7475 /www/nrich/html/content/id/7475/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <ul id="buttonBar"> <li><a href="http://nrich.maths.org/841">Warm-up problem</a></li> <li><a href="http://nrich.maths.org/1408">Try this next</a></li> <li><a href="https://nrich.maths.org/discus/messages/27/27.html?1274115048">Ask NRICH</a></li> <li><a href="http://nrich.maths.org/1412">Read all about it</a></li> <li><a href="http://nrich.maths.org/7474&amp;part=solution">Last week&#39;s solution</a></li> </ul> <div> <div> </div> <mdo:image alt="onion4" src="onion4.jpg" style="width: 326px; height: 338px; float: left; margin: 10px;"></mdo:image> <div>Imagine the circle filled with infinitely many concentric rings. </div> <div> </div> <div>Now focus on the first quadrant. Imagine straightening all the quarter circles into vertical strips. Why do all the tops of the strips lie on the line $y = \frac{1}{2}\pi x$?  Remember there are infinitely many strips so  they fill a triangular area as shown in the diagram.</div> <div> </div> <div>Why is it that the arcs are not stretched in this process, only straightened, so that the area of each strip stays the same when you straighten each arc into a straight line?</div> <div> </div> <div>What does this tell you about the area of the circle?</div> <div> </div> <div>The &#39;strings&#39; could be longer. Suppose instead of straightening quarter circles you straightened arcs which were whole circles. Then on what line would the tops of the straight &#39;stand up arcs&#39; lie? Again the area of the triangle packed with strips made of straightened arcs of circles is equal to the area of the circle and this gives the formula for the area of the circle. </div> <div> </div> </div> <div class="framework"><span style="font-style: italic;">Did you know ... ?</span><br></br> Radians are a more natural measure for angles than degrees. One radian is the angle subtended at the centre of a circle by an arc that is equal in length to the radius of the circle. The length $s$ of an arc is proportional to the radius $r$ and also to the angle $\theta$ that the arc subtends at the centre of the circle, that is $s = r\theta$.  It follows that the magnitude in radians of one complete revolution (360 degrees) is the length of the entire circumference divided by the radius, or $\frac{2\pi r }{r}$ or $2\pi$. Thus $2\pi$ radians is equal to 360 degrees, meaning that one radian is equal to $\frac{180}{\pi}$ degrees.</div> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> The triangle made up of the stand up arcs has area $\frac{1}{2}\times r \times \frac{1}{2}\pi r$ which is $\frac{1}{4} \pi r^2$. This is equal to the area of a quarter of the unit disc so the unit disc has area $\pi r^2$.<br></br>  <br></br> Suppose you straightened complete circuar arcs.Then the tops would lie on the line $y = 2 \pi x$ and the area of the triangle made from these stand up arcs would be $\frac{1}{2}\times r \times2 \pi r$ which is $\pi r^2$. This gives the area of the whole unit disc.<br></br> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> The triangle made up of the stand up arcs has area $\frac{1}{2}\times r \times \frac{1}{2}\pi r$ which is $\frac{1}{4} \pi r^2$. This is equal to the area of a quarter of the unit disc so the unit disc has area $\pi r^2$.<br></br>  <br></br> Suppose you straightened complete circuar arcs.Then the tops would lie on the line $y = 2 \pi x$ and the area of the triangle made from these stand up arcs would be $\frac{1}{2}\times r \times2 \pi r$ which is $\pi r^2$. This gives the area of the whole unit disc.<br></br> <br></br></mdoxml> 2 3 0 0 0 1 1 A weekly challenge: these are shorter problems aimed at Post-16 students or enthusiastic younger students. Collections Weekly Challenge Admin Stage 5 - Core Mapping Stage 5 Core Mapping Document Trigonometry AS Secondary Mapping Document DisplayCabinet