few_shots: 
Example:
Problem/activity:
Task
Cruz and Erica were both getting ready for soccer.
Cruz ran 1 lap around the school.
Erica ran 3 laps around the playground.
Erica said,
I ran more laps, so I ran farther.
Cruz said,
4 laps around the school is 1 mile, but it takes 12 laps around the playground to go 1 mile. My laps are much longer, so I ran farther.
Who is right? Draw a picture to help you explain your answer.


Concept/skill: 
Explain why a fraction a/b is equivalent to a fraction (n × a)/(n × b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions.
Grade 4 expectations in this domain are limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100.

Answer: 
Yes. This task asks students to compare two fractions that are equal. The students need to explain why the two fractions are equal by drawing a picture, which involves using visual fraction models.

Now, assess whether the following problem or activity is 'aligned' with the given Common Core standard in Mathematics.

shot_info: im_task_000212_pos_0

4
standard_description: Recognize and draw shapes having specified attributes, such as a given number of angles or a given number of equal faces. (Sizes are compared directly or visually, not compared by measuring.) Identify triangles, quadrilaterals, pentagons, hexagons, and cubes.
problem_activity: Task
Rhianna has learned the SSS  and SAS congruence tests for triangles and she wonders if these tests might work for parallelograms.
Suppose $ABCD$ and $EFGH$ are two parallelograms all of whose corresponding sides are congruent, that is $|AB| = |EF|, |BC| = |FG|, |CD| = |GH|,$ and $|DA| = |HE|$.  Is it always true that $ABCD$ is congruent to $EFGH$?
Suppose $ABCD$ and $EFGH$ are two parallelograms with a pair of congruent corresponding sides, $|AB| = |EF|$ and $|BC| = |FG|$.  Suppose also that the included angles are congruent,
$m(\angle ABC) = m(\angle EFG)$.  Are $ABCD$ and $EFGH$ congruent?


aligns: no