SHSAT Sunset

Practice Test

Trusting the picture

Change this testNew TestYour practice tests

Question 1 of 4

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?

  • A12
  • B15
  • C17
  • D19
Show the answer

Correct answer: D

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

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

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

x+5=24x + 5 = 24

x=19x = 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.
What makes it hard
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Radius and diameter
d=2rd = 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.

  1. 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.

  2. 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.

  3. Find the diameter RT

    RT passes through the center, so it is a diameter. Use the formula with r = 12:

    RT=2r=2×12=24\mathrm{RT} = 2r = 2 \times 12 = 24

    RT is 24 long. It is the radius RP plus the radius PT.

  4. Solve for x

    RT is also split into x and 5, so:

    x+5=24x + 5 = 24

    Subtract 5 from both sides:

    x=24−5=19x = 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.

  5. 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 2 of 4

2018 Test B Math, question 5

Question

Four straight lines cross at point P: one horizontal, one vertical and two slanted. Above the horizontal line, from right to left: a 60° angle between the horizontal line and the slanted line rising to the right, a 30° angle between that line and the vertical line, an unlabeled angle between the vertical line and the slanted line rising to the left, and the angle y° between that line and the horizontal line. Below the horizontal line, a 25° angle lies between the vertical line and the slanted line falling to the right.

Four straight lines intersect at point P as shown above. What is the value of yy ?

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

Show the answer

Correct answer: 65

Call the missing angle in the top half of the figure xx.

The sum of the four angles on the top of the figure is equal to 180°.

x+y+30+60=180x+y+30+60=180

Since xx is a vertical angle with the 25° angle, then xx is also 25°. Use that to solve for yy.

25+y+30+60=18025+y+30+60=180

y+115=180y+115=180

y=65y=65

Question analysis

This grid-in angle question shows four straight lines crossing at a single point P and asks for the size of an unknown angle y. Students combine vertical angles with the fact that angles along a straight line add to 180 degrees. One angle they need is unlabeled and must be found from its vertical partner.

NYS standards
  • 7.G.5 angle relationships and equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Using vertical angles and straight angles to write an equation for an unknown angle is a core grade 7 geometry task. Four lines create more angles than usual, but the reasoning stays at grade 7.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 4, 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 arithmetic is one subtraction; the work is in seeing the figure correctly. Students have to notice that a fourth angle sits between the 30° angle and y, recognize it as the vertical partner of the 25° angle, and then use the straight line. The drawing is not to scale: y is drawn close to the 60° angle across the vertical line, which rewards anyone who computes rather than eyeballs.

  1. Reading a figure, table or graph major Understanding the question

    Four lines make eight angles around P but only three are labeled; above the horizontal line sit 60°, 30°, an unlabeled angle and y°, and that unlabeled angle is easy to overlook. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas moderate Working it out

    The solution joins two facts: the unlabeled angle is vertical to the 25° angle below P, and the four angles above the horizontal line make a straight angle of 180°. (Embretson, 1998; Daniel & Embretson, 2010)

  3. An answer in an unexpected form minor Choosing the answer

    As a grid-in, nothing confirms the result, and the drawing shows y at about 60°, so a reading of the picture produces a number that looks right. (Enright, Morley & Sheehan, 2002)

Strategy: Pick a straight line through P and list the angles on one side of it; replace the unlabeled one with the labeled angle vertical to it, then subtract from 180.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line
a+b+c+d=180a + b + c + d = 180
a, b, c and d are angles that sit side by side along one side of a straight line, all meeting at one point. Together they make a half turn, which is 180 degrees. This works for any number of angles, not just four.
Vertical angles
∠1=∠2\angle 1 = \angle 2
When two lines cross, they make an X. Angle 1 and angle 2 are opposite each other across the crossing point. Angles like these are called vertical angles. They are always equal.

  1. Look closely at the figure

    Four lines cross at point P. Look above the flat, or horizontal, line. From right to left there are four angles: 60°, 30°, an angle with no label, and y°.

    Below the horizontal line there is a 25° angle. You need to find y.

  2. Find the unlabeled angle

    The unlabeled angle sits between the up-and-down, or vertical, line and the slanted line rising to the left.

    That slanted line keeps going through P and down to the right. The 25° angle sits between that same line and the vertical line, below P.

    So the two angles are opposite each other across P. They are vertical angles, so they are equal. The unlabeled angle is 25°.

  3. Add the angles above the line

    The four angles above the horizontal line fill one side of a straight line. So they add up to 180°.

    a+b+c+d=180a + b + c + d = 180

    y+25+30+60=180y + 25 + 30 + 60 = 180

    y+115=180y + 115 = 180

  4. Solve for y

    Take 115 away from 180 to get y alone.

    y=180−115=65y = 180 - 115 = 65

    Watch out: Do not skip the unlabeled angle. Using only 30 and 60 gives 90. Do not stop at the sum, 115, either. And do not guess from the picture. Angle y looks like 60°, but the drawing is not exact.

    So you would grid 65.

Check your answer

Add all four angles above the horizontal line: 65+25+30+60=18065 + 25 + 30 + 60 = 180. They fill a straight line exactly, so 65 fits.

More questions like this

The original DOE question
Question 62 as printed on page 153 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 153, question 62. Open page 153 of the handbook (PDF)

Question 3 of 4

2018 Test B Math, question 7

Question

Parallelogram PQRS with Q and R along the top and P and S along the bottom. Point T is on the bottom side PS, and segment QT cuts off triangle PQT. Angle PQT, at Q, is marked 50°; angle PTQ, at T, is marked 70°; angle QRS, at R, is marked with an arc and no measure.

In the figure above, PQRS is a parallelogram. The measure of ∠PQT is 50°, and the measure of ∠PTQ is 70°. What is the measure of ∠QRS?

  • A60°
  • B70°
  • C80°
  • D120°
Show the answer

Correct answer: A

Find the missing angle, angle QPT, of triangle PQT:

180° − 70° − 50° = 60°

In parallelogram PQRS, angle QPT is congruent to angle QRS, so the measure of angle QRS is also 60°.

Question analysis

This geometry question uses parallelogram PQRS and a triangle PQT inside it, giving two of the triangle's angles and asking for angle QRS. Students need the triangle angle sum and the fact that opposite angles of a parallelogram are equal. The challenge is connecting the angles found in the triangle to the angles of the parallelogram.

NYS standards
  • 8.G.5 angle sums, transversals, angle-angle similarity
  • 7.G.5 angle relationships and equations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 angle work uses vertical, adjacent and supplementary angles. This also needs the triangle angle sum and the opposite angles of a parallelogram, ideas New York places in grade 8, though the arithmetic is short.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 4, 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

Few steps are needed, but the figure misleads. It draws triangle PQT as isosceles, so the angle at P looks the same as the marked 70°, and the angle at R looks the same again; a student who trusts the drawing never computes. Those who do compute the angle at P must still choose the opposite angle of the parallelogram, not the angle next to it, which is where 120° comes from.

  1. Reading a figure, table or graph major Understanding the question

    The figure carries facts the stem leaves out: T lies on side PS, the 50° is only part of the angle at Q, and the 70° is at T, not at a corner of the parallelogram. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas moderate Working it out

    Two separate facts link up: the angles of triangle PQT sum to 180°, and the angle at P, once known, equals the opposite angle at R of parallelogram PQRS. (Embretson, 1998; Daniel & Embretson, 2010)

  3. A very close wrong answer moderate Choosing the answer

    60° and 70° are close enough that judging the angle at R by eye cannot separate them, and the drawing even favors 70°; only computing the triangle's third angle does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  4. Something you must already know minor Working it out

    Neither the triangle angle sum nor the rule that opposite angles of a parallelogram are equal is given; students must bring both to the figure. (Anderson & Krathwohl, 2001)

Strategy: Find the third angle of triangle PQT from the angle sum, work out which angle of the parallelogram it matches, then use the parallelogram's angle facts to reach angle QRS.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles in a triangle
a+b+c=180a + b + c = 180
a, b and c are the three angles inside any triangle. They always add up to 180 degrees. If you know two of the angles, subtract them from 180 to find the third.
Opposite angles of a parallelogram
∠P=∠R\angle P = \angle R
A parallelogram is a four-sided shape whose opposite sides are parallel, meaning they never meet. In parallelogram PQRS, the corners at P and R are opposite each other. Opposite angles of a parallelogram are always equal.

  1. Sort out the figure

    PQRS is a parallelogram. That is a four-sided shape whose opposite sides are parallel, meaning they never meet.

    Point T is on the bottom side PS. Segment QT cuts off triangle PQT on the left. In that triangle, angle PQT is 50° and angle PTQ is 70°.

    You need angle QRS, the corner at R.

  2. Find the third angle of the triangle

    Triangle PQT has angles at Q, T and P. You know two of them. The angle at P is angle QPT. All three must add to 180°.

    a+b+c=180a + b + c = 180

    ∠QPT+50+70=180\angle QPT + 50 + 70 = 180

    ∠QPT=180−50−70=60\angle QPT = 180 - 50 - 70 = 60

    So angle QPT is 60°.

  3. Match angle P to angle R

    T lies on side PS. So angle QPT is also the corner of the parallelogram at P. Angle P and angle R are opposite corners. Opposite angles are equal, so angle QRS is also 60°.

    Watch out: Angle Q is next to angle P, not opposite it. Angle Q is 120°, choice D, but the question asks about angle R. Also, do not trust the drawing. Angle R may look like 70°, choice B.

    So the answer is A, 60°.

Check your answer

The four corners of the parallelogram are 60°, 120°, 60° and 120°. They add to 360°, as the angles of any four-sided shape must. Also, the 50° angle at Q fits inside the 120° corner at Q.

More questions like this

The original DOE question
Question 64 as printed on page 154 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 154, question 64. Open page 154 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 4 of 4

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 yy in terms of xx ?

  • Ax−1x-1
  • Bx+1x+1
  • Cx+12\frac{x+1}{2}
  • D(x+1)22\sqrt{\frac{\left ( x+1 \right )^{2}}{2}}
Show the answer

Correct answer: A

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

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

y+2=x+1y+2=x+1

y=x−1y=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.
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.

  1. 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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Radii of a circle are equal
r1=r2r_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.

  1. 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+1x+1, goes from the center to the curve. The bottom edge is split into two parts, yy and 2.

  2. Spot the two radii

    The slanted segment is a radius, so the radius is x+1x+1. The whole bottom edge also runs from the center to the curve. Its length is y+2y+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.

  3. Set the radii equal

    Use the rule that all radii of a circle are equal.

    y+2=x+1y + 2 = x + 1

    Now you have one equation with xx and yy.

  4. Solve for y

    Get yy alone by taking 2 from both sides.

    y=x+1−2=x−1y = x + 1 - 2 = x - 1

    Watch out: yy is only part of the bottom radius. Setting yy equal to x+1x+1 gives choice B. Dividing by 2 instead of subtracting 2 gives choice C.

  5. Give the answer

    yy is 2 less than the radius, and the radius is x+1x+1. So the answer is A, x−1x-1.

Check your answer

Pick a number. If x=5x = 5, the radius is 6. Then y=5−1=4y = 5 - 1 = 4, and 4+2=64 + 2 = 6, which matches the radius. Also, yy is a short side of the right triangle. It should be shorter than the slanted side, and 4 is less than 6.

More questions like this

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).