Quarter circle problem | srcav

Here is a problem I have had on my classroom wall for a long time. I have a large display of problems on there that sometimes the students present me solutions to. This is one no one yet had done and I had not attempted.

Last week I was discussing the problem wall with a colleague and this jumped out at me, so I thought I would give it a try. It took rather longer than I’d care to admit, to be honest. I set off on a few false starts and came up with some incorrect solutions due to an incorrect solution I’d made. After a while I gave up and left it a few days before tackling it afresh….

Continue reading at:

http://ift.tt/2F79Eyr

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Quarter circle problem | srcav

Here is a problem I have had on my classroom wall for a long time. I have a large display of problems on there that sometimes the students present me solutions to. This is one no one yet had done and I had not attempted.

Last week I was discussing the problem wall with a colleague and this jumped out at me, so I thought I would give it a try. It took rather longer than I’d care to admit, to be honest. I set off on a few false starts and came up with some incorrect solutions due to an incorrect solution I’d made. After a while I gave up and left it a few days before tackling it afresh….

Continue reading at:

http://ift.tt/2F79Eyr

A lovely circle problem – two ways | srcav

So, I was working with some year 12s on a few problems around circles out of the new Pearson A Level textbook. (Incidentally, it’s this book, and I think it’s probably the best textbook I’ve come across. I would certainly recommend it.)

This question appears in a mixed exercise on circles:

 

 

It’s a lovely question. Before reading on, have a go at it – or at least have a think about what approach you’d take –  as I’m going to discuss a couple of methods and I’d be interested to know how everyone else approached it.

Method 1:

I looked at this problem and saw right angled triangles…

Continue reading at:

http://ift.tt/2BeLQWR

A lovely circle problem – two ways | srcav

So, I was working with some year 12s on a few problems around circles out of the new Pearson A Level textbook. (Incidentally, it’s this book, and I think it’s probably the best textbook I’ve come across. I would certainly recommend it.)

This question appears in a mixed exercise on circles:

 

 

It’s a lovely question. Before reading on, have a go at it – or at least have a think about what approach you’d take –  as I’m going to discuss a couple of methods and I’d be interested to know how everyone else approached it.

Method 1:

I looked at this problem and saw right angled triangles…

Continue reading at:

http://ift.tt/2BeLQWR

Angle problem | srcav

Today has been quite a geometric based day for me. I spent a couple of hours solving non-RAT trigonometry problems with year 10 and then a while with year 11 looking at various algebra angle problems. Then I went on Twitter and saw this from Ed Southall (@solvemymaths):

A couple of nice parallel lines questions that I might grow at y11 tomorrow.

Both are fairly straight forward to solve. I looked at the first one, imagines a third parallel line through the join if x and saw x must be the sum of 40 and 60 hence 100.

The second I saw an alternate angle to the 50 in the top triangle and…

Continue reading at:

http://ift.tt/2BASEua

Angle problem | srcav

Today has been quite a geometric based day for me. I spent a couple of hours solving non-RAT trigonometry problems with year 10 and then a while with year 11 looking at various algebra angle problems. Then I went on Twitter and saw this from Ed Southall (@solvemymaths):

A couple of nice parallel lines questions that I might grow at y11 tomorrow.

Both are fairly straight forward to solve. I looked at the first one, imagines a third parallel line through the join if x and saw x must be the sum of 40 and 60 hence 100.

The second I saw an alternate angle to the 50 in the top triangle and…

Continue reading at:

http://ift.tt/2BASEua

Saturday puzzle | srcav

One of the first things I saw this morning when I awoke was this post from solve my maths on facebook:

That’s interesting I thought, so I thought it have a go.

The radius is given to be 2. So we have an equilateral triangle side length 2. Using my knowledge of triangle and exact trig ratios I know the height of such a triangle is root 3 and as such so is the area. 

Similarly, as the diagonal of the rectangle is 2 and the short side is 1 we can work out from Pythagoras’s Theorem that the longer side is root 3. And again it follows that so is the area.

Lastly we have the square, the…

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