DP^2=DB^2+(BC/2)^2 = 4^2 +(4*(3)^(1/2)/2)^2=16 +(2*(3)^(1/2))^2 =16 +4*3=16+12=28
a)
BC^2=(4*(3)^(1/2))^2=16*3=48 ==> 36<48<49 ==> 6^2 <48 < 7^2 ==> 6 < BC < 7
угол DCB=180-CBD-СВИ=180-90-60=30
BC/DC=cosDCB=cos30=(3)^(1/2)/2 ==> BC=DC*(3)^(1/2)/2=8*(3)^(1/2)/2=4*(3)^(1/2)
DB/DC=sinDCB=sin30=1/2 ==> DC=DB/(1/2)=DB*2=4*2=8
из теоремы пифагора получим
b)
PD=(28)^(1/2) = (4*7)^(1/2) = 2*7^(1/2)