DKB =∪KB/2 =KAB (угол между касательной и хордой)
CKA =∪KA/2 =KBA
△KAH~△BKD, AK/BK =KH/BD
△KBH~△AKC, BK/AK =KH/AC
KH/BD=AC/KH => KH=√(AC*BD)=√77
BK =AK*KH/AC =13*√(11/7)
По теореме Пифагора
AH =√(AK^2 -KH^2) =√(13^2 -77) =√92
BH =√(BK^2 -KH^2) =√(13^2 -7^2)*√(11/7) =√120*√(11/7)
S(AKB) =1/2 (AH+BH) KH =1/2 (√92 +√120*√11/√7)*√77 =√23*√77 +11√30
Ответ: 102