# In Figure 5, a ci

Given;

PQ = 10 cm

OR = 8 cm

PR = 12 cm

Let’s take;

PL = PN = x cm

QM = QL = y cm

RN = RM = z cm

[ Tangents drawn from an external point are equal]

So we get;

PQ = 10 cm x + y = 10 ………. (i)

QR = 8 cm y + z = 8 …………..(ii)

PR = 12 cm x + z = 12 ………….(iii)

By adding all the three equations we get;

2x + 2y + 2z = 10 + 8 + 12

2(x + y + z) = 30

x + y + z = 15

By Solving the we get;

From (i)

10 + z = 15

z = 15 – 10 = 5

From (ii)

x + 8 = 15

x = 15 – 8 = 7

From (iii)

12 + y = 15

y = 15 – 12 = 3

So,

PL = PN = x = 7 cm

RN = RM = z = 5 cm

QM = QL = y = 3 cm

AM = 3 cm, RN = 5 cm, PL = 7 cm

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