In the figure above, PQRS is a parallelogram

2018 Handbook

2018 Test A Math · Question 1 of 57

Also in 2026 Practice Test A Math, question 9; 2025 Practice Test A Math, question 10.

Question

$Parallelogram PQRS with side PS on a horizontal line that extends past P and S, and side QR at the top. At S, the angle between side SR and the line extended to the right of S is marked 72°. At Q, a vertical line rises above the figure at a right angle to side QR (right-angle mark), and angle x° is marked on the left of Q, between that vertical line and side QP.$

In the figure above, PQRS is a parallelogram. What is the value of $x$ ?

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

Show the answer

Correct answer: 162

First, find the measure of angle PQR. The measure of angle PQR is equal to the measure of angle PSR.

m∠PSR = 180 − 72 = 108.

So, the measure of angle PQR is also 108.

108 + 90 + x = 360

198 + x = 360

x = 162

Question analysis

A grid-in geometry question about angles around a parallelogram PQRS. A 72° angle sits outside corner S on the extended base, and at corner Q a right angle and the unknown angle x complete the turn around the point. Students chain three angle facts, and the angle asked for lies outside the shape, not inside it.

NYS standards: 7.G.5 angle relationships and equations
8.G.5 angle sums, transversals, angle-angle similarity
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 angle work usually applies one or two facts, such as supplementary or vertical angles, to find an angle. This chains three, and needs the equal opposite angles of a parallelogram, which rest on grade 8 parallel-line facts.
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: Reading a figure, table or graph major
Combines ideas moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 14 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 two subtractions; the difficulty is seeing which angles the marks refer to. Students lose points by treating the 72° as the parallelogram's own angle at S, or by stopping at angle PQR, the most natural number to grid. The question rewards tracing a route from S to Q before computing anything, and checking at the end that x is the angle outside the shape that the arc marks.

Reading a figure, table or graph major Understanding the question
The 72° is marked outside the parallelogram, between side SR and the base extended past S, and x sits outside corner Q, between QP and a vertical ray that meets QR at a right-angle mark. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Combines ideas moderate Working it out
Three angle facts have to be chained: a straight line gives the angle inside S, opposite angles of a parallelogram carry it to Q, and the angles around Q make a full 360° turn. (Embretson, 1998; Daniel & Embretson, 2010)
An answer in an unexpected form minor Choosing the answer
As a grid-in there are no choices to check against, so a student who stops at angle PQR, inside the parallelogram, gets no signal that the job is unfinished. (Enright, Morley & Sheehan, 2002)

Strategy: Use the straight line at S to find the parallelogram's angle there, carry it to the opposite corner Q, then take that angle and the right angle away from 360.

Traps

Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

108 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds angle PSR = 180 − 72 = 108 and carries it to angle PQR, then grids that interior angle instead of the angle x outside it.
198 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: moderate
Reads the 72° as the parallelogram's interior angle at S, so takes angle PQR as 72 and grids 360 − 90 − 72 = 198.
252 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Overlooks the right-angle mark at Q and subtracts only the 108° angle PQR from 360, giving 252.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line: $a + b = 180$; a and b are two angles side by side on a straight line. Together they make a half turn, so they always add up to 180 degrees. Use it when an angle you need shares a straight line with an angle you know.
Opposite angles of a parallelogram: $\angle PQR = \angle PSR$; The symbol ∠ means angle. ∠PQR is the angle at corner Q, between sides QP and QR. ∠PSR is the angle at the opposite corner, S. In a parallelogram, opposite corners always have equal angles, because the shape leans the same way at both.
Angles around a point: $a + b + c = 360$; a, b and c are angles that fit around one point with no gaps and no overlaps. Going all the way around is a full turn, which is 360 degrees. So the angles add up to 360.

Sort out what the figure shows
PQRS is a parallelogram. That is a four-sided shape whose opposite sides are parallel, which means they never meet. You need the value of x, in degrees.
Look closely at the marks. The 72° angle is outside the shape, at S. It sits between side SR and the base line, which keeps going past S. At Q, a vertical line meets side QR. The small square means a right angle, 90°. The angle x is outside the shape too. It runs from that vertical line around to side QP.
Find the angle inside corner S
Angle PSR is the shape’s own angle at S. It sits right next to the 72° angle, on the same straight base line. So the two angles add up to 180°.
$a + b = 180$
$\angle PSR + 72 = 180$
$\angle PSR = 180 - 72 = 108$
So angle PSR is 108°. Watch out: 72° is not the inside angle at S. Using 72 as the inside angle leads to the wrong answer 198.
Carry the angle over to Q
Corners Q and S are opposite each other. In a parallelogram, opposite angles are equal. So angle PQR matches angle PSR.
$\angle PQR = \angle PSR = 108$
So angle PQR is 108°. Watch out: this is not x. Angle PQR is inside the shape, and x is outside it. Stopping here and gridding 108 is the most common mistake.
Add the angles around Q
Three angles fill the space all the way around point Q. They are the right angle (90°), angle PQR (108°) and x. Together they make a full turn.
$a + b + c = 360$
$90 + 108 + x = 360$
$198 + x = 360$
Watch out: don’t skip the right angle at Q. Leaving out the 90 gives 252.
Solve for x
Subtract 198 from 360 to find what is left for x.
$x = 360 - 198 = 162$
So x is 162°. So you would grid 162.

Check your answer

Add the three angles at Q: $90 + 108 + 162 = 360$ . That is a full turn. Also, the arc for x is wider than a right angle but not as wide as a straight line. So x should be between 90° and 180°, and 162° is.

The original DOE question

Question 58 as printed on page 81 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 81, question 58. Open page 81 of the handbook (PDF)