PSAT Math Quiz: Area And Volume

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.

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.

Question 1

1. $44\text{ cm}^2$
2. $66\text{ cm}^2$ (correct answer)
3. $72\text{ cm}^2$
4. $132\text{ cm}^2$

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

1. $75\text{ cm}^2$ (correct answer)
2. $60\text{ cm}^2$
3. $45\text{ cm}^2$
4. $150\text{ cm}^2$

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

1. $21.5\text{ cm}^2$ (correct answer)
2. $78.5\text{ cm}^2$
3. $100\text{ cm}^2$
4. $50\text{ cm}^2$

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

1. $\dfrac{1}{3}$
2. $\dfrac{1}{2}$ (correct answer)
3. $\dfrac{2}{3}$
4. $\dfrac{3}{4}$

Explanation: Hemisphere volume $=\frac{2}{3}\pi r^3$. Cone has radius $r$ and height $r$, volume $=\frac{1}{3}\pi r^3$. Remaining $=\frac{2}{3}\pi r^3-\frac{1}{3}\pi r^3=\frac{1}{3}\pi r^3$. Fraction remaining $=\frac{1/3}{2/3}=\frac{1}{2}$. (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

1. $24$ (correct answer)
2. $30$
3. $36$
4. $48$

Explanation: Drop perpendiculars from $A$ and $B$ to $CD$ at points $F$ and $E$. Then $DF=EC=(22-10)/2=6$. In right triangle $BEC$: $BC=10$, $EC=6$, so $BE=\sqrt{100-36}=8$. Area $=\frac{1}{2}(6)(8)=24$. (B) uses $EC=5$ (half of $10$). (C) uses $BE=12$. (D) computes $\frac{1}{2}(6)(16)$ using total base length.

Question 6

1. $16$
2. $20$
3. $24$ (correct answer)
4. $32$

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

Question 7

1. $54\text{ in}^2$
2. $108\text{ in}^2$ (correct answer)
3. $216\text{ in}^2$
4. $252\text{ in}^2$

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

1. $48+6\pi$ (correct answer)
2. $60+6\pi$
3. $48+12\pi$
4. $54+6\pi$

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

Question 9

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

1. $308\text{ tiles}$
2. $462\text{ tiles}$
3. $616\text{ tiles}$ (correct answer)
4. $154\text{ tiles}$

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

1. $15\text{ units}^2$ (correct answer)
2. $30\text{ units}^2$
3. $12.5\text{ units}^2$
4. $25\text{ units}^2$

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

1. $12\text{ square units}$ (correct answer)
2. $24\text{ square units}$
3. $6\text{ square units}$
4. $10\text{ square units}$

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

1. $162\text{ ft}^2$ (correct answer)
2. $270\text{ ft}^2$
3. $216\text{ ft}^2$
4. $54\text{ ft}^2$

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

1. $\dfrac{50\pi}{3}$ (correct answer)
2. $25\pi$
3. $\dfrac{25\pi}{2}$
4. $100\pi$

Explanation: Sector area is $\dfrac{\theta}{360^{\circ}}\pi r^{2}=\dfrac{60^{\circ}}{360^{\circ}}\pi(10)^{2}=\dfrac16\cdot100\pi=\dfrac{50\pi}{3}$.
B. $25\pi$ uses half the radius (5 cm) erroneously.
C. $\dfrac{25\pi}{2}$ uses half the correct area of the full circle.
D. $100\pi$ is the entire circle's area, not just the sector.

Question 15

1. $96\pi$ (correct answer)
2. $144\pi$
3. $72\pi$
4. $288\pi$

Explanation: Volume $=\dfrac13\pi r^{2}h=\dfrac13\pi(6)^{2}(8)=\dfrac13\pi(36)(8)=96\pi$.
B uses $h=12$ instead of 8.
C omits the 1/3 factor's effect incorrectly.
D multiplies without the 1/3 and doubles the height.

Question 16

1. 35 (correct answer)
2. 70
3. 17.5
4. 49

Explanation: Area $=ab\sin\theta=10\times7\times\sin30^{\circ}=70\times0.5=35$.
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

1. $54\sqrt{3}$ (correct answer)
2. $36\sqrt{3}$
3. $18\sqrt{3}$
4. $108\sqrt{3}$

Explanation: For a regular hexagon, $\text{Area}=\dfrac{3\sqrt3}{2}s^{2}=\dfrac{3\sqrt3}{2}(6)^2=\dfrac{3\sqrt3}{2}\cdot36=54\sqrt3$.
B. $36\sqrt3$ uses the formula for an equilateral triangle.
C. $18\sqrt3$ halves that incorrect triangle value.
D. $108\sqrt3$ doubles the correct hexagon area.

Question 18

1. $\dfrac{686\pi}{3}$ (correct answer)
2. $343\pi$
3. $\dfrac{343\pi}{3}$
4. $\dfrac{1372\pi}{3}$

Explanation: Hemisphere volume $=\dfrac{2}{3}\pi r^{3}=\dfrac{2}{3}\pi(7)^{3}=\dfrac{2}{3}\pi(343)=\dfrac{686\pi}{3}$.
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

1. $66\text{ m}^2$ (correct answer)
2. $132\text{ m}^2$
3. $44\text{ m}^2$
4. $36\text{ m}^2$

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

1. $144\pi$ (correct answer)
2. $72\pi$
3. $128\pi$
4. $288\pi$

Explanation: Volume is $\pi r^{2}h=\pi(4)^{2}(9)=\pi\cdot16\cdot9=144\pi$.
B halves the height.
C uses $r=4$ but multiplies by 8 instead of 9.
D doubles the correct height.