<prompt> Acha ABCDEF iwe heksagoni ya kawaida. Acha G, H, I, J, K, na L ziwe sehemu za kati za pande AB, BC, CD, DE, EF, na AF, mtawalia. Sehemu overlineAH, overlineBI, overlineCJ, overlineDK, overlineEL, na overlineFG amefungwa hexagon ndogo ya kawaida. Acha uwiano wa eneo la heksagoni ndogo na eneo la ABCDEF ubainishwe kama sehemu frac mn ambapo m na n ni nambari kamili chanya. Tafuta m n.  <response>: asy defaultpen0.8ptfontsize12pt; jozi A,B,C,D,E,F; jozi G,H,I,J,K,L; Adir0; Bdir60; Cdir120; Ddir180; Edir240; Fdir300; kuchoraA--B--C--D--E--F--mzunguko,bluu; GAB2; HBC2; ICD2; JDE2; KEF2; LFA2; int i; kwa i0; i6; i1 tekazungusha60iA--H,iliyowekwa alama; unganisha M,N,O,P,Q,R; MkiendeleziA,H,B,I; NkiendeleziB,I,C,J; OkiendeleziC,J,D,K; PkiendeleziD,K,E,L; QkiendeleziE,L,F,G; RkiendeleziF,G,A,H; kuchoraM--N--O--P--Q--R--mzunguko,nyekundu; leboA,A,1,0; leboB,B,NE; leboC,C,NW; leboD,D, W; leboE,E,SW; leboF,F,SE; leboG,G,NE; leboH,H, 0,1; leboI,I,NW; leboJ,J,SW; leboK,K, S; leboL,L,SE; leboM,M; leboN,N; leboO,0,0,NE; nukta 0,0; asyAcha M iwe makutano ya overlineAH na overlineBIna N iwe makutano ya overlineBI na overlineCJ.Acha O iwe kitovu.Acha BC2 bila hasara ya jumla.Kumbuka kuwa angle BMH ni pembe ya wima kwa pembe ya heksagoni ya kawaida, na pia ina digrii 120circ.Kwa sababu pembetatu ABH na pembetatu BCI ni picha za mzunguko za kila mmoja, tunapata angleMBHpembeHAB na hivyo pembetatu ABH sim pembetatu BMH sim pembetatu BCI.Kwa kutumia hoja sawa, NIMH, naMNBI-NI-BMBI-BMMH.Utumiaji wa Sheria ya cosines kwenye pembetatu BCI, BIsqrt2212-221cos120circsqrt7 anzaalignfracBCCIBIfrac3sqrt7fracBMMHBH BMMHfrac 3BHsqrt7frac3sqrt7 MNBI-BMMHsqrt7-frac3sqrt7 frac4sqrt7 fractextEneo la heksagoni ndogotextEneo la heksagoni kubwaleftfracMN BCkulia2kushotofrac2sqrt7kulia2frac47endalignHivyo, jibu ni 4 7 11. 