few_shots: 
Example 1:
Concept/skill: 
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.

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


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:
Concept/skill: 
Solve quadratic equations in one variable.

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


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

Example 3:
Concept/skill: 
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.

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 $P$.


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, determine whether the following problem or activity can enable students to understand the concepts or skills in 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: Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b.
problem_activity: Activity
In this activity, students find the missing dimensions of cylinders when given the volume and the other dimension. A volume equation representing the cylinder is given for each problem.
Identify students who use these strategies: guess and check, divide each side of the equation by the same value to solve for missing variable, or use the structure of the volume equation to reason about the missing variable
Launch
Arrange students in groups of 2. Give students 2–3 minutes of quiet work time followed by time to share their explanation for the first problem with their partners.
Representation: Internalize Comprehension.
Activate or supply background knowledge of strategies students can use to solve for unknown variables in equations. Allow students to use calculators to ensure inclusive participation in the activity.
Supports accessibility for: Memory; Conceptual processing
Writing, Conversing: MLR1 Stronger and Clearer Each Time.
Use this routine to give students a structured opportunity to revise a response to one or both of the questions. Ask each student to meet with 2–3 other partners in a row for feedback. Provide student with prompts for feedback that will help students strengthen their ideas and clarify their language (e.g., “What did you do first?”, “How do you know
\(h\)
is...?”, and “How did you use the volume formula?”, etc.). Students can borrow ideas and language from each partner to strengthen their final explanation.
Design Principle(s): Optimize output (for explanation)
Student Facing
The volume
\(V\)
of a cylinder with radius
\(r\)
is given by the formula
\(V=\pi r^2h\)
.
The volume of this cylinder with radius 5 units is
\(50\pi\)
cubic units. This statement is true:
\( 50\pi = 5^2 \pi h\)
[IMG]
What does the height of this cylinder have to be? Explain how you know.
The volume of this cylinder with height 4 units is
\(36\pi\)
cubic units. This statement is true:
\(36\pi = r^2 \pi 4\)
[IMG]
What does the radius of this cylinder have to be? Explain how you know.
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?
Suppose a cylinder has a volume of
\(36\pi\)
cubic inches, but it is not the same cylinder as the one you found earlier in this activity.
What are some possibilities for the dimensions of the cylinder?
How many different cylinders can you find that have a volume of
\(36\pi\)
cubic inches?
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Extension Student Response.
Activity Synthesis
Select previously identified students to explain the strategies they used to find the missing dimension in each problem. If not brought up in students’ explanations. Discuss the following strategies and explanations:
Guess and check: plug in numbers for
\(h\)
, a value that make the statements true. Since the solutions for these problems are small whole numbers, this strategy works well. In other situations, this strategy may be less efficient.
Divide each side of the equation by the same value to solve for the missing variable: for example, divide each side of
\(36\pi=r^2 \pi 4\)
by the common factor,
\(4\pi\)
. It’s important to remember
\(\pi\)
is a number that can be multiplied and divided like any other factor.
Use the structure of the equation to reason about the missing variable: for example,
\(50\pi\)
is double
\(25\pi\)
, so the missing value must be 2.


aligns: no