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Each book has a matching audio narration mp3. You can also read the books online with the online reader. This site is brought to you by Dream English 2 CD's of original kids songs for free to sing with the whole family! Click here to read Shapes online.

Your Answer

Click here to download the audio narration mp3 of Shapes. Additional Free Resources: Use these free resources together with the I Like Shoes book to make an even better learning experience:. Dream English has a great free mp3 download of a shapes song for kids! Do this systematically. This can only be done with 1 and 2. This can only be done with 1 and 3.

Area and circumference of circles

This can only be done with 2 and 3. So there are only three possibilities:. There are only three sides to the triangle so we have them now. The only question is, which numbers go in the corners?


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But that is easy. These are the numbers that occur twice in the three sums. So we quickly get the four answers that we got by trial and error in the first session. This method of proving that there are precisely four answers may be the nicest one of the lot. It certainly has a nice argument that shows that the sums are restricted to between 9 and 12 inclusive. Here we try to use algebra to see what progress can be made. We start by putting the letters a, b, c, d, e, f into the circles. As shown in the diagram below.

The sum in the second bracket is just the sum of the three corners. So we now have. So we have. So the sum has to lie between 9 and 12, just as we discovered in Method 4 above. But now we have a bonus, because the equation here tells us what the corners sum to. Whatever, we still end up with the four answers that we keep getting. It does rely on knowing a bit of algebra but once over that hurdle it gives the corner circle numbers pretty quickly and the rest of the numbers fall into place.

Find all its answers and show there are no others. Background Here we look at a problem that is similar to the Six Circle Problem. This problem is an extension of the Six Circle Problem. The Eight Circle Problem: Is it possible to put the numbers 1, 2, 3, 4, 5, 6, 7, 8 in the circles so that the sums of the three numbers on either side of the square are the same? The techniques that we have used with the Six Circle Problem can be used again here.

Any of the Methods that we noted in the proof of the four answer conjecture see Background for Session 2 will work here. Clearly though, some are more efficient. For instance, if your students have a reasonable algebraic ability, we suggest that they try Method 5 to get the side sums.

Circles and using a Compass

This method cuts down the number of options more effectively than looking at the biggest sum that 1 can be in. So, your class should find that the only possible sums are 12, 13, 14 and This problem takes a little more work than the Six Circle Problem but at the end, because your students will have approached it systematically, they will know that they have found all possible answers.

Because of the extra work we suggest that your class works in groups of at least 4 and splits up the work between them. Generalise the Six Circle Problem by finding which sets of six numbers can replace 1, 2, 3, 4, 5, 6 and in balancing the sums on either side of the triangle. Problem: What other sets of 6 numbers can be put into the six circles below so that the sums of the numbers on each of the sides are the same? This is a generalisation of the original Six Circle Problem since when we find the answer, it will give us the answer to the original problem too.

What can you come up with? Probably the fact that any six consecutive numbers will work is not too hard to see. Then maybe any six numbers that are the same distance apart like 5, 8, 11, 14, 17, and 20 will work. These can be thought of as any linear combination of 1, 2, 3, 4, 5, 6. But are these the only possibilities? Surprisingly the last one of these works to give two answers. The trick here is to experiment and see what conjectures you can come up with. And when you have done that you will start to see how many answers you get for each set. So then you can ask some more questions.

Can you find sets of numbers for which there is precisely one way to have the sums all equal? What can be said about such sets? Can you find sets of numbers for which there are precisely two ways to have the sums all equal? Can you find sets of numbers for which there are precisely three ways to have the sums all equal?


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  5. Can you find sets of numbers for which there are precisely four ways to have the sums all equal? We know already for the four answers that we got in Session 1, that.

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    You should find that this is true for all possible correct ways to fill the six circles. The difference between opposite corner and middle numbers is the same. So take this and work with it a bit and turn it into a Conjecture. Conjecture: A set of six numbers will correctly fill the six circles if and only if the set consists of three distinct pairs that have a common difference. But how to prove this?

    Well, if we put a, b and c in the corners and their pairs in the opposite middle circle we get the configuration below. But what if we have a legitimate set? Can we show that it must be possible to arrange them in three pairs all with a common difference? Because we have three equal sums, then. So the set consists of three distinct pairs that have a common difference. Discover and prove that in the six Circle problem there are only zero, two or four answers for any set of six numbers.

    Background Here we take up the questions of what sets give 1, 2, 3, or 4 answers. Again experimenting should be done to start with but the clue to the problem is to be found in the list of properties of the answers from Session 1. Remember, number 3 on the list says:. Changing corner numbers to the middle and vice-versa changes one answer into another. Theorem 2. A set of six numbers produces four answers if and only if the set is a linear combination of the numbers 1, 2, 3, 4, 5, 6.

    You can check that the side sums are still equal. Look at the possible 5 cases.

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    This is not possible. These all lead to contradictions or no progress. These all lead to contradictions. First of all it is worth noting that there is a lot of work in this unit and you may want to drop some things out. We suggest that the proof of the four answer case may be one thing that could go as it is conceptually difficult. However, they can get at least as far as the conjecture that there are at most four answers for any set of six numbers. Start the lesson by getting them to think about how many answers any set might produce.


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    This might be done in a whole class setting or in their groups. As usual they should come up with a conjecture first and then try to find a proof of that conjecture. When a proof has been found get a student to present it to the rest of the class. Let them criticise it. If no one is able to find a proof, even with your help, give a proof yourself. Let them criticise that. It might be useful if you are doing this to out in an odd error to keep them on their toes. Let them all write up a proof.

    There are many answers to this question. You might want to restrict the different sums to being consecutive. Your student will tell you what this is. It could be assigned in class to individuals or small groups or given as a homework exercise to generate interesting discussions the following day. The relatively high levels of complexity and technical demand enhance its instructional value.

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    Seventh grade student may require some teacher guidance to be successful with this task. If we take the area of the large circle and subtract the area of the seven small circles, we will be left with all of the area contained in the large circle that is not contained in a small circle, that is, the area around the small circles.