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<h4>Activity (20 minutes)</h4>
<p>The purpose of this task is to motivate the need to write and solve a quadratic equation to solve a problem.</p>
<p>Students are asked to frame a picture by cutting up a rectangular piece of “framing material” into strips and arranging it around a picture such that they create a frame with a uniform width. The framing material has a different length-to-width ratio and is smaller than the picture, so students cannot simply center the framing material on the back of the picture.</p>
<p>Students are not expected to succeed in the task through trial and error. They are meant to struggle, just enough to want to know a way to solve the problem that is better than guessing and checking. Some students may try writing an equation to help them figure out the right measurement for the frame, but they don’t yet have the knowledge to solve the quadratic equation. The solutions to this equation are irrational, so it is also unlikely that students will find them by chance.</p>
<h4>Launch</h4>
<p>Ask students if they ever had an artwork or a picture framed at a frame shop. If students are unfamiliar with custom framing, explain that it is very expensive. The framing materials, which are cut to exact specifications, can be costly. The time and labor needed to properly frame a picture further increase the cost.</p>
<p>Tell students that they are now going to frame a picture, using a sheet of paper as their framing material. The sheet is to be cut such that:</p>
<ul class="os-raise-noindent">
<li> All of it is used. </li>
<li> The framing material does not overlap. </li>
<li> The frame does not cover any part of the picture. </li>
<li> The resulting frame has uniform width all the way around. </li>
</ul>
<p>Distribute to each student:</p>
<ul class="os-raise-noindent">
<li> Scissors </li>
<li> A ruler </li>
<li> The “picture” measuring 7 inches by 4 inches </li>
<li> The “framing material” (colored paper measuring 4 inches by 2.5 inches) </li>
</ul>
<p>Inform students that if they should need it, there will be additional pieces of framing material, in case they wish to start over. Remind them to remove all pieces of the old framing material before cutting the new framing material.</p>
<div class="os-raise-extrasupport">
<div class="os-raise-extrasupport-header">
<p class="os-raise-extrasupport-title">Support for English Language Learners</p>
<p class="os-raise-extrasupport-name">MLR 7 Compare and Connect: Representing, Conversing</p>
</div>
<div class="os-raise-extrasupport-body">
<p> Use this routine to prepare students for the whole-class discussion. Invite students to quietly circulate and examine at least two other “frames” in the room. Give students quiet think time to consider what is the same and what is different about the constructed frames. Next, ask students to find a partner to discuss what they noticed. Listen for and amplify observations that include connections across approaches, challenges faced identifying dimensions of framing material, and the need for more sophisticated strategies for solving this problem using quadratic equations. </p>
<p class="os-raise-text-italicize">Design Principle(s): Cultivate conversation</p>
<p class="os-raise-extrasupport-title">Learn more about this routine</p>
<p>
<a href="https://www.youtube.com/watch?v=PF8fRA107OA;&rel=0" target="_blank">View the instructional video</a>
and
<a href="https://k12.openstax.org/contents/raise/resources/94a1159e7b81493c647515711f325771076d99b8" target="_blank">follow along with the materials</a>
to assist you with learning this routine.
</p>
<p class="os-raise-extrasupport-title">Provide support for students</p>
<p>
<a href="https://k12.openstax.org/contents/raise/resources/0b8a1a4ac3425e84a1d5452b3a5dffa38deb6b13" target="_blank">Distribute graphic organizers</a>
to the students to assist them with participating in this routine.
</p>
</div>
</div>
<br>
<div class="os-raise-extrasupport">
<div class="os-raise-extrasupport-header">
<p class="os-raise-extrasupport-title">Support for Students with Disabilities</p>
<p class="os-raise-extrasupport-name">Representation: Access for Perception</p>
</div>
<div class="os-raise-extrasupport-body">
<p>
Display and read the directions aloud. Students who both listen to and read the information will benefit from additional processing time. Check for understanding by inviting students to rephrase directions in their own words. </p>
<p class="os-raise-text-italicize">Supports accessibility for: Language; Memory</p>
</div>
</div>
<br>
<h4>Student Activity</h4>
<p>Your teacher will give you a picture that is 7 inches by 4 inches, a piece of colored paper that measures 4 inches by 2.5 inches representing framing material, a ruler, and a pair of scissors.</p>
<p>Cut the framing material to create a rectangular frame for the picture. The frame should have the same width all the way around and have no overlaps. All of the framing material should be used with no leftover pieces. The frame should not cover any part of the picture.</p>
<p>Let’s review the information we know.</p>
<ul class="os-raise-noindent">
<li> The picture measures 7 inches by 4 inches. </li>
<li> There are 10 square inches of framing material. </li>
<li> The frame must be uniform in width. </li>
<li> All of the framing material must be used. </li>
<li> The frame should not cover any part of the picture. </li>
</ul>
<p>Describe the process you used to create the frame.</p>
<p>Write down your answer, then select the <strong>solution</strong> button to compare your work.</p>
<p><strong>Answer:</strong> Here are some examples of the process shown in images. Only some of the attempts are correct while others are not.</p>
<p><img height="351" src="https://k12.openstax.org/contents/raise/resources/90d0e858216b9989415bed43da1b63267ae9e45c" width="624"></p>
<p>Now that you have completed the activity, can you think of any mathematical method to solve the problem?</p>
<h4>Student Facing Extension</h4>
<h5>Are you ready for more?</h5>
<p>Juan says, “The perimeter of the picture is 22 inches. If I cut the framing material into 9 pieces, each one being 2.5 inches by \(\frac {4}{9}\) inch, I’ll have more than enough material to surround the picture because those pieces would yield \(9 \cdot 2.5\), or 22.5 inches for the frame.”</p>
<p>Do you agree with Juan? Be prepared to show your reasoning.</p>
<p><strong>Answer:</strong> No, Juan is not correct because 9 pieces would not be enough to go around the frame. While the total length of those pieces is enough to surround the picture, he will need at least two pieces for each short side to cover 4 inches and at least 3 pieces for each long side to cover 7 inches. This adds up to 10 pieces, as shown below.</p>
<p><img height="416" src="https://k12.openstax.org/contents/raise/resources/5fae4a0aa1dabe1a82fbdbd38464f6c6a71d60b5" width="412"></p>
<h4>Activity Synthesis</h4>
<p>Ask a few students to show their “frames.” Consider asking questions such as:</p>
<ul class="os-raise-noindent">
<li>“How did you decide how thick the frame should be?” (Sample answer: I tried a couple of different widths to see if they would work. I first tried \( \frac {1}{2}\) inch, but that didn’t give enough to frame the entire picture. Then, I tried \( \frac {1}{4}\) inch, but that was too thin.) </li>
<li>“How did you know what width would be too large or too small?” (Sample answer: It is hard to tell, but I know that the framing needs to have a linear measurement of at least 22 inches, enough for the entire perimeter of the picture, plus some more length for the corners. If the frame is too thin, there will be extra material. If it is too thick, there won’t be enough to get the 22-plus-some-inch length.) </li>
<li>“How did you determine the width such that no framing materials were left unused? Is it possible to determine this?” (Sample answer: I tried to find a way to evenly spread the area of the framing material around the picture, but I wasn’t quite sure how to do that.) </li>
<li>“What frame width did you end up with?”</li>
<li>“Were all the strips or pieces the same size?”</li>
</ul>
<p>Students are likely to share the challenges they encountered along the way.</p>
<p>Tell students that they will learn strategies that are more effective than trial and error to solve problems such as this one. Using the information they have now, they will be able to write a quadratic equation to represent the given situation, which they can use to calculate the exact width of the picture frame before even cutting anything!</p>
<p>For students who are curious about the solution to the exact width of the picture frame, the solution to the problem is approximately 0.342 inches in width.</p>
<h3>8.1.2: Self Check</h3>
<p class="os-raise-text-bold"><em>After the activity, students will answer the following question to check their understanding of the concepts explored in the activity.</em></p>
<p class="os-raise-text-bold">QUESTION:</p>
<p>Suppose you want to frame a picture that measures 8 inches by 3 inches. The frame is created from 12 square inches of framing material. The frame must be uniform in width, and all of the framing material must be used. Which statement is true?</p>
<table class="os-raise-textheavytable">
<thead>
<tr>
<th scope="col">
Answers
</th>
<th scope="col">
Feedback
</th>
</tr>
</thead>
<tbody>
<tr>
<td>
The frame will measure exactly 8 inches by 3 inches.
</td>
<td>
Incorrect. Let’s try again a different way: The frame cannot have the same dimensions as the picture. Its dimensions will be longer to include the width of the frame. <span>The true statement is that the total area of the framed picture will be 36 square inches: 24 square inches for the picture and 12 square inches for the frame. \(24 + 12 = 36\) square inches.</span>
</td>
</tr>
<tr>
<td>
The total area of the framed picture will be 24 square inches.
</td>
<td>
Incorrect. Let’s try again a different way: The area of the picture by itself is \(8 \cdot 3\), or 24 square inches. The area of the framing material is 12 square inches. To find the total area, these values must be added together. The answer is the total area of the framed picture will be \(24 + 12\), or 36 square inches.
</td>
</tr>
<tr>
<td>
The perimeter of the framed picture will be 22 inches.
</td>
<td>
Incorrect. Let’s try again a different way: The perimeter of the picture by itself is \(8 + 8 + 3 + 3\), or 22 inches. The framed picture will have a larger perimeter than 22 inches. <span>The true statement is that the total area of the framed picture will be 36 square inches: 24 square inches for the picture and 12 square inches for the frame. \(24 + 12 = 36\) square inches.</span>
</td>
</tr>
<tr>
<td>
The total area of the framed picture will be 36 square inches.
</td>
<td>
That’s correct! Check yourself: The area of the picture is \(8 \cdot 3\), or 24 square inches. The area of the frame material is 12 square inches. So, the total area of the framed picture will be \(24 + 12\), or 36 square inches.
</td>
</tr>
</tbody>
</table>
<br>
<h3>8.1.2: Additional Resources</h3>
<p class="os-raise-text-bold"><em>The following content is available to students who would like more support based on their experience with the self check. Students will not automatically have access to this content, so you may wish to share it with those who could benefit from it.</em></p>
<h4>Modeling a Quadratic Problem</h4>
<p>This activity challenges you to complete a very difficult task!</p>
<p>Let’s review the information we know from the picture in the Self Check.</p>
<ul class="os-raise-noindent">
<li> The picture measures 8 inches by 3 inches. </li>
<li> There are 12 square inches of framing material. </li>
<li> The frame must be uniform in width. </li>
<li> All of the framing material must be used. </li>
<li> The frame should not cover any part of the picture. </li>
</ul>
<p>What other information can you learn from the situation?</p>
<ul class="os-raise-noindent">
<li> The perimeter of the picture is \(8 + 8 + 3 + 3\), or 22 inches. The framed picture will have a perimeter greater than this. </li>
<li> If the picture is 8 inches by 3 inches, then the area of the picture is \(8 \cdot 3\), or 24 square inches. </li>
<li> Since there are 12 square inches of framing material and all of it must be used, the framed picture will have an area of \(24 + 12\), or 36 square inches. </li>
</ul>
<p>The more you know about the situation, the closer you are to being able to solve this problem mathematically instead of guessing and checking to see if you are correct.</p>
<p>We will use this information in later activities to set up a quadratic equation that can be used to solve the problem.</p>
<h4>Try It: Modeling a Quadratic Problem</h4>
<p>Suppose you want to frame a different picture that measures 10 inches by 6 inches. The frame is created from 36 square inches of framing material. The frame must be uniform in width, and all of the framing material must be used.</p>
<ol class="os-raise-noindent">
<li>What is the perimeter of the picture?</li>
</ol>
<p>Write down your answers, then select the <strong>solution</strong> button to compare your work.</p>
<h5 class="os-raise-text-bold">Solution</h5>
<p>The perimeter of the picture is \(10 + 10 + 6 + 6 = 32\) inches.</p>
<ol class="os-raise-noindent" start="2">
<li>What is the area of the picture?</li>
</ol>
<p>Write down your answers, then select the <strong>solution</strong> button to compare your work.</p>
<h5>Solution</h5>
<p>The area of the picture is \(10 \cdot 6 = 60\) square inches.</p>
<ol class="os-raise-noindent" start="3">
<li>What will be the total area of the framed picture?</li>
</ol>
<p>Write down your answers, then select the <strong>solution</strong> button to compare your work.</p>
<h5>Solution</h5>
<p>The total area of the framed picture will be \(60 + 36 = 96\) square inches.</p>