few_shots: 
Example 1:
Problem/activity:
Find the general solution for:
$${\mathrm{cos}\left({\pi\over2}+x\right)=\mathrm{tan}{\pi\over4}}$$


Description: 
Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles.

Answer: 
No. This problem does not directly teach how side ratios in right triangles lead to definitions of trigonometric ratios. Instead, it involves solving a trigonometric function.

Example 2:
Problem/activity:
Task
Solve the quadratic equation
$$ x^2 = (2x - 9)^2 $$
using as many different methods as possible.


Description: 
Solve quadratic equations in one variable.

Answer: 
Yes. This task shows one quadratic equation with one variable, x.

Example 3:
Problem/activity:
Task
Suppose $A$ is an angle with vertex $P$, as pictured below:
[IMG]
Draw a circle with center $P$ and with  radius $r \gt 0$.  Explain why the circle
meets each ray of angle $A$ in a single point.  Label these points $Q$ and $R$ respectively.
Draw circles with centers $Q$ and $R$ respectively and radius $r$.  These circles meet at $P$ and a second point to be labelled $B$.
Show that ray $\overrightarrow{PB}$  bisects angle $P$.


Description: 
Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line.

Answer: 
Yes. This task asks students to draw a ray that bisects an angle. The act of drawing circles and rays may involve compass and straightedge or dynamic geometric software.

Now, assess whether the following problem or activity aligns with the provided description.

shot_info: fl_problem_001894_neg_0

HS_0

im_task_000870_pos_0

HS_1

im_task_001160_pos_0

HS_2
standard_description: Derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation.
problem_activity: Optional activity
In this activity, students compute the length of a figure that is composed of half-circles and straight line segments. For the first question, they are given the length of the line segments and the diameter of the circle. For the second question, students have to compute the diameter of the circle.
As students work, monitor for students who decomposed the figure in different ways and used different approximations of
\(\pi\)
. The discussion focuses on whether the differences lead to meaningful differences in the estimate for the distance around the track.
Launch
If desired, introduce the context of a running track. Allow students to choose what approximation to use for
\(\pi\)
. Quiet work time followed by partner discussion.
Action and Expression: Internalize Executive Functions.
Provide students with a graphic organizer to organize the information provided in the problem and to structure their problem-solving strategy. The graphic organizer should include the prompts: “What do I need to find out?”, “What do I need to do?”, and “How I solved the problem.”
Supports accessibility for: Language; Organization
Speaking, Writing: MLR5 Co-Craft Questions.
To help students understand the situation, display the image of the running track, and ask students to write down possible mathematical questions they could ask. Listen for, and amplify language students use to describe different parts of the track (inside and outside), half-circles, straight line segments, etc. Then, invite pairs to share their questions with the class. This will help students produce the language of mathematical questions, and support common understanding of the situation.
Design Principle(s): Maximize meta-awareness; Support sense-making
Student Facing
The field inside a running track is made up of a rectangle that is 84.39 m long and 73 m wide, together with a half-circle at each end.
[IMG]
What is the distance around the inside of the track? Explain or show your reasoning.
The track is 9.76 m wide all the way around. What is the distance around the outside of the track? Explain or show your reasoning.
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Student Response.
Student Facing
Are you ready for more?
This size running track is usually called a 400-meter track. However, if a person ran as close to the “inside” as possible on the track, they would run less than 400 meters in one lap. How far away from the inside border would someone have to run to make one lap equal exactly 400 meters?
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Extension Student Response.
Anticipated Misconceptions
If students have trouble seeing the circle and rectangle that compose the figure, suggest that they draw additional lines to decompose the figure.
Students might apply the formula for the circumference of a circle to find the circumference of the oval-shaped field and track. If this happens, remind students that the track is not in the shape of a circle. Ask if they see a way to form a circle and a rectangle within the space.
Students who are familiar with the fact that this size track is referred to as a 400-meter track may be confused when their answer does not equal 400 meters. Explain that a runner does not run right on the edge of the track and possibly direct the student to look at the extension.
Activity Synthesis
Most of the discussion will occur in small groups. However, the whole class can debrief on the following questions:
“In what ways did you decompose the figure into different shapes?”
“Why did you choose a particular approximation for
\(\pi\)
, and what was the resulting answer?”
“How are different students’ answers related, and are they all reasonable lengths for this situation?”
Students who use more digits for their approximation of
\(\pi\)
may come up with a slightly different answer, such as 398.1. In general, when making calculations, if only an estimate is desired, then using 3.14 for
\(\pi\)
is usually good enough. In a situation like this, where the given measurements are quite precise, it can be worth trying more digits in the expansion of
\(\pi\)
, but it turns out not to make much difference in this case.


aligns: no