https://artofproblemsolving.com/wiki/index.php?title=Cavalieri%27s_principle&feed=atom&action=history Cavalieri's principle - Revision history 2022-01-20T05:31:13Z Revision history for this page on the wiki MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=Cavalieri%27s_principle&diff=161044&oldid=prev Mag1c: Created page with "If the cross sections of two 2D objects at each height have the same length, the areas of the 2D objects are the same. <asy> label("Equal Area:",(2.25,3)); draw((0,0)--(1,0)-..." 2021-08-28T02:01:51Z <p>Created page with &quot;If the cross sections of two 2D objects at each height have the same length, the areas of the 2D objects are the same. &lt;asy&gt; label(&quot;Equal Area:&quot;,(2.25,3)); draw((0,0)--(1,0)-...&quot;</p> <p><b>New page</b></p><div>If the cross sections of two 2D objects at each height have the same length, the areas of the 2D objects are the same.<br /> <br /> &lt;asy&gt;<br /> label(&quot;Equal Area:&quot;,(2.25,3));<br /> draw((0,0)--(1,0)--(1.5,2)--(0.5,2)--cycle);<br /> draw(&quot;$1$&quot;,(0.25,1)--(1.25,1),grey);<br /> draw((3,0)--(4,0)--(4,2)--(3,2)--cycle);<br /> draw(&quot;$1$&quot;,(3,1)--(4,1),grey);<br /> &lt;/asy&gt;<br /> <br /> If the cross sections of two 3D objects at each height have the same area, the volumes of the 3D objects are the same.<br /> <br /> &lt;asy&gt;<br /> size(300);<br /> import three;<br /> <br /> currentprojection=perspective(1,0,0.5);<br /> <br /> label(&quot;Equal Volume:&quot;,(1.5,2));<br /> <br /> pen bg=paleblue+opacity(0.2);<br /> draw(unitcone,bg);<br /> pen bg2=gray(0.9)+opacity(0.4);<br /> draw(unitcircle3, bg2);<br /> <br /> draw(shift(-sqrt(pi)/2,3,0)*scale(sqrt(pi),sqrt(pi),1)*surface((0,0,0)--(1,0,0)--(0.5,0.5,1)--cycle),bg);<br /> draw(shift(-sqrt(pi)/2,3,0)*scale(sqrt(pi),sqrt(pi),1)*surface((1,1,0)--(1,0,0)--(0.5,0.5,1)--cycle),bg);<br /> draw(shift(-sqrt(pi)/2,3,0)*scale(sqrt(pi),sqrt(pi),1)*surface((1,1,0)--(0,1,0)--(0.5,0.5,1)--cycle),bg);<br /> draw(shift(-sqrt(pi)/2,3,0)*scale(sqrt(pi),sqrt(pi),1)*surface((0,0,0)--(0,1,0)--(0.5,0.5,1)--cycle),bg);<br /> draw(shift(-sqrt(pi)/2,3,0)*scale(sqrt(pi),sqrt(pi),1)*unitsquare3,bg2);<br /> <br /> label(&quot;$\pi$&quot;,(0,0,0),bg2);<br /> label(&quot;$\pi$&quot;,(0,3+sqrt(pi)/2,0));<br /> &lt;/asy&gt;</div> Mag1c