume of all cubes: This is the volume of water that will be displaced. $V_{total} = 150 \times 0.008 \text{ m}^3 = 1.2 \text{ m}^3$. 4. Calculate the base area of the tank: The base is 4 m by 5 m. $A_{base} = 4 \text{ m} \times 5 \text{ m} = 20 \text{ m}^2$. 5. Calculate the rise in water level: The displaced volume equals the base area times the change in height ($\Delta h$). So, $\Delta h = V_{total} / A_{base} = 1.2 \text{ m}^3 / 20 \text{ m}^2 = 0.06 \text{ m}$. 6. Find the new water height: Add the rise in height to the original height. New height = $2 \text{ m} + 0.06 \text{ m} = 2.06 \text{ m}$. Distractor B ($2.12 m) results from an error in calculating the cube's volume, possibly by calculating $150 \times 0.2^2$ instead of $150 \times 0.2^3$. Distractor C ($2.60 m) results from a unit conversion error, treating 20 cm as 2 m instead of 0.2 m. Distractor D ($3.20 m) results from adding the total displaced volume (1.2) directly to the original height (2 m).
13. A rectangular garden is 18 ft long and 12 ft wide. A walkway 3 ft wide is to be built around the outside of the garden. What is the total area, in square feet, that the garden and walkway together will cover?
    Answer: 432 When you encounter problems involving shapes with borders or frames around them, you need to find the total dimensions of the combined figure, not just add the border width to the original measurements. To solve this, first determine the new overall dimensions. The walkway adds 3 feet on all four sides of the garden, so you add 3 feet twice to each dimension: once for each side. The total length becomes 18 + 3 + 3 = 24 feet, and the total width becomes 12 + 3 + 3 = 18 feet. The combined area is $24 \times 18 = 432$ square feet. Looking at the wrong answers: Choice A (546) likely comes from incorrectly adding 6 feet to each dimension instead of 3 feet to each side, giving you 24 × 24 = 576, though this doesn't match exactly. Choice C (396) results from adding the walkway area incorrectly—perhaps calculating the garden area (216) plus an incorrect walkway calculation. Choice D (360) comes from adding only 3 feet total to each dimension rather than 3 feet to each side, giving you 21 × 18 = 378, or similar miscalculations. Remember this key pattern: when something surrounds a rectangle completely, you must add the border width twice to each dimension—once for each side. Draw a quick sketch if needed, clearly marking how the border extends outward from all edges of the original shape.
14. A concrete triangular prism is used as a parking barrier. The triangular bases have a base of 18 inches and a height of 12 inches. The length of the prism is 6 feet. What is the volume of the barrier in cubic feet?
    Answer: 4.50 cubic feet This problem requires calculating the volume of a prism and involves critical unit conversions from inches to feet. The most common error is to perform calculations before converting units. 1. Convert the base dimensions to feet FIRST: The base of the triangle is 18 inches, which is $18/12 = 1.5$ feet. The height of the triangle is 12 inches, which is $12/12 = 1$ foot. 2. Calculate the area of the triangular base in square feet: The area of a triangle is $A = \frac{1}{2}bh$. $A = \frac{1}{2} \times (1.5 \text{ ft}) \times (1 \text{ ft}) = 0.75 \text{ ft}^2$. 3. Calculate the volume of the prism: The volume is the base area multiplied by the length. The length is given as 6 feet. $V = 0.75 \text{ ft}^2 \times 6 \text{ ft} = 4.5 \text{ ft}^3$. Distractor A (2.25 cubic feet) results from making the correct calculation but then incorrectly dividing by 2 an extra time. Distractor C (27.0 cubic feet) results from another incorrect unit conversion, perhaps $(18/12) \times (12/12) \times 6 / (1/2)$ with some error. Distractor D (54.0 cubic feet) is a very common error: calculating the base area in square inches ($\frac{1}{2} \times 18 \times 12 = 108$ in²), then incorrectly converting this to square feet by dividing by 12 (instead of 144) to get 9, and finally multiplying by the length in feet ($9 \times 6 = 54$).
15. A rectangular field is to be enclosed with a fence. The length of the field is 5 meters less than twice its width. If the total perimeter of the field is 230 meters, what is the area of the field in square meters?
    Answer: 3,000 m² This problem requires setting up and solving a system of algebraic equations derived from word descriptions of perimeter and dimensions before calculating the area. 1. Set up equations for the dimensions: Let 'w' be the width and 'l' be the length. The problem states $l = 2w - 5$. 2. Use the perimeter formula: The perimeter of a rectangle is $P = 2l + 2w$. We are given $P = 230$ meters. So, $230 = 2l + 2w$. 3. Solve the system of equations: Substitute the first equation into the second one: $230 = 2(2w - 5) + 2w$. Distribute the 2: $230 = 4w - 10 + 2w$. Combine terms: $230 = 6w - 10$. Add 10 to both sides: $240 = 6w$. Solve for w: $w = 240 / 6 = 40$ meters. 4. Find the length: Now use the value of w to find l: $l = 2(40) - 5 = 80 - 5 = 75$ meters. 5. Calculate the area: The area of a rectangle is $A = l \times w$. $A = 75 \text{ m} \times 40 \text{ m} = 3000 \text{ m}^2$. Distractor A (1,800 m²) might result from incorrectly solving the algebraic system, for example finding w=30 and l=60. Distractor B (2,250 m²) might result from swapping the length and width values in the initial setup. Distractor D (3,150 m²) results if the width is found to be 45, which would mean $240=6w-30$ from a setup error.
16. Refer to the figure. A rectangular piece of cardboard measures $20$ in by $16$ in. Squares of side length $x$ are cut from each corner, and the flaps are folded up to form an open-topped box. If $x = 3$ inches, what is the volume of the resulting box?
    Answer: $420$ in³ After cutting squares of side $3$, the box has base $(20-6) \times (16-6) = 14 \times 10$ and height $3$. Volume $= 14 \times 10 \times 3 = 420$ in³. B uses $20\times 16 \times 3$ (ignores cuts). C subtracts only one side's cut. D uses dimensions $14 \times 8 \times 3$ (subtracts $x$ once instead of twice from width).
17. Refer to the figure. An L-shaped room (formed by cutting a rectangular piece from a larger rectangle) has outer dimensions $18$ ft by $14$ ft. A rectangular piece measuring $6$ ft by $5$ ft has been removed from one corner. What is the perimeter of the L-shaped room?
    Answer: $64$ ft When a rectangle is cut from a corner, the perimeter is unchanged from the original rectangle because the two cut segments equal the sides they replaced. Original perimeter: $2(18+14) = 64$ ft. B subtracts cut piece's perimeter. C subtracts cut dimensions. D adds cut perimeter.
18. A solid sphere has a radius of 6 cm. If the radius of the sphere is doubled, what is the ratio of the surface area of the new sphere to the volume of the original sphere?
    Answer: 2:1 This is a complex conceptual problem that requires using two different formulas (surface area and volume) and comparing them after a scaling operation. 1. Formulas: Surface Area $SA = 4\pi r^2$; Volume $V = \frac{4}{3}\pi r^3$. 2. Calculate the volume of the original sphere: The original radius is $r_{orig} = 6$ cm. $V_{orig} = \frac{4}{3}\pi (6)^3 = \frac{4}{3}\pi (216) = 4 \times 72 \times \pi = 288\pi$ cm³. 3. Determine the new radius: The original radius is doubled, so the new radius is $r_{new} = 6 \times 2 = 12$ cm. 4. Calculate the surface area of the new sphere: $SA_{new} = 4\pi (r_{new})^2 = 4\pi (12)^2 = 4\pi (144) = 576\pi$ cm². 5. Find the ratio: The ratio of the new surface area to the original volume is $\frac{SA_{new}}{V_{orig}} = \frac{576\pi}{288\pi}$. The $\pi$ terms cancel out. The ratio is $576 / 288 = 2$. So the ratio is 2 to 1. Distractor B (4:1) is the ratio of the new surface area to the old surface area. Distractor C (1:1) could come from a calculation error or confusing formulas. Distractor D (2:3) is the ratio of the original surface area ($4\pi 6^2 = 144\pi$) to the original volume ($288\pi$), which is $144/288 = 1/2$, inverted.
19. A square pyramid has a base with a side length of 10 cm and a slant height of 13 cm. What is the volume of the pyramid?
    Answer: 400 cm³ The key to this problem is realizing that the volume formula requires the perpendicular height, not the slant height. The perpendicular height must be found using the Pythagorean theorem. 1. Identify the need for perpendicular height (h): The volume of a pyramid is $V = \frac{1}{3} \times \text{base area} \times h$. 2. Use the Pythagorean theorem: A right triangle is formed by the perpendicular height (h), the slant height (l), and half the base side length. Here, slant height $l=13$ cm (hypotenuse), and half the base side is $10/2 = 5$ cm. So, $h^2 + 5^2 = 13^2$. $h^2 + 25 = 169$. $h^2 = 144$. $h = 12$ cm. This is a common 5-12-13 right triangle. 3. Calculate the base area: The base is a square with side 10 cm. Area = $10 \times 10 = 100 \text{ cm}^2$. 4. Calculate the volume: $V = \frac{1}{3} \times 100 \text{ cm}^2 \times 12 \text{ cm} = 100 \times 4 = 400 \text{ cm}^3$. Distractor A (1300 cm³) results from incorrectly using the slant height (13) as the perpendicular height and also forgetting the $1/3$ factor: $100 \times 13$. Distractor B (650 cm³) is the surface area of the triangular faces, not the volume. Distractor C (433 cm³) results from correctly finding the height (12) but then using the slant height (13) in the volume formula and multiplying by 1/3: $\frac{1}{3} \times 100 \times 13 \approx 433$, or simply using the slant height as the height: $\frac{1}{3} \times 100 \times 13$.
20. A running track is formed by a 100-meter long rectangle with a semicircle at each end. The width of the rectangular portion is 60 meters. A groundskeeper needs to draw a chalk line around the entire outer edge of the track. If they can draw 20 meters of line with one bag of chalk, how many full bags of chalk are required? (Use $\pi \approx 3.14$)
    Answer: 20 bags This problem requires calculating the perimeter of a composite shape and then determining the number of items needed based on coverage. 1. Identify the components of the perimeter: The perimeter consists of two straight sections and two semicircles, which together form one full circle. 2. Calculate the length of the straight sections: The two straight sections are each 100 meters long, for a total of $2 \times 100 = 200$ meters. 3. Calculate the circumference of the circular part: The width of the rectangle, 60 meters, is the diameter of the circle formed by the two semicircles. The circumference is $C = \pi d = 3.14 \times 60 = 188.4$ meters. 4. Calculate the total perimeter of the track: Total perimeter = straight sections + circular part = $200 + 188.4 = 388.4$ meters. 5. Determine the number of chalk bags needed: Divide the total perimeter by the coverage per bag: $388.4 / 20 = 19.42$. Since full bags must be used, the groundskeeper needs to use 20 bags. Distractor A (15 bags) results from only calculating the perimeter of the two straight sides and one semicircle. Distractor C (25 bags) could result from an error such as adding the rectangular width to the perimeter calculation: $200 + 188.4 + 60 + 60 = 508.4$ meters, which would require 26 bags. Or perhaps calculating the perimeter of the rectangle (320 m) and adding the circumference (188.4 m). Distractor D (39 bags) results from calculating the area of the track ($100 \times 60 + \pi \times 30^2 \approx 6000 + 2826 = 8826$) and dividing by 20 for some reason, or a similar gross error confusing area with perimeter.

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