R, S, and T are midpoints of the sides of a square

2018 Handbook

2018 Test B Math · Question 35 of 57

Also in 2026 Practice Test B Math, question 29; 2025 Practice Test B Math, question 35.

Question

$Square MNPQ with N at the top left, P at the top right, Q at the bottom right and M at the bottom left; side PQ is labeled 6 cm. S is on the top side NP, T is on the bottom side MQ, and R is on the left side MN. Segments ST, RS and RT are drawn, and the two triangles RNS (top left corner) and TMR (bottom left corner) are shaded.$

R, S, and T are midpoints of the sides of square MNPQ, as shown above. What is the sum of the areas of the shaded triangles?

A9 sq cm
B12 sq cm
C18 sq cm
D36 sq cm

Show the answer

Correct answer: A

The length of one side of the square is 6 cm. Since $R$ , $S$ , and $T$ are midpoints, then $\overline{TM}$ , $\overline{MR}$ , $\overline{RN}$ , and $\overline{NS}$ are all equal to 3 cm. Triangles $TMR$ and $RNS$ are both right triangles, so the area of one of the triangles is

$\frac{1}{2}\times 3\times 3=\frac{9}{2}$

The triangles are congruent, so the sum of the areas is

$\frac{9}{2}+\frac{9}{2}=9\text{ sq cm}$

Question analysis

Points R, S and T are midpoints of the sides of square MNPQ, whose sides are 6 cm long, and students find the total area of the two shaded triangles in its corners. It tests the area of a triangle. The common slips are forgetting the one-half or measuring the wrong region.

NYS standards: 6.G.1 area of polygons
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Triangle area is grade 6 content, and the corner triangles here are right triangles with easy legs. Typical grade 7 area problems are similar or harder; the only extra step is using midpoints to get the lengths.
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
Solution: 3 steps
Reading level: Question and choices: about grade 1, 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 shaded triangles are right triangles tucked into the corners, so their legs lie along the square's sides and are easy to measure once the midpoints are used. Most lost points come from the one half in the area formula, or from reporting a larger region of the figure, such as the left half or the whole square. Labeling each 3 cm leg on the figure before computing pays off here.

Reading a figure, table or graph moderate Understanding the question
Only one length is labeled, the 6 cm side PQ; the legs NR, NS, MR and MT must each be read as half a side, 3 cm, from the word midpoints. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The area of a triangle, one half times base times height, is not given, and leaving out the one half doubles each 4.5 sq cm triangle. (Anderson & Krathwohl, 2001)

Strategy: Use the midpoints to get each triangle's legs as half the square's side, apply one-half base times height, and remember there are two shaded triangles.

Traps

A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
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 Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Adds the four 3 cm legs of the two shaded triangles, 3 + 3 + 3 + 3 = 12, treating a total length as an area.
C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 3 by 3 for each triangle without the one half, 9 + 9 = 18, which also equals the area of the left half, rectangle NSTM.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Squares the only labeled length, 6 × 6 = 36, giving the area of the whole square instead of the two shaded corners.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, the side the triangle sits on. h is the height, the straight-up distance from the base to the top corner. You take half because a triangle is half of a rectangle with the same base and height.

Read the figure
MNPQ is a square, and side PQ is 6 cm. All sides of a square are equal, so every side is 6 cm. A midpoint is the point exactly halfway along a side. R is the midpoint of side NM. S is the midpoint of NP, and T is the midpoint of MQ.
The shaded triangles are RNS at the top and RMT at the bottom.
Find the short sides
Each midpoint cuts a 6 cm side into two equal parts.
$6 \div 2 = 3$
So NR, NS, RM and MT are each 3 cm long.
Find the area of one triangle
Triangle RNS has a square corner at N. A square corner is a right angle. So NS can be the base and NR the height.
$A = \frac{1}{2} \times 3 \times 3 = 4.5$
One triangle is 4.5 sq cm. Triangle RMT has the same sides, so it is also 4.5 sq cm.
Watch out: do not forget the one half. Without it, each triangle is 9 sq cm and the total is 18. That is choice C.
Add the two triangles
There are two shaded triangles, so add their areas to get the total.
$4.5 + 4.5 = 9$
So the answer is A, 9 sq cm.

Check your answer

The left half of the square is rectangle NSTM, 3 cm by 6 cm. Its area is $3 \times 6 = 18$ sq cm. The white triangle RST fills half of it. So the shaded part is the other half, 9 sq cm. Choice D, 36, is the whole square.

The original DOE question

Question 92 as printed on page 159 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 159, question 92. Open page 159 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).