Question

In the given figure find AD, BE, CF where AB = 12 cm, BC = 8 cm and AC = 10 cm.

Answer 1

GIVEN:
AB = 12 cm ,BC = 8 cm, CA = 10 cm

We know that the tangents drawn from an external point to a circle are equal in length.

Let, AD = AF = x ,  BD = BE = y & CE = CF = z .

AB = 12 cm
AD + DB = 12 cm
x + y = 12 cm………….(1)

BC = 8 cm
BE + EC = 8 cm
y + z = 8 cm……………..(2)

AC = 10 cm
AF + FC = 10 cm
x + z = 10 cm……………(3)

On Adding eq 1, 2 & 3
(x + y) + (y + z) + (x + z)= 12 + 8+ 10
2(x + y+z) = 30
x + y+z = 30/2
x + y+z  = 15…………….(4)

On subtracting eq 4 & 1
(x + y+z)  - (x + y )= 15 - 12
x + y + z  - x - y = 15 - 12
z = 3

On subtracting eq 4 & 2
(x + y+z)  - (y + z )= 15 - 8
x + y + z  - y - z = 15 - 8
x = 7

On subtracting eq 4 & 3
(x + y+z)  - (x + z )  = 15 - 10
x + y + z  - x - z = 15 - 10
y = 5

Hence, AD = x = 7 cm, BE= y = 5cm, CF = z = 3 cm.

HOPE THIS WILL HELP YOU...

Answer 2

Let AD be x, CF be y and BE be z.

AD = AF
CF = CE
BE = BD


AD + BD = x + z
x + z = 12 eq. 1

Similarly,
x + y = 10 eq. 2
y + z = 8 eq. 3


Using 1 and 2....

x + z = 12
x + y = 10
(-) (-) (-)
________
z - y = 10
________

z + y = 8
z - y = 10
_______
2z = 18


z = 9 [ Substituting values in eq 1, 2 and 3]
x = 3
y = 7


Therefore,
AD = 3 cm
CF = 7 cm
BE = 9 cm