511 /www/nrich/html/content/97/06/six2/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <p>Draw two circles, each of radius $1$ unit, so that each circle goes through the centre of the other one. What is the area of the overlap?</p> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> The required area is shown below in fig. $1$.<br></br> <mdo:image align="texttop" alt="" border="0" src="fig1.gif"></mdo:image><br></br> <p>To find the area it may help to consider just one of the circles and the sector created.</p> <p>The angle in the sector is $120$ degrees. (See fig. $2$.)</p> <mdo:image align="texttop" alt="" border="0" src="fig2.gif"></mdo:image><br></br> Hence the area of the sector is $\frac{120}{360} \times\pi \times1 \times1 = \frac{\pi}{3}$.<br></br> <br></br> Now by considering the area of the triangle in fig. $3$, we can find the area of the shaded segment.<br></br> <br></br> <mdo:image align="texttop" alt="" border="0" src="fig3.gif"></mdo:image><br></br> The area of the triangle $= 0.5 \times\sin 120^{\circ} = \frac{\sqrt{3}}{4}$ So the area of the shaded segment $= \left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right)$ square units. The area of the overlap is twice this amount which is : $ 2\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$ square units.<br></br> <br></br></mdoxml> 2 4 0 0 0 1 0 Draw two circles, each of radius 1 unit, so that each circle goes through the centre of the other one. What is the area of the overlap? Measures and Mensuration Area Admin Short problems 2D Geometry, Shape and Space Pythagoras' theorem Measures and Mensuration Arcs, sectors and segments