<prompt> Katika pembetatu ABC, AB 13, AC 15, na BC 14. Acha I iwe kitovu. Mzingo wa pembetatu ABC hugusa pande BC, AC, na AB kwa D, E, na F, mtawalia. Pata eneo la quadrilateral AEIF.asykuagiza jiometri;ukubwa wa kitengo 2 cm;jozi A, B, C, D, E, F, I;A 1,2;B 0,0;C 3,0;I kitovuA,B,C;D I tafakariB,CI2;E I tafakariC,AI2;F I tafakariA,BI2;kuchoraA--B--C--mzunguko;choraduaraA,B,C;kuchoraE--I--F;leboA, A, N;leboB, B, SW;leboC, C, SE;leboD, D, S;leboE, E, NE;leboF, F, NW;leboI, I, S;asy  <response>: Kwa kuwa AE na AF ni tanji kutoka sehemu moja hadi kwenye duara moja, AE AF. Acha x AE AF. Vile vile, hebu y BD BF na z CD CE .asykuagiza jiometri;ukubwa wa kitengo 2 cm;jozi A, B, C, D, E, F, I;A 1,2;B 0,0;C 3,0;I kitovuA,B,C;D I tafakariB,CI2;E I tafakariC,AI2;F I tafakariA,BI2;kuchoraA--B--C--mzunguko;choraduaraA,B,C;leboA, A, N;leboB, B, SW;leboC, C, SE;leboD, D, S;leboE, E, NE;leboF, F, NW;lebox, A E2, NE;lebox, A F2, NW;leboy, B F2, NW;leboy, B D2, S;leboz, C D2, S;leboz, C E2, NE;asyKisha x y AB 13, x z AC 15, na y z BC 14. Kuongeza milinganyo hii yote, tunapata 2x 2y 2z 42, kwa hivyo x y z 21. Tukiondoa mlinganyo y z 14, tunapata x 7.Kwa fomula ya Heron, eneo la pembetatu ABC ni K sqrt2121 - 1421 - 1521 - 13 84,kwa hivyo inradius ni r K s 8421 4.Tunaweza kugawanya AEIF katika pembetatu mbili za kulia AEI na AFI.asykuagiza jiometri;ukubwa wa kitengo 2 cm;jozi A, B, C, D, E, F, I;A 1,2;B 0,0;C 3,0;I kitovuA,B,C;D I tafakariB,CI2;E I tafakariC,AI2;F I tafakariA,BI2;kuchoraA--B--C--mzunguko;choraduaraA,B,C;kuchoraE--I--F;kuchoraA--I;leboA, A, N;leboB, B, SW;leboC, C, SE;leboD, D, S;leboE, E, NE;leboF, F, NW;leboI, I, S;asyEneo la pembetatu AEI ni frac12 cdot AE cdot IE frac12 cdot x cdot r frac12 cdot 7 cdot 4 14,na eneo la pembetatu AFI pia ni 14. Kwa hiyo, eneo la quadrilateral AEIF ni 14 14 28. 