diketahui jarak antara dua titik pusat lingkaran 26 cm. panjang jari-jari lingkaran yang besar 14 cm dan panjang garis singgung persekutuan luar 24 cm. panjang

AB = 26cm
AD = 14 cm
CD = EB = 24
BC = r = ...
AE^2 = AB^2 - EB^2
= 26^2 - 24^2
= 676-576 = 100
AE = √100 = 10cm

ED = BC = 14-10=4cm (((((C)))))

dik : op = 26 cm
         R = 14 cm
         l   = 24 cm
dit : r....?
jawab : op2 - (R - r)2 = l2
              26 kuadrat - (14 - r)2 = 24 kuadrat
               676 - (14 - r)2 = 576
                      - (14 - r)2 = 576 - 676
                      - (14 - r)2 = - 100
                         (14 - r)2 = 100
                        (14 - r)2 = akar dari 100
                         (14- r) = 10
                          -r = 10 - 14
                          -r = -4
                           r = 4