PSAT Math Quiz: Area And Volume
20 questions · exam conditions
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Area And VolumeQuestion 1 of 20

A trapezoid has bases of lengths 14 cm14\text{ cm} and 8 cm8\text{ cm}. Its height is 6 cm6\text{ cm}. What is the area of the trapezoid?

44 cm244\text{ cm}^2
66 cm266\text{ cm}^2
72 cm272\text{ cm}^2
132 cm2132\text{ cm}^2
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PSAT Math Quiz

PSAT Math Quiz: Area And Volume

Practice Area And Volume in PSAT Math 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 Area And Volume, giving you a quick way to practice the rules, question types, and explanations that matter most for PSAT Math.

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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.

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Question 1

A trapezoid has bases of lengths 14 cm14\text{ cm} and 8 cm8\text{ cm}. Its height is 6 cm6\text{ cm}. What is the area of the trapezoid?

  1. 44 cm244\text{ cm}^2
  2. 66 cm266\text{ cm}^2 (correct answer)
  3. 72 cm272\text{ cm}^2
  4. 132 cm2132\text{ cm}^2
Explanation: The question asks for the area in square centimeters of a trapezoid with bases 14 cm and 8 cm, and height 6 cm. The formula for the area of a trapezoid is (1/2) × (base1 + base2) × height. Substitute the values: (1/2) × (14 + 8) × 6 = (1/2) × 22 × 6. Calculate 22 × 6 = 132, then (1/2) × 132 = 66 cm², with squared units for area. A common error is using only one base or forgetting the 1/2 factor. When working with trapezoids, emphasize selecting the correct formula and verifying unit consistency.

Question 2

A right triangle has legs of length 9 cm9\text{ cm} and 12 cm12\text{ cm}. A rectangle is formed using the triangle's hypotenuse as the rectangle's length and using a width of 5 cm5\text{ cm}. What is the area of the rectangle?

  1. 75 cm275\text{ cm}^2 (correct answer)
  2. 60 cm260\text{ cm}^2
  3. 45 cm245\text{ cm}^2
  4. 150 cm2150\text{ cm}^2
Explanation: First, we need to find the hypotenuse of the right triangle with legs 9 cm and 12 cm using the Pythagorean theorem: c² = a² + b². So c² = 9² + 12² = 81 + 144 = 225, giving c = 15 cm. The rectangle uses this hypotenuse (15 cm) as its length and has width 5 cm. Rectangle area = length × width = 15 × 5 = 75 cm². A common mistake is using one of the legs instead of the hypotenuse, or forgetting to apply the Pythagorean theorem.

Question 3

A circle is inscribed in a square so that it touches all four sides of the square. The diameter of the circle is 10 cm10\text{ cm}. What is the area of the region inside the square but outside the circle? Use π=3.14\pi=3.14.

  1. 21.5 cm221.5\text{ cm}^2 (correct answer)
  2. 78.5 cm278.5\text{ cm}^2
  3. 100 cm2100\text{ cm}^2
  4. 50 cm250\text{ cm}^2
Explanation: The question asks for the area inside a square but outside an inscribed circle with diameter 10 cm, in square centimeters, using π=3.14. The formulas needed are square area (side²) and circle area (πr²), with side equal to diameter, 10 cm. Square area: 10 × 10 = 100 cm². Circle radius is 5 cm, area: 3.14 × 25 = 78.5 cm². Remaining area: 100 - 78.5 = 21.5 cm². A common error is using radius for square side, leading to 25 cm² square minus smaller area. When finding regions between shapes, calculate each area separately and subtract, verifying units like cm².

Question 4

The solid shown is formed by removing a cone from a hemisphere so that the cone's base coincides with the hemisphere's circular base and the cone's apex touches the top of the hemisphere. If the radius is rr, what fraction of the hemisphere's volume remains?

  1. 13\dfrac{1}{3}
  2. 12\dfrac{1}{2} (correct answer)
  3. 23\dfrac{2}{3}
  4. 34\dfrac{3}{4}
Explanation: Hemisphere volume =23πr3=\frac{2}{3}\pi r^3. Cone has radius rr and height rr, volume =13πr3=\frac{1}{3}\pi r^3. Remaining =23πr3−13πr3=13πr3=\frac{2}{3}\pi r^3-\frac{1}{3}\pi r^3=\frac{1}{3}\pi r^3. Fraction remaining =1/32/3=12=\frac{1/3}{2/3}=\frac{1}{2}. (A) is the remaining absolute volume as a fraction of full sphere (wrong denominator). (C) is cone/hemisphere incorrectly. (D) is an estimation error.

Question 5

In the figure, trapezoid ABCDABCD has AB∥CDAB\parallel CD, with AB=10AB=10, CD=22CD=22, and legs AD=BC=10AD=BC=10. Point EE lies on CDCD such that BE⊥CDBE\perp CD. What is the area of triangle BECBEC?

  1. 2424 (correct answer)
  2. 3030
  3. 3636
  4. 4848
Explanation: Drop perpendiculars from AA and BB to CDCD at points FF and EE. Then DF=EC=(22−10)/2=6DF=EC=(22-10)/2=6. In right triangle BECBEC: BC=10BC=10, EC=6EC=6, so BE=100−36=8BE=\sqrt{100-36}=8. Area =12(6)(8)=24=\frac{1}{2}(6)(8)=24. (B) uses EC=5EC=5 (half of 1010). (C) uses BE=12BE=12. (D) computes 12(6)(16)\frac{1}{2}(6)(16) using total base length.

Question 6

In the figure, ABCDABCD is a square with side length 88. Points EE and FF are midpoints of sides BCBC and CDCD, respectively. What is the area of triangle AEFAEF?

  1. 1616
  2. 2020
  3. 2424 (correct answer)
  4. 3232
Explanation: Place A=(0,8)A=(0,8), B=(8,8)B=(8,8), C=(8,0)C=(8,0), D=(0,0)D=(0,0). Then E=(8,4)E=(8,4), F=(4,0)F=(4,0). Using the shoelace formula: Area =12∣0(4−0)+8(0−8)+4(8−4)∣=12∣0−64+16∣=12(48)=24=\frac{1}{2}|0(4-0)+8(0-8)+4(8-4)|=\frac{1}{2}|0-64+16|=\frac{1}{2}(48)=24. (A) subtracts only two triangles. (B) is an arithmetic error. (D) omits one of the three corner triangles.

Question 7

A right triangular sign has legs of lengths 9 in9\text{ in} and 12 in12\text{ in}. The sign is painted on both sides. What is the total painted area of the sign?

  1. 54 in254\text{ in}^2
  2. 108 in2108\text{ in}^2 (correct answer)
  3. 216 in2216\text{ in}^2
  4. 252 in2252\text{ in}^2
Explanation: The question asks for the total painted area in square inches of a right triangular sign with legs 9 in and 12 in, painted on both sides. The formula for the area of a triangle is (1/2) × base × height, applied to one side and doubled for both. For one side, area = (1/2) × 9 × 12 = (1/2) × 108 = 54 in². For both sides, total = 2 × 54 = 108 in², with units squared for area. A common error is forgetting to account for both sides or using the hypotenuse incorrectly. In problems involving surfaces, confirm if multiple faces are included and handle units appropriately.

Question 8

In the figure, a rectangle with dimensions 12×1812 \times 18 has a semicircle removed from one of its longer sides (diameter along the side). The diameter of the semicircle equals the shorter dimension of the rectangle. What is the perimeter of the resulting shape?

  1. 48+6π48+6\pi (correct answer)
  2. 60+6π60+6\pi
  3. 48+12π48+12\pi
  4. 54+6π54+6\pi
Explanation: Rectangle perimeter would be 2(12+18)=602(12+18)=60. The semicircle is removed from an 1818-length side; its diameter is 1212, so it takes up 1212 units of that side. Remaining flat portion on that side =18−12=6=18-12=6. The two other long sides contributions become: 1818 (opposite side) +6+ 6 (remaining on cut side) +12+12+ 12 + 12 (short sides) =48= 48. Add the semicircle arc =12(2π⋅6)=6π=\frac{1}{2}(2\pi\cdot 6)=6\pi. Total =48+6π=48+6\pi. (B) forgets to remove the diameter. (C) uses full circumference. (D) uses arc of wrong radius.

Question 9

On a coordinate plane, a rectangle has vertices (−2,1)(-2,1), (4,1)(4,1), (4,6)(4,6), and (−2,6)(-2,6). What is the area of the rectangle?

  1. 20 square units
  2. 25 square units
  3. 30 square units (correct answer)
  4. 36 square units
Explanation: The question asks for the area of a rectangle with vertices at (-2,1), (4,1), (4,6), and (-2,6) on a coordinate plane, with the answer in square units. The formula for the area of a rectangle is length × width, determined from the differences in x- and y-coordinates, ensuring units are consistent as unitless here. The width is the change in x: 4 - (-2) = 6 units, and the height is the change in y: 6 - 1 = 5 units. Therefore, area = 6 × 5 = 30 square units. A key error is miscalculating the height as 6 units by subtracting y-coordinates incorrectly, leading to 6 × 6 = 36. Another mistake might be confusing the shape with a non-rectangle and using a different formula. As a test-taking strategy, plot the points mentally to confirm the shape and double-check coordinate differences.

Question 10

A company is building a rectangular room that is 14 ft14\text{ ft} by 11 ft11\text{ ft}. The floor will be covered with square tiles that are 6 in6\text{ in} on each side. If tiles cannot be cut and must cover the floor exactly, how many tiles are needed?

  1. 308 tiles308\text{ tiles}
  2. 462 tiles462\text{ tiles}
  3. 616 tiles616\text{ tiles} (correct answer)
  4. 154 tiles154\text{ tiles}
Explanation: The question asks for the number of 6-inch square tiles needed to cover a 14 ft by 11 ft room exactly without cutting. First, convert feet to inches: 14 ft = 168 in, 11 ft = 132 in, noting no area formula is directly needed but division for tiling. Along the length, 168 / 6 = 28 tiles; along the width, 132 / 6 = 22 tiles. Total tiles = 28 × 22 = 616. A key error is forgetting to convert units, leading to incorrect divisions like 14 / 0.5 = 28 but mismatching width. For tiling problems, ensure dimensions are in the same units as tile size and verify exact fit by checking divisibility.

Question 11

On a coordinate plane, a triangle has vertices A(1,1)A(1,1), B(7,1)B(7,1), and C(1,6)C(1,6). What is the area of triangle ABCABC in square units?

  1. 15 units215\text{ units}^2 (correct answer)
  2. 30 units230\text{ units}^2
  3. 12.5 units212.5\text{ units}^2
  4. 25 units225\text{ units}^2
Explanation: The question asks for the area of triangle ABC with vertices at (1,1), (7,1), and (1,6), in square units. The formula needed is the area of a triangle, (1/2) × base × height. Identify the base as the distance between A and B, which is 7 - 1 = 6 units along the line y=1. The height is the perpendicular distance from C to this base, which is 6 - 1 = 5 units. Thus, area = (1/2) × 6 × 5 = 15 square units. A common error is misidentifying the base or height, such as using vertical distance incorrectly. When working with coordinate geometry, plot points to visualize and select base and height wisely for simple calculations.

Question 12

On the coordinate plane, triangle ABCABC has vertices A(1,1)A(1,1), B(7,1)B(7,1), and C(1,5)C(1,5). What is the area of triangle ABCABC?

  1. 12 square units12\text{ square units} (correct answer)
  2. 24 square units24\text{ square units}
  3. 6 square units6\text{ square units}
  4. 10 square units10\text{ square units}
Explanation: To find the area of triangle ABC with vertices at A(1,1), B(7,1), and C(1,5), we first identify that this is a right triangle since AB is horizontal and AC is vertical. The area formula for a triangle is A = ½ × base × height. The base AB = 7 - 1 = 6 units (horizontal distance), and the height AC = 5 - 1 = 4 units (vertical distance). Therefore, Area = ½ × 6 × 4 = 12 square units. A common error is forgetting the factor of ½ in the triangle area formula, which would give 24 square units.

Question 13

A rectangular room is 18 ft18\text{ ft} long and 12 ft12\text{ ft} wide. A rectangular rug measuring 9 ft9\text{ ft} by 6 ft6\text{ ft} is placed on the floor. What is the area of the floor that is not covered by the rug?

  1. 162 ft2162\text{ ft}^2 (correct answer)
  2. 270 ft2270\text{ ft}^2
  3. 216 ft2216\text{ ft}^2
  4. 54 ft254\text{ ft}^2
Explanation: This problem asks for the area of the floor NOT covered by the rug, requiring us to find the difference between the total floor area and the rug area. The area formula for a rectangle is A = length × width. The floor area is 18 ft × 12 ft = 216 ft², and the rug area is 9 ft × 6 ft = 54 ft². Therefore, the uncovered area is 216 ft² - 54 ft² = 162 ft². A common error is calculating only the rug area (54 ft²) or only the floor area (216 ft²) instead of finding their difference.

Question 14

Refer to the figure. A sector of a circle with radius 10 cm10\text{ cm} and central angle 60∘60^{\circ} is shaded. What is the area of the shaded sector, in square centimeters?

  1. 50π3\dfrac{50\pi}{3} (correct answer)
  2. 25π25\pi
  3. 25π2\dfrac{25\pi}{2}
  4. 100π100\pi
Explanation: Sector area is θ360∘πr2=60∘360∘π(10)2=16⋅100π=50π3\dfrac{\theta}{360^{\circ}}\pi r^{2}=\dfrac{60^{\circ}}{360^{\circ}}\pi(10)^{2}=\dfrac16\cdot100\pi=\dfrac{50\pi}{3}.
B. 25π25\pi uses half the radius (5 cm) erroneously.
C. 25π2\dfrac{25\pi}{2} uses half the correct area of the full circle.
D. 100π100\pi is the entire circle's area, not just the sector.

Question 15

Refer to the figure. A right circular cone has a base radius of 6 ft6\text{ ft} and a height of 8 ft8\text{ ft}. What is the volume of the cone, in cubic feet?

  1. 96π96\pi (correct answer)
  2. 144π144\pi
  3. 72π72\pi
  4. 288π288\pi
Explanation: Volume =13πr2h=13π(6)2(8)=13π(36)(8)=96π=\dfrac13\pi r^{2}h=\dfrac13\pi(6)^{2}(8)=\dfrac13\pi(36)(8)=96\pi.
B uses h=12h=12 instead of 8.
C omits the 1/3 factor's effect incorrectly.
D multiplies without the 1/3 and doubles the height.

Question 16

Refer to the figure. Parallelogram ABCDABCD has side lengths AB=10 cmAB=10\text{ cm} and AD=7 cmAD=7\text{ cm} with an included angle of 30∘30^{\circ} at vertex AA. What is the area of the parallelogram, in square centimeters?

  1. 35 (correct answer)
  2. 70
  3. 17.5
  4. 49
Explanation: Area =absin⁡θ=10×7×sin⁡30∘=70×0.5=35=ab\sin\theta=10\times7\times\sin30^{\circ}=70\times0.5=35.
B multiplies the sides but omits the sine factor.
C halves only one side before multiplying.
D multiplies the shorter side by itself.

Question 17

Refer to the figure. ABCDEFABCDEF is a regular hexagon with side length 6 cm6\text{ cm}. What is the area of the hexagon, in square centimeters?

  1. 54354\sqrt{3} (correct answer)
  2. 36336\sqrt{3}
  3. 18318\sqrt{3}
  4. 1083108\sqrt{3}
Explanation: For a regular hexagon, Area=332s2=332(6)2=332⋅36=543\text{Area}=\dfrac{3\sqrt3}{2}s^{2}=\dfrac{3\sqrt3}{2}(6)^2=\dfrac{3\sqrt3}{2}\cdot36=54\sqrt3.
B. 36336\sqrt3 uses the formula for an equilateral triangle.
C. 18318\sqrt3 halves that incorrect triangle value.
D. 1083108\sqrt3 doubles the correct hexagon area.

Question 18

Refer to the figure. A bowl is shaped like a hemisphere with radius 7 cm7\text{ cm}. What is the volume of the bowl, in cubic centimeters?

  1. 686π3\dfrac{686\pi}{3} (correct answer)
  2. 343π343\pi
  3. 343π3\dfrac{343\pi}{3}
  4. 1372π3\dfrac{1372\pi}{3}
Explanation: Hemisphere volume =23πr3=23π(7)3=23π(343)=686π3=\dfrac{2}{3}\pi r^{3}=\dfrac{2}{3}\pi(7)^{3}=\dfrac{2}{3}\pi(343)=\dfrac{686\pi}{3}.
B is the volume of a full sphere with radius 5, not this hemisphere.
C omits the factor 2.
D is the volume of a full sphere of radius 7.

Question 19

A trapezoid has bases of lengths 14 m14\text{ m} and 8 m8\text{ m}. Its height is 6 m6\text{ m}. What is the area of the trapezoid?

  1. 66 m266\text{ m}^2 (correct answer)
  2. 132 m2132\text{ m}^2
  3. 44 m244\text{ m}^2
  4. 36 m236\text{ m}^2
Explanation: The question asks for the area of a trapezoid with bases 14 m and 8 m, and height 6 m, in square meters. The formula needed is the area of a trapezoid, (1/2) × (sum of bases) × height. Substitute the values: (1/2) × (14 + 8) × 6 = (1/2) × 22 × 6 = 11 × 6 = 66 m². This calculation directly gives the area without further steps. A common error is forgetting the (1/2) factor, resulting in 132 m², or using only one base. For trapezoid areas, always average the bases before multiplying by height, and confirm units like meters are squared for area.

Question 20

Refer to the figure. The right circular cylinder shown has radius 4 in.4\text{ in.} and height 9 in.9\text{ in.} What is the volume of the cylinder, in cubic inches?

  1. 144π144\pi (correct answer)
  2. 72π72\pi
  3. 128π128\pi
  4. 288π288\pi
Explanation: Volume is πr2h=π(4)2(9)=π⋅16⋅9=144π\pi r^{2}h=\pi(4)^{2}(9)=\pi\cdot16\cdot9=144\pi.
B halves the height.
C uses r=4r=4 but multiplies by 8 instead of 9.
D doubles the correct height.