Dear Student , Please find below the solution to the asked query : We know that the length of tangents drawn from an external point of circle are equal So , BD=BM=8 cmCD=CN=6 cmLet AM=x cmSo,AM=AN=x cmThusAB=x+8 cmBC=6+8=14 cmCA=x+6 cmNow in ∆ABC ar∆ABC=ar∆OAB+ar∆OBC+ar∆OCA84=12×OM×AB+12×OD×BC+12×ON×CA84=12×4×x+8+12×4×14+12×4×x+684=12 4x+32+56+4x+24168=8x+1128x=56x=568=7 cm So,AB=x+8 cm=7+8 cm=15 cm AC= (x+6) cm= (7+6) cm= 13 cm Hope this information will clear your doubts about the topic If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible Regards

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Dear Student ,

Please find below the solution to the asked query :

We know that the length of tangents drawn from an external point of circle are equal . So, BD = BM = 8 cm CD = CN = 6 cm Let AM = x cm So, AM = AN = x cm Thus AB = (x + 8) cm BC = (6 + 8) = 14 cm CA = (x + 6) cm Now in ∆ ABC . ar (∆ ABC) = ar (∆ OAB) + ar (∆ OBC) + ar (∆ OCA) 84 = \frac{1}{2} \times OM \times AB + \frac{1}{2} \times OD \times BC + \frac{1}{2} \times ON \times CA 84 = \frac{1}{2} \times 4 \times (x + 8) + \frac{1}{2} \times 4 \times 14 + \frac{1}{2} \times 4 \times (x + 6) 84 = \frac{1}{2} [4 x + 32 + 56 + 4 x + 24] 168 = 8 x + 112 8 x = 56 x = \frac{56}{8} = 7 cm . So, AB = (x + 8) cm = (7 + 8) cm = 15 cm .

AC= (x+6) cm= (7+6) cm= 13 cm

Hope this information will clear your doubts about the topic .

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible .

Regards

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