Question

3 mark question pls solve

Answer 1

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Here given,

x+y =  8 cm   ....(1) 

y+z = 12 cm .....(2)

x+z = 10 cm .....(3)

Adding (1), (2) and (3), we get 

2(x+y+z) = 30

x+y+z = 30/2

x+y+z = 15  .....(4)

(4) - (2), we get

x+12 = 15

x = 15 - 12

x =3

(4) - (3),  we get

10+y = 15

y = 15 - 10

y = 5

(4) - (1), we get 

8+z = 15

z = 15 - 8

z = 7

So,

AD = 3 cm

BE = 5 cm

CF = 7 cm  Answer.

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Answer 2

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drawn from an external point to a circle are equal. ⇒ AD = AF, BD = BE, CE = CF.  Let AD = AF = a  BD = BE = b  CE = CF = c   AB = AD + DB = a + b = 8 -------- (1) BC = BE + EC = b + c = 10 -------- (2) AC = AF + FC = a + b = 12 -------- (3)   Adding (1), (2) and (3), we get 2 (a + b + c) = 30 ⇒ (a + b + c) = 15 -------- (4)   Subtracting (1) from (4), we get c = 7 Subtracting (2) from (4), we get a = 5 Subtracting (3) from (4), we get b = 3   Therefore, AD = a = 5 cm, BE = b = 3 cm, CF = c = 7 cm