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A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.
CBSE | Circles | Class 10 | Maths | NCERT
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

Answer:

The given figure:

Considering the △ABC,

We know that any two tangents drawn from the same point to the circle have the same length.

As a result,

(i) BE = BD = 8 cm

(ii) CF = CD = 6 cm

(iii) AE = AF = x

It is observed that,

(i) CA = CF+FA = 6+x

(ii) BC = DC+BD = 6+8 = 14

(iii) AB = EB+AE = 8+x

Now compute the semi perimeter “s” as follows:

2s = AB+CA+BC

Putting the values respectively we get,

2s = 28+2x

s = 14+x

Area of triangle ABC

On solving the above equation we get,

= √(14+x)48x ……… (i)

Again, the area of triangle ABC = 2 × area of (△AOF + △COD + △DOB)

= 2×[(½×OF×AF)+(½×CD×OD)+(½×DB×OD)]

= 2×½(4x+24+32) = 56+4x …………..(ii)

Now from eq.(i) and eq.(ii) we get,

√(14+x)48x = 56+4x

On squaring both the sides,

48x(14+x) = (56+4x)2

48x = [4(14+x)]2/(14+x)

48x = 16(14+x)

48x = 224+16x

32x = 224

x = 7 cm

As a result, AB = 8+x Therefore, AB = 15 cm and CA = x+6 =13 cm.

i

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