<prompt> Tufe imeandikwa katika tetrahedron ambayo vipeo ni A 6,0,0,B 0,4,0, C 0,0,2, na D 0,0,0. Tafuta eneo la tufe.  <response>: Kumbuka kuwa AD 6, BD 4, na CD 2. Kisha na Pythagoras, AB 2 sqrt13, AC 2 sqrt10, na BC 2 sqrt5. Na Theorem ya Heron,anzaalignABC2 sqrt5 sqrt10 sqrt13-sqrt5 sqrt10 sqrt13sqrt 5 - sqrt10 sqrt13sqrt5 sqrt10 - sqrt13 sqrt10 sqrt132 - sqrt52sqrt52 - sqrt10 - sqrt 132 18 2 sqrt1302 sqrt130 - 18 196,endalignhivyo ABC 14.asykuagiza tatu;kuagiza yabisi;ukubwa 200;makadirio ya sasa mtazamo 6,3,2;mara tatu A 6,0,0, B 0,4,0, C 0,0,2, D 0,0,0;kuchoraA--D,dashed;kuchoraB--D,dashed;kuchoraC--D,dashed;chorakuhama23,23,23usotufe23,kijivu0.8;kuchoraA--B--C--mzunguko;leboA, A, SW;leboB, B, E;leboC, C, N;leboD, D, NW;asyAcha r iwe kipenyo cha tufe, na I iwe kitovu cha tufe. Tunaona hiloABCD frac13 cdot ABD cdot CD frac13 cdot frac12 cdot 4 cdot 6 cdot 2 8.Tunaweza pia kuandikaABCD ABCI ABDI ACDI BCDI.Tunaweza kuangalia tetrahedron yenye msingi ABCI na msingi ABC na urefu r, hivyoABCI frac13 cdot 14 cdot r frac143 r.Vile vile,anzaalignABDI frac13 cdot 12 cdot r 4r, ACDI frac13 cdot 6 cdot r 2r, BCDI frac13 cdot 4 cdot r frac43 r.mwishoalignHivyo,frac143 r 4r 2r frac43 r 8.Tunatatua kwa r, tunapata r frac23. 