ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Perimeter And Area
20 questions · exam conditions
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Perimeter And AreaQuestion 1 of 20

The area of a rectangle is 96 square centimeters. If its width is 8 centimeters, what is its perimeter in centimeters?

12
20
40
48
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ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz

ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Perimeter And Area

Practice Perimeter And Area in ACCUPLACER Quantitative Reasoning, Algebra & Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Perimeter And Area, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Quantitative Reasoning, Algebra & Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The area of a rectangle is 96 square centimeters. If its width is 8 centimeters, what is its perimeter in centimeters?

  1. 12
  2. 20
  3. 40 (correct answer)
  4. 48
Explanation: The formula for the area of a rectangle is A=l×wA = l \times w. Given A=96A = 96 and w=8w = 8, we can find the length: 96=l×896 = l \times 8, so l=96÷8=12l = 96 \div 8 = 12 cm. The formula for the perimeter is P=2(l+w)P = 2(l + w). Substituting the length and width, we get P=2(12+8)=2(20)=40P = 2(12 + 8) = 2(20) = 40 cm.

Question 2

A square has an area of 64 square inches. A rectangle has the same perimeter as the square. If the length of the rectangle is 10 inches, what is its width in inches?

  1. 6 (correct answer)
  2. 8
  3. 16
  4. 22
Explanation: First, find the side length of the square: 64=8\sqrt{64} = 8 inches. The perimeter of the square is 4×8=324 \times 8 = 32 inches. The rectangle has the same perimeter, 32 inches. The formula for the perimeter of a rectangle is P=2(l+w)P = 2(l + w). We have 32=2(10+w)32 = 2(10 + w). Divide by 2: 16=10+w16 = 10 + w. Solving for ww gives w=16−10=6w = 16 - 10 = 6 inches.

Question 3

A triangular garden plot has a side that measures 12 meters, and the height corresponding to this side is 8 meters. The other two sides of the plot measure 10 meters and 14 meters. What is the area of the garden plot in square meters?

  1. 36
  2. 40
  3. 48 (correct answer)
  4. 56
Explanation: The area of a triangle is calculated by the formula A=12×base×heightA = \frac{1}{2} \times \text{base} \times \text{height}. The problem provides a base of 12 meters and its corresponding height of 8 meters. The lengths of the other two sides (10 and 14 meters) are extra information not needed to find the area. The area is 12×12×8=6×8=48\frac{1}{2} \times 12 \times 8 = 6 \times 8 = 48 square meters.

Question 4

A rectangular garden is 25 meters long and 10 meters wide. Fencing for the garden costs $12 per meter. A 2-meter wide gate will be installed, which does not require fencing material. What is the total cost for the fencing material required?

  1. $792
  2. $816 (correct answer)
  3. $840
  4. $3,000
Explanation: First, calculate the perimeter of the garden: P=2(25+10)=2(35)=70P = 2(25 + 10) = 2(35) = 70 meters. The total length of material needed is the perimeter minus the width of the gate: 70−2=6870 - 2 = 68 meters. The total cost is the length of the fencing material multiplied by the cost per meter: 68 \text{ m} \times \12/\text{m} = $816$.

Question 5

A large rectangular room measures 20 feet by 15 feet. A rectangular section measuring 8 feet by 10 feet is removed from one of the corners. What is the area, in square feet, of the remaining L-shaped room?

  1. 80
  2. 220 (correct answer)
  3. 300
  4. 380
Explanation: First, calculate the area of the original large rectangle: 20 ft×15 ft=30020 \text{ ft} \times 15 \text{ ft} = 300 square feet. Next, calculate the area of the section that was removed: 8 ft×10 ft=808 \text{ ft} \times 10 \text{ ft} = 80 square feet. To find the area of the remaining L-shaped room, subtract the removed area from the total area: 300−80=220300 - 80 = 220 square feet.

Question 6

A piece of land is in the shape of a trapezoid. Its two parallel sides measure 60 feet and 100 feet. The height, which is the perpendicular distance between the parallel sides, is 50 feet. What is the area of the land in square feet?

  1. 80
  2. 4,000 (correct answer)
  3. 6,000
  4. 8,000
Explanation: The formula for the area of a trapezoid is A=12(b1+b2)hA = \frac{1}{2}(b_1 + b_2)h, where b1b_1 and b2b_2 are the lengths of the parallel sides and hh is the height. Substituting the given values: A=12(60+100)×50=12(160)×50=80×50=4,000A = \frac{1}{2}(60 + 100) \times 50 = \frac{1}{2}(160) \times 50 = 80 \times 50 = 4,000 square feet.

Question 7

A window is shaped like a rectangle with a semicircle on top. The rectangular part is 4 feet wide and 5 feet high. The diameter of the semicircle is the 4-foot width of the rectangle. What is the approximate perimeter of the entire window?

  1. 14+2π14 + 2\pi feet (correct answer)
  2. 14+4π14 + 4\pi feet
  3. 18+2π18 + 2\pi feet
  4. 18+4π18 + 4\pi feet
Explanation: The perimeter of the window consists of three sides of the rectangle and the curved part of the semicircle. The sides are the two 5-foot heights and the 4-foot bottom width, for a total of 5+5+4=145 + 5 + 4 = 14 feet. The top of the rectangle is not part of the window's perimeter. The diameter of the semicircle is 4 feet, so its radius is 2 feet. The circumference of a full circle is C=2πrC = 2\pi r, so the length of the semicircle arc is 12(2πr)=πr=2π\frac{1}{2}(2\pi r) = \pi r = 2\pi feet. The total perimeter is the sum of the straight sides and the arc: 14+2π14 + 2\pi feet.

Question 8

The perimeter of a rectangle is 50 feet. The length of the rectangle is 5 feet greater than its width. What is the area of the rectangle in square feet?

  1. 125
  2. 150 (correct answer)
  3. 154
  4. 156
Explanation: Let ww be the width and ll be the length. We are given l=w+5l = w + 5. The perimeter formula is P=2(l+w)P = 2(l + w). Substituting the given values, we get 50=2((w+5)+w)50 = 2((w + 5) + w), which simplifies to 50=2(2w+5)50 = 2(2w + 5), then 25=2w+525 = 2w + 5. Solving for ww gives 2w=202w = 20, so w=10w = 10 feet. The length is l=10+5=15l = 10 + 5 = 15 feet. The area is A=l×w=15×10=150A = l \times w = 15 \times 10 = 150 square feet.

Question 9

The circumference of a circular pond is 50π50\pi feet. What is the area of the pond in square feet?

  1. 25π25\pi
  2. 50π50\pi
  3. 2500π2500\pi
  4. 625π625\pi (correct answer)
Explanation: The formula for the circumference of a circle is C=2πrC = 2\pi r. We are given C=50πC = 50\pi, so 50π=2πr50\pi = 2\pi r. Dividing both sides by 2π2\pi gives the radius, r=25r = 25 feet. The formula for the area of a circle is A=πr2A = \pi r^2. Substituting the radius, we get A=π(25)2=625πA = \pi (25)^2 = 625\pi square feet.

Question 10

A rectangular garden has a length that is 3 feet more than twice its width. If the perimeter of the garden is 42 feet, what is the area of the garden in square feet?

  1. 108 square feet (correct answer)
  2. 126 square feet
  3. 144 square feet
  4. 162 square feet
Explanation: Let w = width. Then length = 2w + 3. Perimeter = 2(w) + 2(2w + 3) = 2w + 4w + 6 = 6w + 6 = 42. Solving: 6w = 36, so w = 6 feet. Length = 2(6) + 3 = 15 feet. Area = 6 × 15 = 108 square feet. Choice B results from using the wrong perimeter formula. Choice C comes from incorrectly setting up the length relationship. Choice D results from calculation errors in the algebra.

Question 11

A circular track has a radius of 50 meters. A rectangular warm-up area is constructed inside the track such that the rectangle's length is 60 meters and its width is 40 meters. What is the difference between the area of the circular track and the area of the rectangular warm-up area? (Use π≈3.14\pi \approx 3.14)

  1. 3,250 square meters
  2. 5,450 square meters (correct answer)
  3. 7,850 square meters
  4. 10,250 square meters
Explanation: Area of circular track = πr² = 3.14 × 50² = 3.14 × 2,500 = 7,850 m². Area of rectangular area = 60 × 40 = 2,400 m². Difference = 7,850 - 2,400 = 5,450 m². Choice A results from calculation errors with the circle area. Choice C gives only the circular area without subtracting. Choice D results from adding the areas instead of finding the difference.

Question 12

The length of a rectangle is 7 more than its width, ww. If the perimeter of the rectangle is 46, what is its area?

  1. 120 (correct answer)
  2. 132
  3. 154
  4. 176
Explanation: Let the width be ww and the length be w+7w+7. The perimeter is given by P=2(l+w)P = 2(l+w), so 46=2((w+7)+w)46 = 2((w+7)+w). Simplifying gives 46=2(2w+7)46 = 2(2w+7), which leads to 23=2w+723 = 2w+7. Solving for ww gives 2w=162w = 16, so the width w=8w=8. The length is 8+7=158+7=15. The area is A=l×w=15×8=120A = l \times w = 15 \times 8 = 120.

Question 13

The length of each side of a rectangular field is tripled. The area of the new, larger field is how many times the area of the original field?

  1. 3
  2. 6
  3. 9 (correct answer)
  4. 12
Explanation: Let the original length and width be ll and ww, respectively. The original area is Aoriginal=l×wA_{original} = l \times w. The new length is 3l3l and the new width is 3w3w. The new area is Anew=(3l)×(3w)=9(l×w)=9×AoriginalA_{new} = (3l) \times (3w) = 9(l \times w) = 9 \times A_{original}. Therefore, the new area is 9 times the original area.

Question 14

A rectangular painting measures 10 inches by 16 inches. It is placed in a wooden frame that has a uniform width of 2 inches. What is the area of the frame only, in square inches?

  1. 56
  2. 104
  3. 120 (correct answer)
  4. 280
Explanation: First, find the area of the painting itself: 10×16=16010 \times 16 = 160 square inches. The frame adds 2 inches to each side, so the total width becomes 10+2+2=1410 + 2 + 2 = 14 inches, and the total length becomes 16+2+2=2016 + 2 + 2 = 20 inches. The total area of the painting plus the frame is 14×20=28014 \times 20 = 280 square inches. To find the area of the frame only, subtract the painting's area from the total area: 280−160=120280 - 160 = 120 square inches.

Question 15

A circular garden has a radius of 10 meters. A path of uniform width is to be built around it. If the area of the path is 44π44\pi square meters, what is the width of the path in meters?

  1. 2 (correct answer)
  2. 4
  3. 11
  4. 12
Explanation: Let ww be the width of the path. The radius of the garden is rinner=10r_{inner} = 10. The radius of the garden plus the path is router=10+wr_{outer} = 10+w. The area of the path is the difference between the outer area and the inner area: Apath=π(router)2−π(rinner)2A_{path} = \pi(r_{outer})^2 - \pi(r_{inner})^2. We are given Apath=44πA_{path} = 44\pi, so 44π=π(10+w)2−π(10)244\pi = \pi(10+w)^2 - \pi(10)^2. Divide by π\pi: 44=(10+w)2−10044 = (10+w)^2 - 100. Add 100 to both sides: 144=(10+w)2144 = (10+w)^2. Take the square root: 12=10+w12 = 10+w. Solving for ww gives w=2w=2 meters.

Question 16

The area of a circle is 36π36\pi square units. What is the circumference of the circle?

  1. 6π6\pi
  2. 9π9\pi
  3. 18π18\pi
  4. 12π12\pi (correct answer)
Explanation: The formula for the area of a circle is A=πr2A = \pi r^2. Given the area is 36π36\pi, we have 36π=πr236\pi = \pi r^2. Dividing by π\pi gives r2=36r^2 = 36, so the radius r=6r = 6 units. The formula for the circumference is C=2πrC = 2\pi r. Substituting the radius, we get C=2π(6)=12πC = 2\pi(6) = 12\pi units.

Question 17

A parallelogram has a base of 15 cm and a corresponding height of 6 cm. The length of the adjacent, non-parallel side is 8 cm. What is the area of the parallelogram in square centimeters?

  1. 46
  2. 48
  3. 90 (correct answer)
  4. 120
Explanation: The area of a parallelogram is given by the formula A=base×heightA = \text{base} \times \text{height}. The length of the adjacent side (8 cm) is extra information not needed to calculate the area. Using the given base and height, the area is 15 cm×6 cm=9015 \text{ cm} \times 6 \text{ cm} = 90 square centimeters.

Question 18

A rectangular poster has a length of 4124\frac{1}{2} feet and a width of 2132\frac{1}{3} feet. What is the area of the poster in square feet?

  1. 6566\frac{5}{6}
  2. 8168\frac{1}{6}
  3. 132313\frac{2}{3}
  4. 101210\frac{1}{2} (correct answer)
Explanation: To find the area, multiply the length by the width. First, convert the mixed numbers to improper fractions: 412=924\frac{1}{2} = \frac{9}{2} and 213=732\frac{1}{3} = \frac{7}{3}. Now multiply the fractions: A=92×73=636A = \frac{9}{2} \times \frac{7}{3} = \frac{63}{6}. Simplify the fraction by dividing the numerator and denominator by 3: 212\frac{21}{2}. Convert back to a mixed number: 101210\frac{1}{2} square feet.

Question 19

The area of a square plot of land is 81 square meters. A second square plot has a perimeter that is double the perimeter of the first plot. What is the area of the second square plot in square meters?

  1. 162
  2. 324 (correct answer)
  3. 648
  4. 1,296
Explanation: First, find the dimensions of the first square. If the area is 81, the side length is 81=9\sqrt{81} = 9 meters. The perimeter of the first square is 4×9=364 \times 9 = 36 meters. The perimeter of the second square is double this, so 2×36=722 \times 36 = 72 meters. The side length of the second square is 72÷4=1872 \div 4 = 18 meters. The area of the second square is 182=32418^2 = 324 square meters.

Question 20

A rectangular swimming pool that is 30 feet long and 15 feet wide is surrounded by a concrete walkway of a uniform width of 3 feet. What is the area of the walkway in square feet?

  1. 144
  2. 270
  3. 306 (correct answer)
  4. 756
Explanation: First, calculate the area of the pool: 30×15=45030 \times 15 = 450 square feet. The walkway adds 3 feet to each side, so the new total length is 30+3+3=3630 + 3 + 3 = 36 feet and the new total width is 15+3+3=2115 + 3 + 3 = 21 feet. The total area of the pool and walkway is 36×21=75636 \times 21 = 756 square feet. To find the area of the walkway, subtract the pool's area from the total area: 756−450=306756 - 450 = 306 square feet.