<prompt> Katika ABCD ya pembe nne, BC8, CD12, AD10, na mangle A mangle B 60circ. Kwa kuzingatia kwamba AB p sqrtq , ambapo p na q ni nambari kamili chanya, pata pq.  <response>: asychora0,0--20.87,0--15.87,8.66--5,8.66--mzunguko; chora5,8.66--5,0; kuchora15.87,8.66--15.87,0; kuchora 5,8.66--16.87,6.928; leboA,0,0,SW; leboB,20.87,0,SE; leboE,15.87,8.66,NE; leboD,5,8.66,NW; leboP,5,0,S; leboQ,15.87,0,S; leboC,16.87,7,E; lebo12,10.935,7.794,S; lebo10,2.5,4.5,W; lebo10,18.37,4.5,E; asyChora sehemu ya mstari DE hivi kwamba mstari DE unaambatana na mstari BC. Kisha, ABED ni trapezoid ya isosceles hivyo ADBE10, na BC8 na EC2. Tumepewa hiyo DC12. Kwa kuwa angle CED 120circ, kwa kutumia Sheria ya Cosines kwenye bigtriangleup CED inatoa122DE24-22DEcos 120 circambayo inatoa144-4DE22DE. Kuongeza 1 kwa pande zote mbili kunatoa 141DE12, kwa hivyo DEsqrt141-1. bigtriangleup DAP na bigtriangleup EBQ zote ni 30-60-90, kwa hivyo AP5 na BQ5. PQDE, na kwa hivyo ABAPPQBQ5sqrt141-159sqrt141 rightarrow p,q9,141 rightarrow 150 . 