Practice Test

7.G.4 circle area and circumference

Question 1 of 13

2026 Practice Test B Math, question 28

Question

A circle has a radius of 5 inches. What is the area of the circle in terms of $\pi$ ?

Use $\pi$ to express your answer.

Enter your answer in the space provided. Enter only your answer.

On the test you build the answer on an on-screen keypad; here, type it. Use ^ for a power, sqrt( ) or √ for a root, pi for π, and <= or >= for ≤ and ≥.

Show the answer

Correct answer: 25π

Solution:

$25\pi$

The area of a circle is its radius squared times $\pi$ .

$A = \pi r^2$

$r = 5$

$A = \pi(5)^2$

$A = 25\pi$

Question analysis

This circle question gives a radius of 5 inches and asks for the area in terms of π, typed into a box rather than chosen. Students recall and apply the formula for the area of a circle. The formula is not printed on the test, and the area and circumference formulas are easy to mix up.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the area of a circle from its radius is the grade 7 circle standard applied directly, in one routine calculation. Typing the answer in terms of π, with no choices to compare, adds a little risk.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is one formula and one square, so the question comes down to recall and to the form of the answer. The area and circumference formulas both pair π with the radius, and 2πr gives a tidy value with π in it that looks just as right. The other losses are about form: replacing π with 3.14, which the instructions rule out twice, or dropping π after squaring 5.

Something you must already know moderate Working it out
The formula A = πr² is not printed with the question, and the circumference formula 2πr uses the same r and π, so recalling the wrong one gives a value that looks just as finished. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
The answer is built on a keypad with a π key and must keep π as a symbol; a decimal worked out with 3.14 is not accepted, and no choices show the expected form. (Enright, Morley & Sheehan, 2002)

Strategy: Recall the area formula rather than the circumference formula, square the radius before multiplying, and leave π as a symbol instead of replacing it with 3.14.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

10π Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Uses the circumference formula 2πr, or doubles the radius instead of squaring it, so 2 × 5 gives 10π, the distance around the circle.
78.5 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the area correctly but replaces π with 3.14, 25 × 3.14 = 78.5, though the question asks for the area in terms of π.
25 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Squares the radius, 5 × 5 = 25, and stops before multiplying by π, so the entry leaves out the π the instructions ask for.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a circle: $A = \pi r^2$; A is the area, the space inside the circle. r is the radius, the distance from the center to the edge. π (pi) is a fixed number, a little more than 3. r² means r times r. Square the radius first, then multiply by π. The area is in square units.
Circumference of a circle: $C = 2 \pi r$; C is the circumference, the distance around the circle. r is the radius, and π is the same number as above. This formula measures a length, not the space inside. It is here so you can tell the two formulas apart.

Sort out the question
The circle has a radius of 5 inches. The radius is the distance from the center to the edge. You need the area, the space inside the circle. In terms of π means π stays in the answer as a symbol. Do not change it to 3.14.
Choose the area formula
The area formula is $A = \pi r^2$ . It uses the radius squared.
Watch out: the circumference formula, $C = 2\pi r$ , also uses π and the radius. It gives $2 \times \pi \times 5 = 10\pi$ , the distance around the circle in inches. That is not the area.
Square the radius
Squaring a number means multiplying it by itself:
$r^2 = 5^2 = 5 \times 5 = 25$
Watch out: squaring is not doubling. $5 \times 2 = 10$ would lead to 10π again.
Multiply by π
Put 25 into the formula in place of r²:
$A = \pi \times 25 = 25\pi$
The area is 25π square inches. Write the number first and then π, the way you write 3x for 3 times x.
Watch out: leave π as π. Using 3.14 gives 78.5, which is not accepted. Dropping π leaves 25, which is not the area.
Enter it with the π key
On the test, the keypad has a π key in its bottom left corner. Press 2, then 5, then the π key. You do not need a times sign. Enter only the answer, with no A = and no units. On this site, type 25π or 25pi.
So you would enter 25π.

Check your answer

Estimate. π is a little more than 3, so 25π is a little more than 75 square inches. A square drawn around the circle is 10 inches on each side, so its area is 100 square inches. The circle fits inside that square, so its area must be less than 100. It is.

More questions like this

The original DOE question

Question 28 as shown on the DOE’s digital test, page 100 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test B Math, page 100, question 28.

Question 2 of 13

2026 Practice Test C Math, question 20

Question

Two circles, with centers at A and C, are shown. Line segment AB is congruent to line segment DE.

$Two circles. The larger circle has center A, and a radius is drawn from A to point B on the circle, labeled 5. The smaller circle has center C, and a segment is drawn from point D on the circle, through C, to point E on the circle.$

Determine the area, A, and the circumference, C, of each circle to complete the table.

Select one answer in each row.

Choose one answer in each row, then check. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

Show the answer

Correct answer: A = 25π: Circle A; C = 5π: Circle C

$A = 25\pi$ : Circle A

$C = 5\pi$ : Circle C

As shown in the illustration, AB (the radius of Circle A) is 5 units in length. Since DE (the diameter of Circle C) is congruent to AB, we know it is also 5 units in length.

Circle A:

$A = \pi r^{2}$

$A = 25\pi$

$r = 5$

We know the radius of Circle A is 5 so the answer to row 1 is A.

Circle C:

$C = \pi \cdot d$

$C = 5\pi$

We have to look for the circle with a diameter of 5. That is circle C.

Question analysis

Two circles centered at A and C are shown: radius AB of Circle A is labeled 5, and segment DE, drawn through C, is congruent to AB. Students decide whether an area of 25π and a circumference of 5π each belong to Circle A, Circle C or neither. Keeping radius and diameter apart, and area and circumference, is the challenge.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Applying the area and circumference formulas is the grade 7 circle standard. This gives a radius for one circle and, for the other, a segment known only through congruence, and both rows of the table must be right.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Something you must already know moderate
Much to keep track of minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Circle A is direct: its labeled radius of 5 gives an area of 25π. Circle C is where points are lost, since its size comes from a sentence rather than a label, and the segment through its center is a diameter, so its circumference is 5π while its area is only 6.25π. The row labels favor guessers, because the A and C in A = 25π and C = 5π match the circles' names, though they stand for area and circumference.

Reading a figure, table or graph moderate Understanding the question
Circle C's size is shown only by segment DE, drawn through center C with no label, and only the stem's sentence that DE is congruent to AB makes it a diameter of 5. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know moderate Working it out
Neither A = πr² nor C = πd is given, and one circle is measured by a radius and the other by a diameter, so each formula needs the right length. (Anderson & Krathwohl, 2001)
Much to keep track of minor Working it out
Four values are in play, areas of 25π and 6.25π and circumferences of 10π and 5π, and both rows must be matched correctly for the item to count. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Write down the radius and the diameter of each circle, compute both circles' areas and circumferences in terms of π, then match each value in the table.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

1:B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Squares the diameter instead of the radius, so Circle C's area seems to be π × 5² = 25π while Circle A's becomes 100π.
1:C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Uses 2πr for the area, so Circle A seems to have an area of 10π and Circle C one of 5π, and no circle seems to have 25π.
2:A Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Uses π times the radius for the circumference, so Circle A's radius of 5 seems to give a circumference of 5π.
2:C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Overlooks that DE is congruent to AB, so Circle C has no known size, and with Circle A's circumference at 10π picks Neither.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a circle: $A = \pi r^{2}$; A is the area, the space inside the circle. r is the radius, the distance from the center to the edge. π (pi) is a fixed number, a little more than 3. Square the radius, then multiply by π.
Circumference of a circle: $C = \pi d$; C is the circumference, the distance around the circle. d is the diameter, the distance across the circle through its center. A diameter is twice as long as a radius. Multiply the diameter by π.

Sort out the question
Circle A has its center at A. Its radius, AB, is 5. Circle C has its center at C. Segment DE goes through C, so it is a diameter. Congruent means the same length, so DE is 5 too.
Careful: in the table, A means area and C means circumference. Circle A and Circle C are just names.
Find each radius and diameter
Circle A: the radius is 5, so the diameter is $2 \times 5 = 10$ .
Circle C: the diameter is 5, so the radius is $5 \div 2 = 2.5$ .
Watch out: without the sentence about DE, Circle C seems to have no size. Then you might pick Neither for the second row.
Check the area row
Use the area formula for each circle. For Circle A:
$A = \pi \times 5^{2} = 25\pi$
For Circle C:
$A = \pi \times 2.5^{2} = 6.25\pi$
Only Circle A has an area of 25π.
Watch out: squaring Circle C’s diameter instead of its radius gives 25π by mistake.
Check the circumference row
Use the circumference formula for each circle. For Circle A:
$C = \pi \times 10 = 10\pi$
For Circle C:
$C = \pi \times 5 = 5\pi$
Only Circle C has a circumference of 5π.
Watch out: π times the radius is not the circumference. It would give Circle A 5π.
Fill in the table
Choose one button in each row. In the row for A = 25π, choose the Circle A column. In the row for C = 5π, choose the Circle C column. Neither stays empty.
So the answers are A = 25π: Circle A and C = 5π: Circle C.

Check your answer

Compare sizes. Circle A’s radius, 5, is twice Circle C’s radius, 2.5. So the distance around Circle C should be half of Circle A’s 10π. Half of 10π is 5π, which matches. And the bigger circle has the bigger area, 25π.

More questions like this

The original DOE question

Question 20 as shown on the DOE’s digital test, page 93 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 93, question 20.

Question 3 of 13

2026 Practice Test C Math, question 50

Question

Terry wants to pour cement around the edge of the circular patio in her backyard. The patio has a radius of 5 feet. What is the distance, in feet, around the edge of the patio? Use 3.14 for $\pi$ .

A15.7
B31.4
C49.3
D78.5

Show the answer

Correct answer: B

To find the distance around a circle if the radius is known, use the formula for circumference, $C = 2\pi r$ .

$C = 2(3.14)(5) = 31.4$

Question analysis

Terry wants to pour cement around the edge of a circular patio with a radius of 5 feet. Students find the distance around the edge, using 3.14 for π. It tests the circumference of a circle, and because the formulas for a circle's area and circumference both use the radius, they are easy to mix up.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Applying a circle formula is a grade 7 standard, and this is a one-formula problem with a given value for π. The only demand beyond routine is recalling the right formula, which the test does not print.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know moderate
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One formula and one multiplication finish the problem once the right formula is chosen. The cement in the story pulls toward area, and the area, 78.5, is listed, as is half the circumference, from multiplying 3.14 by the radius alone. Remembering that the circumference is π times the diameter, here 10 feet, and that the answer is in feet, not square feet, settles it.

Something you must already know moderate Working it out
The circumference formula is not printed, and the area formula uses the same radius and π, so recalling the wrong one gives a number that looks just as finished. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
Pouring cement suggests covering a surface, but the question asks for “the distance, in feet, around the edge”, which is the circumference, not the area. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Decide which circle measurement the distance around the edge is, recall its formula, and substitute the radius and 3.14, doubling the radius wherever the formula calls for the diameter.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

A Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies π by the radius, 3.14 × 5 = 15.7, using the radius where C = πd needs the diameter, so it finds half the distance around.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Starts from the area formula and squares π instead of the radius, 3.14 × 3.14 × 5 ≈ 49.3, mixing the wrong formula with a slip in squaring.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Finds the patio’s area, 3.14 × 5² = 78.5 square feet, using the area formula instead of the circumference for the distance around.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle: $C = 2 \pi r$; C is the circumference, the distance around the circle. r is the radius, the distance from the center to the edge. π (pi) is a fixed number, about 3.14. 2r is the diameter, the distance all the way across. The distance around is always about 3.14 times that.
Area of a circle: $A = \pi r^2$; A is the area, the space inside the circle, in square units. r² means r times r. It is here so you can tell the two formulas apart. This question does not use it.

Sort out the question
The patio is a circle with a radius of 5 feet. The radius is the distance from the center to the edge. Terry wants cement around the edge.
You need the distance around the edge, in feet. The distance around a circle is called its circumference. Use 3.14 for π.
Choose the right formula
The distance around is a length, so use the circumference formula, $C = 2\pi r$ .
Watch out: pouring cement may make you think of area. The area, $3.14 \times 5 \times 5 = 78.5$ square feet, is choice D. That is the space inside the patio, not the distance around it.
Put in the numbers
Put 3.14 in for π and 5 in for r. Multiply 2 by 5 first, because that is easy:
$C = 2 \times 3.14 \times 5 = 10 \times 3.14 = 31.4$
The distance around the patio is 31.4 feet.
Watch out: $3.14 \times 5 = 15.7$ , choice A, forgets the 2. It goes only halfway around.
Give the answer
The distance around the edge of the patio is 31.4 feet.
So the answer is B, 31.4.

Check your answer

Estimate. The diameter is 10 feet. The distance around a circle is a little more than 3 times its diameter. $3 \times 10 = 30$ , so about 31 feet makes sense. Choice D, 78.5, is far too big, and it is in square feet.

More questions like this

The original DOE question

Question 50 as shown on the DOE’s digital test, page 123 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 123, question 50.

Question 4 of 13

2026 Practice Test D Math, question 8

Question

A square metal plate has a side length of 3 inches. A circular hole with a diameter of 2 inches is cut out from the center of the plate.

$A square, shaded gray, with a white circle cut out of its center. A side of the square is labeled 3 in., and the diameter of the circle is labeled 2 in. The shaded region is the part of the square outside the circle.$

What is the area of the shaded region, to the nearest hundredth of a square inch?

Enter your answer in the space.

Fill-in question: work out the answer and type it as a number.

Show the answer

Correct answer: 5.86

To determine the area of a square, use $A = s^{2}$ .

Begin by finding the length of the side of the square, which is 3 inches. The area of the square is $(3\text{ in})^{2} = 9\text{ square inches}$ .

To determine the area of the circle, use $A = \pi r^{2}$ .

The radius of the circle (the hole) is one-half the diameter or $2\text{ in.} \div 2 = 1\text{ in.}$ , so the area of the entire circle is $\pi(1\text{ in.})^{2} \approx 3.14\text{ sq in.}$

To determine the area of the shaded region, subtract the area of the hole in the square from the area of the square:

$9\text{ sq in.} - \pi\text{ sq in.} =$

$9\text{ sq in.} - (3.14)\text{ sq in.} =$

$5.86\text{ sq in.}$

The shaded area to the nearest hundredth of a square inch $= 5.86$ sq in.

Question analysis

A 3-inch square metal plate has a circular hole 2 inches across cut from its center, and students find the area of the metal that is left, to the nearest hundredth of a square inch. It tests the areas of a circle and a square. Using the radius, not the diameter, and subtracting the right areas are the keys.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a circle's area is a grade 7 standard, and subtracting a hole from a shape is common grade 7 practice. The radius must come from the diameter and the result must be rounded, but the steps are routine.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Combines ideas minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 49 words.
Corrected from the DOE handbook: A wrong symbol in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every number is small, so the work is choosing the right pieces: the area of the square, the area of the hole, and the difference. A radius of 1 makes the circle's area simply π, which is easy to confuse with the circumference, 2π, when the formulas are half remembered. The region asked for is the gray metal, not the hole, and the final rounding goes up, since the third decimal place is 8.

Something you must already know moderate Working it out
Neither area formula is given, and the hole is labeled by its 2-inch diameter, so students must recall A = πr² and halve the diameter to get the radius 1 first. (Anderson & Krathwohl, 2001)
Combines ideas minor Working it out
The shaded metal is a square with a circle removed, so two area formulas are combined by subtraction rather than either one used alone. (Embretson, 1998; Daniel & Embretson, 2010)
An answer in an unexpected form minor Choosing the answer
The result is typed to the nearest hundredth with no choices to compare, so the long decimal 5.858… must be rounded up, not cut off after two places. (Enright, Morley & Sheehan, 2002)

Strategy: Halve the diameter to get the radius, find the area of the square and of the circle, subtract the circle from the square, and round only at the end.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

2.72 Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Uses the circumference, 2π × 1 ≈ 6.28, in place of the circle's area and subtracts it from the square's 9 square inches, getting about 2.72.
3.14 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the area of the circular hole, π × 1² ≈ 3.14 square inches, and enters it instead of the area of the shaded metal left around it.
5.85 Rounds too early (King et al., 2004)
Reveals a slip Expected pull: weak
Works the area out correctly as about 5.858 but cuts off the extra digits instead of rounding up to the nearest hundredth.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation gives the area of the hole as “π(1 in.)² ≅ 3.14 sq in.” The symbol ≅ means “is congruent to”, which an area and a number cannot be; the line means “is approximately equal to”, written ≈. The site now says “π(1 in.)² ≈ 3.14 sq in.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square: $A = s^{2}$; A is the area, the amount of flat space inside. s is the length of one side. s² means s × s. It works because a square with side s holds s rows of s little unit squares.
Area of a circle: $A = \pi r^{2}$; A is the area. r is the radius, the distance from the center to the edge. It is half the diameter, the distance straight across through the center. π (pi) is a number a little more than 3, about 3.14. r² means r × r.

Sort out the question
The plate is a square with 3-inch sides. A circle with a 2-inch diameter is cut out of the middle. The diameter is the distance straight across the circle through its center. The shaded region is the metal that is left. You need its area to the nearest hundredth of a square inch.
Find the area of the square
Use the formula for the area of a square, with s = 3 inches:
$A = s^{2} = 3^{2} = 3 \times 3 = 9$
The whole plate, before the hole is cut, is 9 square inches.
Find the radius of the hole
The circle formula needs the radius, not the diameter. The figure labels the 2 in. distance across the hole. The radius is half of it:
$r = 2 \div 2 = 1$
The radius is 1 inch.
Find the area of the hole
Use the formula for the area of a circle, with r = 1 inch:
$A = \pi r^{2} = \pi \times 1^{2} = \pi \approx 3.14$
The sign ≈ means is about equal to. The hole is about 3.14 square inches.
Watch out: 2πr is the circumference, the distance around the circle, not its area. Using it gives 9 − 6.28 = 2.72. And 3.14 is the hole, not the metal, so do not stop there.
Subtract the hole from the square
The shaded metal is the square with the circle taken away:
$9 - \pi \approx 9 - 3.14 = 5.86$
The shaded region is about 5.86 square inches.
Round and enter the answer
With more digits of π, 3.1416, you get 9 − 3.1416 = 5.8584. The digit after the hundredths place is 8, which is 5 or more, so round up to 5.86. Chopping it to 5.85 is wrong. Type only the number, with no units.
So you would enter 5.86.

Check your answer

The hole is a bit more than a third of the plate, since 3.14 is a bit more than 9 ÷ 3 = 3. So the metal left should be a bit less than two thirds of 9, which is 6. The answer 5.86 is just under 6, so it makes sense.

More questions like this

The original DOE question

Question 8 as shown on the DOE’s digital test, page 81 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 81, question 8. This is the DOE’s version, before the corrections listed in the analysis.

Question 5 of 13

2026 Practice Test D Math, question 12

Question

$A circle with center P. Points R and T are on the circle at the ends of a horizontal diameter through P, and point S is on the circle above it. Segment SP is labeled 12. A vertical segment from S meets RT at a right angle, dividing RT into a part labeled x, from R to that point, and a part labeled 5, from that point to T.$

P is the center of the circle above, and R, S, and T are points on the circle. What is the value of x?

Show the answer

Correct answer: D

$\overline{\mathrm{SP}}$ represents the radius of the circle and has length 12.

Both $\overline{\mathrm{RP}}$ and $\overline{\mathrm{PT}}$ are radii and also have length 12.

Therefore, the length of $\overline{\mathrm{RT}}$ , which is a diameter, is 24 and is also represented by $x + 5$ .

$x + 5 = 24$

$x = 19$

Question analysis

A circle with center P has points R, S and T on it, a segment SP labeled 12, and a vertical segment from S that splits RT into a piece labeled x and a piece labeled 5. Students find the value of x. It tests basic circle facts, and the challenge is seeing which lengths in the figure must be equal.

NYS standards: 7.G.4 circle area and circumference
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 circle work centers on area and circumference formulas. This needs only the fact that every radius has the same length, but spotting which segments are radii in a figure with extra markings is less routine than using a formula.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
No obvious method minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 24 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The figure gives one measurement on a slanted radius and two labels on the diameter, and the right angle at the foot of the vertical segment is a decoy. Once RP and PT are seen as radii, RT is 24, and x is that diameter less the 5. Students who never connect the slanted 12 to the horizontal line fall back on the labeled numbers, adding 12 and 5, or take x to end at the center.

Reading a figure, table or graph moderate Understanding the question
The 12 labels the slanted segment SP, not the horizontal one, and the arrow for x runs from R past the center P to the foot of the vertical segment, so x is neither a radius nor a whole diameter. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
No obvious method minor Working it out
The right-angle mark and the triangle under S suggest the Pythagorean theorem, which is no help; the way in is seeing that RP and PT are radii as well. (Schoenfeld, 1985)

Strategy: Mark every segment from P to the circle as 12, notice that R and T are the ends of a diameter, then write an equation for that diameter using x and the 5.

Traps

Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
Information you don't need: The question gives numbers or facts that play no part in the answer.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: moderate
Reads x as the radius RP, as if its arrow ended at the center P, though it runs past P to the foot of the vertical segment.
B Trusting the picture (Lesh, Post & Behr, 1987)
Reveals a surface strategy Expected pull: weak
Judges x from the drawing, where it looks about three times as long as the piece labeled 5, instead of working from the radius of 12.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Adds the two labeled numbers, 12 + 5 = 17, without placing the 12 on the diameter, as if x were a radius plus the 5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Radius and diameter: $d = 2r$; r is the radius, the distance from the center to any point on the circle. d is the diameter, the distance straight across the circle through the center. A diameter is two radii laid end to end, one on each side of the center, so it is twice as long.

Sort out the question
P is the center of the circle, and R, S and T are on the circle. Segment SP is 12. Segment RT goes straight across through P. A short line from S meets RT at a right angle. It splits RT into x, from R to that point, and 5, from that point to T. You need x.
See that SP is a radius
SP runs from the center P to the point S on the circle. So SP is a radius, and the radius is 12. Every radius of a circle has the same length. RP and PT also run from the center to the circle, so RP = 12 and PT = 12.
Find the diameter RT
RT passes through the center, so it is a diameter. Use the formula with r = 12:
$\mathrm{RT} = 2r = 2 \times 12 = 24$
RT is 24 long. It is the radius RP plus the radius PT.
Solve for x
RT is also split into x and 5, so:
$x + 5 = 24$
Subtract 5 from both sides:
$x = 24 - 5 = 19$
Watch out: x runs from R past the center P, so it is longer than a radius. Reading it as 12, choice A, stops at P. Adding 12 + 5 = 17, choice C, puts the 12 in the wrong place.
Work from the numbers, not the picture
The figure is not drawn to scale. In it, x looks about three times as long as 5, which suggests 15, choice B. Trust the labels instead. The right triangle under S is not needed at all.
So the answer is D, 19.

Check your answer

The point where the short line meets RT is 19 from R. P is 12 from R, so that point is 19 − 12 = 7 past the center. Then 7 + 5 = 12, which is PT, a radius. Everything fits.

More questions like this

The original DOE question

Question 12 as shown on the DOE’s digital test, page 85 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 85, question 12.

Question 6 of 13

2024 Test A Math, question 8

Question

$Square PQRS, with Q at the top left, R at the top right, P at the bottom left and S at the bottom right, and a right-angle mark at S. Point S is marked as the center of a circle that passes through P and R. The quarter of the circle inside the square, bounded by sides SP and SR and the arc from P to R, is shaded.$

In the figure above, PQRS is a square. Point S is the center of the circle, and points P and R are on the circle. If the area of the square is 4 square centimeters, what is the area, in square centimeters, of the shaded quarter of the circle?

A $\frac{\pi}{4}$
B $\pi$
C $2\pi$
D $4\pi$

Show the answer

Correct answer: B

The area of square PQRS is 4 square centimeters. So

$s^{2} = 4$

$\sqrt{s^{2}} = \sqrt{4}$

$s = 2\text{ cm}$

Sides PS and RS are each a radius of the circle. So the radius of the circle is 2 cm.

The shaded region is $\frac{1}{4}$ of the circle. The area of a circle is $A = \pi r^{2}$ . So the area of the shaded region, in square centimeters, is

$\frac{1}{4}A = \frac{1}{4}\pi r^{2} =$

$\frac{1}{4}\pi(2^{2}) =$

$\frac{1}{4}\pi(4) = \pi$

Question analysis

A geometry question with a figure: square PQRS has one corner at the center of a circle, and the quarter of the circle inside the square is shaded. Given the area of the square, students find the area of the shaded quarter circle. The key is connecting the square's side to the circle's radius.

NYS standards: 7.G.4 circle area and circumference
6.G.5 perfect squares and perfect cubes
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Circle area is grade 7 content, and this question adds only one linking step from the square's area to the radius. Typical grade 7 circle work gives the radius directly; here it takes one short extra inference.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Something you must already know minor
Combines ideas minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 52 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question chains three short facts: from the square's area to its side, from the side to the radius, and from the whole circle to a quarter of it. None is hard, but each wrong choice marks a broken link, most often the whole circle's area, which looks like a finished answer. Pointing to S as the center before computing anything settles the radius.

Reading a figure, table or graph moderate Understanding the question
The radius is never labeled: students must see that S is both a corner of the square and the circle's center, so sides PS and RS are radii. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
Neither the circle area formula, π times the radius squared, nor the step from a square's area to its side length is given. (Anderson & Krathwohl, 2001)
Combines ideas minor Working it out
Square and circle are linked in one chain: the area 4 gives a side of 2, the side is the radius, and the shaded part is a quarter of π × 2². (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Take the square root of the square's area to get its side, which is also the radius, then take one quarter of pi times the radius squared.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Stops one step early: A choice is a number from partway through the solution.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

A Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Halves the square's side as if it were a diameter, so it uses a radius of 1, and one quarter of π × 1² gives π/4.
C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Misremembers the circle's area as 2πr², blending it with the circumference 2πr, so one quarter of 2π × 2² gives 2π.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the whole circle's area, π × 2² = 4π, and stops before taking the quarter that is shaded.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square: $A = s^2$; A is the area, the space inside. s is the length of one side, and s² means s × s. All four sides of a square are equal, so its area is side times side.
Area of a circle: $A = \pi r^2$; A is the area of the whole circle. r is the radius: the distance from the center to the edge. π, called pi, is a number a little more than 3, about 3.14. Multiply π by r × r. This works for every circle.

Sort out the figure
PQRS is a square with an area of 4 sq cm. Point S is the center of the circle. Points P and R are on the circle. The shaded part is a quarter of the circle, inside the square. You need the shaded area, in square centimeters. The answer will have π in it.
Find the side of the square
Use the square formula. The area is 4, so find the number that times itself gives 4.
$A = s^2$
$4 = s^2$
$s = 2$
Each side is 2 cm long, because $2 \times 2 = 4$ .
See that the side is the radius
Side SP goes from the center, S, to point P on the circle. So SP is a radius. So is side SR. The radius is 2 cm. Watch out: the side is a radius, not a diameter, the full width of the circle. Using a radius of 1 leads to $\frac{\pi}{4}$ , choice A.
Find the area of the whole circle
$A = \pi r^2$
$A = \pi \times 2^2 = \pi \times 4 = 4\pi$
The whole circle has an area of 4π sq cm. Watch out: this is the whole circle. Stopping here gives choice D.
Take one quarter
The shaded part is one quarter of the circle, so take one quarter of 4π.
$\frac{1}{4} \times 4\pi = \pi$
The shaded area is π sq cm. So the answer is B, $\pi$ .

Check your answer

π is about 3.14, so the shaded part is about 3.14 sq cm. The whole square is 4 sq cm. In the figure, the shaded part fills most of the square, but not all. So a bit less than 4 makes sense. The choice 4π is about 12.6, far too big.

More questions like this

The original DOE question

Question 65 as printed on page 77 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 77, question 65. Open page 77 of the handbook (PDF)

Question 7 of 13

2024 Test B Math, question 32

Question

Two circular plates have the dimensions shown.

$Two circles side by side. The larger circle has a dashed diameter labeled 18 in., and the smaller circle has a dashed diameter labeled 12 in.$

What is the difference between the areas of the two plates, in square inches?

A $6\pi$
B $9\pi$
C $45\pi$
D $180\pi$

Show the answer

Correct answer: C

First, determine the area of each circle using the formula $A=\pi r^{2}$ :

A circle with a diameter of 18 inches has a radius of 9 inches, so the area is $\pi(9)^{2}=81\pi$ .

A circle with a diameter of 12 inches has a radius of 6 inches, so the area is $\pi(6)^{2}=36\pi$ .

The difference between the two areas is $81\pi-36\pi=45\pi$ .

Question analysis

Two circular plates are drawn with their diameters marked in inches, and students find the difference between their areas in square inches, in terms of pi. It tests the area of a circle. The diameters, not the radii, are marked, so a step comes before the area formula, which the test does not provide.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a circle's area is a grade 7 standard, and this asks for two areas and their difference. It matches typical classwork: the formula must be recalled and each diameter halved first, with no further twist.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two area calculations and a subtraction, kept in terms of π, make this routine for students who know the area formula. Points are lost by using the labeled diameters as radii, by subtracting circumferences, or by subtracting the radii before squaring, and each of these has a listed answer. Writing both areas, 81π and 36π, before subtracting avoids all three.

Something you must already know moderate Working it out
The area formula A = πr² is not given, and the figure labels each plate by its diameter, so 18 and 12 must be halved to radii of 9 and 6 first. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The dashed lines run all the way across each circle, which marks them as diameters; reading the 18 in. and 12 in. as radii leads to a listed choice. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Halve each diameter to get the radius, find each plate's area as pi times the radius squared, and subtract the smaller area from the larger.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

A Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Subtracts the circumferences, 18π − 12π = 6π, using πd, the distance around a circle, in place of the area formula.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Subtracts the radii first, 9 − 6 = 3, and then squares, π(3)² = 9π, as if the difference of two squares were the square of the difference.
D Wrong formula (King et al., 2004)
Reveals a slip Expected pull: strong
Uses the labeled diameters as radii, π(18)² − π(12)² = 324π − 144π = 180π, forgetting to halve them first.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a circle: $A = \pi r^2$; A is the area, the flat space inside the circle. r is the radius, the distance from the center to the edge. It is half the diameter. π (pi) is a fixed number, about 3.14. r² means r × r. Use it for the space inside any circle.

Sort out the question
Each dashed line goes all the way across its circle, through the center. A line like that is a diameter. The big plate has a diameter of 18 inches. The small plate has a diameter of 12 inches.
You need the difference between the two areas. That means the big area minus the small area, in square inches.
Find each radius
The area formula uses the radius, not the diameter. The radius is half the diameter.
$18 \div 2 = 9$
$12 \div 2 = 6$
The radii are 9 inches and 6 inches. Watch out: If you use 18 and 12 as the radii, you get $180\pi$ , choice D.
Find the big plate’s area
$A = \pi r^2 = \pi \times 9^2 = \pi \times 81 = 81\pi$
The big plate’s area is $81\pi$ square inches. Keep π as a symbol, since all the choices use π.
Find the small plate’s area
$A = \pi \times 6^2 = \pi \times 36 = 36\pi$
The small plate’s area is $36\pi$ square inches. Squaring means multiplying a number by itself, so $6^2 = 6 \times 6 = 36$ .
Subtract the two areas
$81\pi - 36\pi = 45\pi$
Subtract the numbers in front of π, just as 81 apples minus 36 apples is 45 apples. The difference is $45\pi$ square inches.
Watch out: Don’t subtract the radii first. Squaring 9 − 6 = 3 gives $9\pi$ , choice B. Subtracting the distances around the plates, $18\pi - 12\pi$ , gives $6\pi$ , choice A. That uses the wrong formula.
So the answer is C, $45\pi$ .

Check your answer

Estimate with π about 3. The big plate is about $3 \times 81 = 243$ square inches. The small one is about $3 \times 36 = 108$ . The difference is about 135, and $45 \times 3 = 135$ . It matches.

More questions like this

The original DOE question

Question 89 as printed on page 208 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 208, question 89. Open page 208 of the handbook (PDF)

Question 8 of 13

2023 Test A Math, question 37

Question

$A circle with a shaded sector between two radii, one pointing straight up and one pointing up and to the right. The central angle of the shaded sector is marked 45°.$

The shaded sector of the circle shown above has an area of $18\pi$ square feet. What is the circumference of the circle?

A $144\pi$ ft
B $24\pi$ ft
C $18\pi$ ft
D $9\pi$ ft

Show the answer

Correct answer: B

The area of the entire circle is $\pi r^{2}$ .

The fraction of the circle that is shaded is $\frac{45}{360} = \frac{1}{8}$ . So, the area of the sector is $\frac{1}{8}\pi r^{2}$ . Use that information to find r:

$\frac{1}{8}\pi r^{2} = 18\pi$

$r^{2} = 144$

$r = 12$

Use r to calculate the circumference of the circle:

$c = 2\pi r = 2\pi(12) = 24\pi$

The circumference is $24\pi$ feet.

Question analysis

This circle question shows a shaded 45° sector with an area of 18π square feet and asks for the circumference of the whole circle. Students work backward: from the sector to the full area, from the area to the radius, and from the radius to the circumference. Mixing up area and circumference, or stopping partway, gives wrong choices.

NYS standards: 7.G.4 circle area and circumference
6.G.5 perfect squares and perfect cubes
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 circle problems usually give the radius and ask for area or circumference. This one runs backward from the area of a 45° sector, needing the sector's share of the circle, a square root and a switch to circumference: four linked steps.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas moderate
Something you must already know moderate
Reading a figure, table or graph minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 30 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question moves from a part of the circle's area to the whole circle's length, so area and circumference both appear and are easy to swap. Many students scale the sector up to 144π and stop, giving an area when a length was asked. Solving r² = 144 also needs a square root, after which 2πr is routine. Asking at the end whether the result is a length, measured in feet rather than square feet, catches the main trap.

Combines ideas moderate Working it out
Three ideas chain together: the 45° sector is 1/8 of the circle, so the whole area is 8 × 18π = 144π, then the radius comes from the area, and the circumference from the radius. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know moderate Working it out
Neither πr² nor 2πr is given, nor that a full turn is 360°, so every link of the chain rests on formulas students must already know. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The stem says only “shaded sector”; its 45° angle is marked at the center in the figure, the one place the fraction of the circle comes from. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Work out what fraction of the circle a 45° sector is, scale its area up to the whole circle, solve for the radius, and only then use the circumference formula.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Scales the sector up to the whole circle, 8 × 18π = 144π, and gives that area, though the question asks for the circumference.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Repeats the sector's area from the stem, 18π, as if the shaded part's area were the circle's circumference.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No step with the 45° angle or the circle formulas gives 9π, which is half the sector's area, so it mostly attracts guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a circle in a sector: $\text{fraction of the circle} = \frac{\text{angle}}{360}$; A sector is a slice of a circle, like a slice of pizza. Its angle is the angle at the center, in degrees. A full turn around the center is 360 degrees. So the slice is that fraction of the whole circle.
Area of a circle: $A = \pi r^{2}$; A is the area, the space inside the circle. r is the radius, the distance from the center to the edge. π, called pi, is a number a little more than 3. r² means r times r.
Circumference of a circle: $C = 2\pi r$; C is the circumference, the distance around the circle. r is the radius and π is pi. Two radii make the diameter, the distance across, so this is the same as π times the diameter.

Sort out the question
The shaded slice has a 45° angle at the center. Its area is $18\pi$ square feet. You need the circumference, the distance around the whole circle. For that you need the radius, so work back to it.
Find what fraction is shaded
$\text{fraction of the circle} = \frac{\text{angle}}{360}$
$\frac{45}{360} = \frac{1}{8}$
The shaded slice is one eighth of the circle. Eight slices like it would fill the circle.
Find the area of the whole circle
$8 \times 18\pi = 144\pi$
The whole circle has an area of $144\pi$ square feet. Watch out: this is an area, not the circumference. Stopping here gives choice A.
Find the radius
$A = \pi r^{2}$
$\pi r^{2} = 144\pi$
$r^{2} = 144$
$r = 12$
Divide both sides by π. Then ask what number times itself makes 144. It is 12, so the radius is 12 feet.
Find the circumference
$C = 2\pi r$
$C = 2 \times \pi \times 12 = 24\pi$
Leave π as it is, because the choices are written with π. The circumference is $24\pi$ feet. So the answer is B, $24\pi$ ft.

Check your answer

Work forward. A radius of 12 feet gives an area of $\pi \times 12 \times 12 = 144\pi$ square feet. One eighth of that is $144\pi \div 8 = 18\pi$ square feet, the shaded area in the question.

More questions like this

The original DOE question

Question 94 as printed on page 85 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 85, question 94. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 9 of 13

2023 Test B Math, question 52

Question

$An equilateral triangle drawn with dashed sides. On each side a semicircle bulges outward, with that side as its diameter. The three curved arcs together form the outline of the garden.$

The garden design shown above is created in the shape of three semicircles whose diameters lie on the sides of an equilateral triangle. If each diameter is 2 meters long, what is the perimeter of the garden, in meters?

A6
B12
C $3\pi$
D $6\pi$

Show the answer

Correct answer: C

The perimeter (or circumference) of one of the three semicircles is $\frac{1}{2}(\text{diameter})\pi=\frac{1}{2}(2\pi)=\pi$ . Since there are 3 semicircles that make up the perimeter, the perimeter is $3\pi$ .

Question analysis

A garden is outlined by three semicircles whose diameters, each 2 meters long, are the sides of an equilateral triangle. Students find the perimeter of the garden in meters. It tests the circumference of a circle applied to half circles, and the figure mixes straight dashed sides with curved ones, so deciding what counts as the outline matters.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Applying the circumference formula is a grade 7 standard. Here it is used for a semicircle and repeated three times, a short extension of routine classwork, with a dashed triangle inside the figure to look past.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph minor
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The perimeter is just three half-circles, and each is half of π × 2. The figure’s dashed triangle invites measuring straight sides instead, and remembering the whole circle’s circumference without halving it gives a choice twice too large. Tracing the garden’s outer edge with a finger before computing shows that only the curves count.

Reading a figure, table or graph minor Understanding the question
The dashed triangle lies inside the garden, so its three 2-meter sides are not part of the perimeter; only the three curved arcs bound the garden. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The circumference formula is not given, and each arc is half of a circle with a 2-meter diameter, so students must take half of π × 2. (Anderson & Krathwohl, 2001)

Strategy: Find the circumference of a full circle with a 2-meter diameter, take half of it for one semicircle, and multiply by the number of semicircles in the outline.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Adds the three dashed sides of the triangle, 3 × 2 = 6, measuring the triangle inside the garden instead of its curved outline.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 12, which is twice the triangle’s perimeter and matches no arc length, so it mostly draws guesses.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Uses a whole circumference, π × 2, for each of the three arcs, forgetting that each one is only half a circle, which doubles the perimeter.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle: $C = \pi \times d$; C is the circumference, the distance around a circle. d is the diameter, the straight line across the circle through its center. π is a number a little more than 3.14. The distance around any circle is π times as long as the distance across it.
Arc of a semicircle: $\text{arc} = \frac{1}{2} \times \pi \times d$; A semicircle is half of a circle. Its arc is the curved part. d is the diameter, the straight side. The arc is half of the full circle’s circumference, so take half of π times d.

Sort out the figure
The garden’s edge is made of three semicircles, or half circles. Each one sits on a side of the dashed triangle. Each side is a diameter, 2 meters long. The perimeter is the distance around the outside. Here that is only the three arcs, the curved parts. The dashed triangle is inside, so it does not count.
Find a full circle’s circumference
Picture the whole circle that one semicircle comes from. Its diameter is 2 meters. Use the circumference formula.
$C = \pi \times 2 = 2\pi$
The full circle would measure $2\pi$ meters around.
Find the length of one arc
Each semicircle’s arc is half of that circle. Use the arc formula.
$\text{arc} = \frac{1}{2} \times \pi \times 2 = \pi$
One arc is $\pi$ meters long.
Watch out: Using the full circle’s $2\pi$ for each arc doubles the answer. That gives $6\pi$ , choice D.
Add the three arcs
The garden’s edge is three equal arcs.
$\pi + \pi + \pi = 3\pi$
The perimeter is $3\pi$ meters.
Watch out: The dashed sides add to 2 + 2 + 2 = 6 meters, choice A. That is the triangle, not the curved edge.
So the answer is C, $3\pi$ meters.

Check your answer

Estimate. $3\pi$ is about 3 × 3.14 = 9.42 meters. A curved path is longer than the straight line across it. So the edge should be longer than the triangle’s 6 meters, and 9.42 is.

More questions like this

The original DOE question

Question 109 as printed on page 206 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 206, question 109. Open page 206 of the handbook (PDF)

Question 10 of 13

2022 Test A Math, question 1

Question

The figure below shows a circle with a hole in the middle. The outer diameter of the circle is $1\frac{1}{4}$ inches. The diameter of the hole is $\frac{1}{4}$ inch.

$A large circle shaded gray, with a smaller white circle, the hole, at its center. A double-headed arrow across the hole is labeled 1/4 in., and a double-headed arrow below the figure, spanning the full width of the large circle, is labeled 1 1/4 in. The shaded region is the part of the large circle outside the hole.$

What is the area of the shaded region, to the nearest hundredth of a square inch?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 1.18

To determine the area of a circle, begin by finding the radius of the circle, which is half of the diameter.

The radius of the entire circle is $1\frac{1}{4}\text{ in.} \div 2 = \frac{5}{8}\text{ in.}$ , so the area of the entire circle is $\pi\left(\frac{5}{8}\right)^{2}\text{ in.}^{2} = \frac{25}{64}\pi\text{ in.}^{2}$

The radius of the hole in the circle is $\frac{1}{4}\text{ in.} \div 2 = \frac{1}{8}\text{ in.}$ , so the area of the hole in the circle = $\pi\left(\frac{1}{8}\right)^{2}\text{ in.}^{2} = \frac{1}{64}\pi\text{ in.}^{2}$

To determine the area of the shaded region, subtract the area of the hole in the circle from the area of the entire circle:

$\left(\frac{25}{64}\right)\pi\text{ in.}^{2} - \left(\frac{1}{64}\right)\pi\text{ in.}^{2} =$

$\left(\frac{24}{64}\right)\pi\text{ in.}^{2} =$

$\left(\frac{3}{8}\right)\pi\text{ in.}^{2} =$

$\left(\frac{3}{8}\right)\text{ in.}^{2}(3.14) =$

$1.1775\text{ in.}^{2}$

1.1775 square inches rounded to the nearest hundredth of a square inch = 1.18 square inches

Question analysis

A gray ring is a circle 1 1/4 inches across with a hole 1/4 inch across at its center, and this grid-in asks for the area of the shaded part to the nearest hundredth of a square inch. Students use the area of a circle twice. Turning diameters given as fractions into radii, and squaring them, is where mistakes happen.

NYS standards: 7.G.4 circle area and circumference
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students find circle areas and often subtract a hole from a shape. This ring gives its diameters as a mixed number and a fraction, so halving, squaring fractions and rounding add work, though no step goes beyond grade 7.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Error-prone arithmetic moderate
Combines ideas minor
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 6, 45 words.
Corrected from the DOE handbook: A missing exponent in the explanation’s area formula was restored.
An arithmetic slip in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The method is the familiar one for a ring, the big circle's area minus the hole's, but the labels are diameters given as fractions, so each one has to be halved and then squared before π enters. Students lose points by skipping the halving, by stopping at the full disk, or by subtracting the radii before squaring. Keeping both areas as fractions of π, 25/64π − 1/64π = 3/8π, leaves one multiplication and one rounding at the very end.

Something you must already know moderate Working it out
Neither area formula is printed, and both circles are labeled by diameter, so students must recall A = πr² and halve 1 1/4 inches and 1/4 inch to radii of 5/8 and 1/8 first. (Anderson & Krathwohl, 2001)
Error-prone arithmetic moderate Working it out
Every value is a fraction: 5/4 ÷ 2 = 5/8, squares of 25/64 and 1/64, then 3/8 × 3.14 worked by hand, so a slip can enter at each stage. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Combines ideas minor Working it out
The gray ring has no formula of its own; it is the large circle's area minus the hole's, so the area formula is used twice and the results subtracted. (Embretson, 1998; Daniel & Embretson, 2010)
An answer in an unexpected form minor Choosing the answer
The grid-in wants the nearest hundredth, so 1.1775 must be rounded to 1.18, and a whole-disk area such as 1.23 gets no warning from a list of choices. (Enright, Morley & Sheehan, 2002)

Strategy: Write 1 1/4 as 5/4 and halve both diameters to get the radii, subtract the hole's area from the area of the full circle, and round to the nearest hundredth only at the very end.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Stops one step early: A choice is a number from partway through the solution.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

4.71 Wrong formula (King et al., 2004)
Reveals a slip Expected pull: strong
Uses the labeled diameters as radii, π(5/4)² − π(1/4)² = 24/16π = 1.5π ≈ 4.71, forgetting to halve them first, which makes the ring four times too big.
1.23 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the area of the whole circle, π(5/8)² = 25/64π ≈ 1.23 square inches, and stops before subtracting the hole's 1/64π.
0.79 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Subtracts the radii first, 5/8 − 1/8 = 1/2, then squares, π(1/2)² ≈ 0.79, as if the difference of two squares were the square of the difference.

What was corrected from the DOE handbook, and why

Explanation: The handbook writes the area of the whole circle as “π(5/8) in.² = 25/64π in.²” and the area of the hole as “π(1/8) in.² = 1/64π in.²” (p. 128, question 58), squaring the unit but not the radius. The area of a circle is πr², and π(5/8) is only 5/8π. The site now reads π(5/8)² in.² = 25/64π in.² and π(1/8)² in.² = 1/64π in.². The areas that follow and the answer, 1.18, are unchanged.
Explanation: The handbook computes (3/8) in.²(3.14) as “1.175 in.²” and then rounds “1.175 square inches” to 1.18 (p. 128, question 58). But 3/8 × 3.14 = 1.1775, so the site now reads 1.1775 in both places. Rounded to the nearest hundredth it is still 1.18 square inches, the answer (with π itself, 3/8 × π = 1.178…, which also rounds to 1.18).

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Radius from diameter: $r = \frac{d}{2}$; d is the diameter, the distance straight across a circle through its center. r is the radius, the distance from the center to the edge. The diameter is two radii laid end to end, so the radius is half of it.
Area of a circle: $A = \pi \times r^{2}$; A is the area, the space inside the circle. π (pi) is a number a little more than 3, about 3.14. r is the radius. r² means r × r. Square the radius first, then multiply by π.

Sort out the question
The gray part is a ring. It is the big circle with the hole cut out. The big circle is $1\frac{1}{4}$ inches across. The hole is $\frac{1}{4}$ inch across. These are diameters. You need the area of the gray ring, in square inches. Round it to the nearest hundredth, two places after the decimal point.
Find both radii
The area formula uses the radius, so halve each diameter first. Write the mixed number as a fraction: $1\frac{1}{4} = \frac{5}{4}$ .
$r = \frac{d}{2}$
$r = \frac{5}{4} \div 2 = \frac{5}{4} \times \frac{1}{2} = \frac{5}{8}$
$r = \frac{1}{4} \div 2 = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$
The big circle’s radius is 5/8 inch. The hole’s radius is 1/8 inch. Watch out: don’t use the diameters as radii. That makes the ring four times too big, and you would grid 4.71.
Find the area of the big circle
$A = \pi \times r^{2}$
$A = \pi \times \left(\frac{5}{8}\right)^{2} = \pi \times \frac{5 \times 5}{8 \times 8} = \frac{25}{64}\pi$
To square a fraction, square the top and square the bottom. The big circle’s area is 25/64 π square inches. Keep π for now. You will use 3.14 at the very end.
Find the area of the hole
$A = \pi \times \left(\frac{1}{8}\right)^{2} = \pi \times \frac{1 \times 1}{8 \times 8} = \frac{1}{64}\pi$
The hole’s area is 1/64 π square inches. Watch out: square each radius on its own. Don’t subtract the radii first. That gives $\frac{5}{8} - \frac{1}{8} = \frac{1}{2}$ , and $\pi \times \left(\frac{1}{2}\right)^{2}$ leads to 0.79.
Subtract the hole
The ring is the big circle minus the hole. Both fractions have 64 on the bottom, so subtract the tops.
$\frac{25}{64}\pi - \frac{1}{64}\pi = \frac{24}{64}\pi = \frac{3}{8}\pi$
To simplify 24/64, divide the top and bottom by 8. Watch out: don’t stop at the big circle. Its area alone is about 1.23 square inches.
Multiply by 3.14 and round
$\frac{3}{8} \times 3.14 = \frac{9.42}{8} = 1.1775$
The ring’s area is 1.1775 square inches. To round to the hundredths place, look at the next digit. It is 7, which is 5 or more, so round up. 1.1775 becomes 1.18. Using the π key instead gives 1.178…, which also rounds to 1.18. So you would grid 1.18.

Check your answer

Estimate. The big circle is about 1.23 square inches. The hole is tiny: $\frac{1}{64} \times 3.14$ is about 0.05 square inch. $1.23 - 0.05 = 1.18$ . That matches. The ring must also be smaller than the whole circle, and it is.

More questions like this

The original DOE question

Question 58 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 79, question 58. Open page 79 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 11 of 13

2018 Test A Math, question 49

Question

A car travels at 4,400 feet per minute. The radius of each tire on the car is 1 foot. How many revolutions does one of these tires make in 1 minute? (Use the approximation $\frac{22}{7}$ for $\pi$ .)

A700
B1,925
C13,828
D15,400

Show the answer

Correct answer: A

One revolution is equal to the circumference of the tire:

$C=2\pi r=2(1)(\frac{22}{7})=\frac{44}{7}$ ft

The car travels at 4,400 ft per minute. To calculate the number of revolutions, divide the speed by the circumference:

$4,400\div \frac{44}{7}=4,400\times \frac{7}{44}=700$

700 revolutions.

Question analysis

This circle word problem says a car travels 4,400 feet per minute on tires with a radius of 1 foot and asks how many revolutions each tire makes in a minute, using 22/7 for pi. Students must see that one revolution covers the tire's circumference, then divide. Dividing by a fraction is the step to check.

NYS standards: 7.G.4 circle area and circumference
6.NS.1 dividing fractions by fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Circumference is grade 7 content. This item adds the idea that each revolution rolls the tire one circumference forward and requires dividing by a fraction, which is typical of a solid grade 7 application problem.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Something you must already know moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 34 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question tests whether students connect distance traveled to turns of a wheel, a link the stem leaves unstated. Every wrong choice is larger than the answer, and two exceed 4,400, which is impossible when each turn covers more than 6 feet. A sense check on size rules those out before any fraction work, leaving only the division by 44/7 to do carefully.

Turning words into math moderate Understanding the question
The stem never says that one revolution rolls the car forward one circumference; students must supply that link to turn 4,400 feet into a count of turns. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know moderate Working it out
The circumference formula, C = 2πr, is not given; with r = 1 and 22/7 for π it gives 44/7 feet per turn. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
Dividing 4,400 by 44/7 means multiplying by 7/44, which is clean only if students see that 4,400 ÷ 44 = 100 before multiplying by 7. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find the distance covered in one turn with the circumference formula, 2πr, then divide the distance traveled in a minute by that circumference.

Traps

A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard path gives 1,925, which equals 4,400 × 7 ÷ 16, so it mostly draws students guessing a mid-sized count.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: moderate
Multiplies 4,400 by π, 22/7, instead of dividing by the circumference, and truncates 13,828.57 to 13,828.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Divides 4,400 by 2/7 instead of 44/7, dropping the 22 of π while flipping 2 × 22/7, which gives 4,400 × 7 ÷ 2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle: $C = 2 \times \pi \times r$; C is the circumference, the distance around a circle. r is the radius, from the center to the edge. π (pi) is about 22/7. The distance around is a bit more than 3 times the distance across, which is 2 × r.
Number of turns: $n = \frac{d}{C}$; n is the number of turns. d is the distance the car travels. C is the circumference. Each full turn rolls the car forward one circumference, so you divide the distance by it.

Sort out what you know
The car goes 4,400 feet in 1 minute. Each tire has a radius of 1 foot. A revolution is one full turn of the tire. You need how many turns a tire makes in 1 minute.
Find how far one turn goes
When a tire turns once, the car rolls forward the distance around the tire. That distance is the circumference. Use $\frac{22}{7}$ for π:
$C = 2 \times \frac{22}{7} \times 1 = \frac{44}{7}$
One turn moves the car $\frac{44}{7}$ feet, about 6.3 feet.
Divide the distance by one turn
$n = 4{,}400 \div \frac{44}{7}$
To divide by a fraction, multiply by its reciprocal (the fraction flipped upside down):
$4{,}400 \times \frac{7}{44} = 100 \times 7 = 700$
Notice that 4,400 ÷ 44 = 100. That makes it easy.
Give the answer
The tire makes 700 turns. Watch out: multiplying 4,400 by π instead of dividing by the circumference gives about 13,828, choice C. So the answer is A, 700.

Check your answer

Estimate. One turn is about 6 feet. 4,400 ÷ 6 is about 730, close to 700. There should be fewer turns than feet, since each turn covers several feet. Choices C and D are bigger than 4,400, so they cannot be right.

More questions like this

The original DOE question

Question 106 as printed on page 90 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 90, question 106. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 12 of 13

2018 Test B Math, question 17

Question

A large circular dinner plate has a radius of 20 centimeters. A smaller circular plate with a circumference of $20\pi$ centimeters is placed in the center of the larger dinner plate. What is the area of the part of the larger dinner plate that is not covered by the smaller plate?

A $20\pi$ sq cm
B $100\pi$ sq cm
C $200\pi$ sq cm
D $300\pi$ sq cm

Show the answer

Correct answer: D

The radius of the large plate is 20 cm. Use that to find the area of the large plate:

$A=\pi r^2=\pi(20^2)=400\pi\text{ sq cm}$

The circumference of the smaller plate is 20π cm. Use that to find the radius, and then the area, of the smaller plate:

$C=2\pi r$

$20\pi=2\pi r$

$r=10$

$A=\pi r^2=\pi(10^2)=100\pi\text{ sq cm}$

Subtract the area of the small plate from the area of the large plate:

400π − 100π = 300π sq cm

Question analysis

This circle question places a smaller plate, described by its circumference, on a larger dinner plate described by its radius, and asks for the area of the large plate left uncovered. Students recall the circumference and area formulas, find the small radius and subtract areas. Because the two plates are described in different ways, circumference and area have to be kept apart.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Applying circle area and circumference formulas is the grade 7 standard, and subtracting one area from another is a common grade 7 exercise. Working backward from a circumference adds one step but no new content.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Something you must already know moderate
Solution: 4 steps
Reading level: Question and choices: about grade 6, 62 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question describes the two plates in different measures, one by radius and one by circumference, so students must convert before they can compare. The trap is a halving argument: the small circumference is half the large one, but the area it covers is only a quarter. Students who keep both areas in terms of π and subtract finish quickly; those who reason from the circumferences alone go wrong.

Many steps moderate Working it out
The chain runs through four stages: the large area 400π, the small radius 10 from the circumference 20π, the small area 100π, then the uncovered difference. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Something you must already know moderate Working it out
Neither formula is given: students need C = 2πr to recover the small plate's radius and A = πr² for both areas. (Anderson & Krathwohl, 2001)

Strategy: Use C = 2πr to get the small plate's radius, find both areas with A = πr², keep everything in terms of π, and subtract.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

A Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Works with circumferences instead of areas, subtracting the small plate's 20π from the large plate's 40π.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Finds the area of the smaller plate, the part that is covered, instead of the part of the large plate left uncovered.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Reasons that a circumference half as long covers half the area, so takes half of 400π, not seeing that area scales with the square.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle: $C = 2 \times \pi \times r$; C is the circumference, the distance around the circle. r is the radius, the distance from the center to the edge. π, said “pie,” is a number a little more than 3. The distance around is always about 6.28 times the radius.
Area of a circle: $A = \pi \times r^2$; A is the area, the space inside the circle. r is the radius. r² means r times r. π is about 3.14. The radius is squared because area spreads out in two directions.

Write down what you know
The big plate has a radius of 20 cm. The radius is the distance from the center to the edge.
The small plate has a circumference of 20π cm. Circumference is the distance around a circle.
You need the area of the big plate that the small plate does not cover. Area is the space inside a shape.
Find the big plate's area
$A = \pi \times r^2$
$A = \pi \times 20^2 = \pi \times 400 = 400\pi$
The big plate is 400π sq cm. Keep π as a symbol, because the choices use it.
Find the small plate's radius
You know the small plate’s circumference, not its radius. Use the circumference formula to work back.
$C = 2 \times \pi \times r$
$20\pi = 2 \times \pi \times r$
Divide both sides by 2π. That leaves $r = 10$ . The small radius is 10 cm.
Find the small plate's area
Use the area formula again, with the small radius.
$A = \pi \times 10^2 = 100\pi$
The small plate covers 100π sq cm.
Watch out: 100π is the covered part. The question asks for the part not covered, so do not stop at choice B.
Subtract to find the uncovered part
$400\pi - 100\pi = 300\pi$
The uncovered area is 300π sq cm.
Watch out: Do not subtract the distances around the plates. That gives 40π − 20π = 20π, choice A.
So the answer is D, 300π sq cm.

Check your answer

The small radius is half the big radius. So the small area is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ of the big area. That leaves three quarters uncovered: $\frac{3}{4} \times 400\pi = 300\pi$ . It matches.

More questions like this

The original DOE question

Question 74 as printed on page 156 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 156, question 74. Open page 156 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 13 of 13

2018 Test B Math, question 43

Question

$A quarter circle whose center, marked with a dot, is at the bottom left corner. A segment labeled x + 1 runs from the center to a point on the arc, and from that point a vertical segment drops to the bottom edge, meeting it at a right angle. Along the bottom edge, the part from the center to the foot of that segment is labeled y, and the part from the foot to the end of the arc is labeled 2.$

In the quarter circle above, what is $y$ in terms of $x$ ?

A $x-1$
B $x+1$
C $\frac{x+1}{2}$
D $\sqrt{\frac{\left ( x+1 \right )^{2}}{2}}$

Show the answer

Correct answer: A

Both $x+1$ and $y+2$ are radii.

So, set them equal to each other and solve for $y$ .

$y+2=x+1$

$y=x-1$

Question analysis

In a quarter circle, one segment is labeled x + 1 and the bottom edge is split into lengths y and 2, and students must write y in terms of x. It tests basic circle facts together with solving a simple equation. The challenge is seeing which marked lengths must be equal.

NYS standards: 7.G.4 circle area and circumference
7.EE.4 equations and inequalities from problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 circle work centers on area and circumference formulas. This needs only the definition of a radius and a one-step equation, but spotting the relationship in a figure is less routine than plugging into a formula.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Reading a figure, table or graph major
Variables instead of numbers minor
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 9, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The figure looks like a right-triangle problem, and that is the trap: the right angle and the slanted radius pull students toward the Pythagorean theorem, which cannot work without the vertical leg. The solution needs only the fact that all radii of a circle are equal, applied to the bottom edge. Students who mark every segment that starts at the center see it at once.

Reading a figure, table or graph major Understanding the question
The labels sit on two different radii: x + 1 on the slanted one, and y + 2 along the bottom edge, which runs from the center all the way to the arc. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Variables instead of numbers minor Understanding the question
The answer is an expression in x, so there is no number to check; trying x = 5, which makes the radius 6 and y = 4, is the fallback. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
x + 1 differs from the right x − 1 by a single sign; only seeing that y plus the 2-unit piece makes a radius, so the 2 is subtracted, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Look for two lengths in the figure that are both radii of the quarter circle, set their expressions equal, and solve for y.

Traps

Trusting the picture: The figure suggests something it does not state; work from the given facts.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Sets y equal to the whole radius x + 1, ignoring the 2-unit piece that completes the bottom radius beyond y.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Treats y and the 2 as factors, y × 2 = x + 1, and divides by 2 where it should subtract 2.
D Trusting the picture (Lesh, Post & Behr, 1987)
Reveals a misconception Expected pull: strong
Trusts the drawing's nearly 45° radius, assumes the triangle's two legs are equal, and solves y² + y² = (x + 1)² for y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Radii of a circle are equal: $r_1 = r_2$; A radius is a segment from the center of a circle to its curved edge. r₁ and r₂ stand for any two radii of the same circle. They are always equal, because every point on the curve is the same distance from the center.

Read the figure
The shape is a quarter circle, one fourth of a whole circle. Its center is the corner with the dot. A radius is a segment from the center to the curved edge. The slanted segment, labeled $x+1$ , goes from the center to the curve. The bottom edge is split into two parts, $y$ and 2.
Spot the two radii
The slanted segment is a radius, so the radius is $x+1$ . The whole bottom edge also runs from the center to the curve. Its length is $y+2$ , so it is a radius too.
Watch out: the slanted radius only looks like it is at 45 degrees. Do not assume the triangle’s two short sides are equal. That idea leads to choice D.
Set the radii equal
Use the rule that all radii of a circle are equal.
$y + 2 = x + 1$
Now you have one equation with $x$ and $y$ .
Solve for y
Get $y$ alone by taking 2 from both sides.
$y = x + 1 - 2 = x - 1$
Watch out: $y$ is only part of the bottom radius. Setting $y$ equal to $x+1$ gives choice B. Dividing by 2 instead of subtracting 2 gives choice C.
Give the answer
$y$ is 2 less than the radius, and the radius is $x+1$ . So the answer is A, $x-1$ .

Check your answer

Pick a number. If $x = 5$ , the radius is 6. Then $y = 5 - 1 = 4$ , and $4 + 2 = 6$ , which matches the radius. Also, $y$ is a short side of the right triangle. It should be shorter than the slanted side, and 4 is less than 6.

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The original DOE question

Question 100 as printed on page 161 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 161, question 100. Open page 161 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

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