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A circle is inscribed in a triangle ABC having sides AB = 8 cm, BC = 10 cm and
CA = 12 cm.Find AD, BE and CF.

Tangents drawn from an external point to a circle are equal.

AD = AF = x (say)

BD = BE = y (say)

and CE = CF = z (say)

Now, AB = AD + DB = x + y = 8  ...(1)

BC = BE + EC = y + z = 10 ...(2)

AC = AF + FC = x + y = 12 ...(3)

 

Adding (1), (2) and (3), we get

2 (x + y + z) = 30

⇒ (x + y + z) = 15  ...(4)

Solving (1) and (4), we get z = 7

Solving (2) and (4), we get x = 5

Solving (3) and (4), we get y = 3

Hence, AD = x = 5 cm

BE = y = 3 cm

CF = z = 7 cm

  • 59
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you didnt mention about points D,E&F

  • -6

AB = AF, BD = BE, CE = CF .   - eqn1

IT IS GIVEN THAT AB = 8cm, BC = 10cm AND CA = 12cm.

NOW,     AD + AF + BD + BE + CE + CF = 8 + 10 + 12

FROM eqn1,

          AD + AD + BD + BD + CF + CF = 30

   2AD + 2BD + 2CF =30

    2 (AD + BD + CF) = 30

       (AD + BD) + CF = 15

                    AB + CF =15

                     10 + CF = 15

                              CF = 15 -10

                              CF = 5cm = CE.

NOW,         BC = BE + CE

                    10 = BE + 5

              10 - 5 = BE

                    BE = 5cm =BD

AND            AB = AD + BD

                      8 = AD + 5

                 8 - 5 =AD

                    AD = 3cm.

I hope the ans is right.

  • 3

AS A CIRCLE IS INSCRIBED IN A TRIANGLE SO, OBVIOUSLY POINTS D, E AND F MUST BE THE INTERSECTING POINTS OF THE CIRCLE AND THE TRIANGLE.

  • -6

thanx a lot saumya..... and yes they are the mid points........ thanxxx

  • -4
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