wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the tangents BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

Open in App
Solution


Length of two tangents drawn from the same point to the circle are equal.
CF = CD = 6 cm
BE = BD = 8 cm
AE = AF = x
We observed that,
AB = AE + EB = x + 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 + x
Let perimeter of triangle ABC be s.
2s = AB + BC + CA
= x + 8 + 14 + 6 + x
= 28 + 2x
s = 14 + x
Area of Δ ABC = s(sa)(sb)(sc)
=(14+x)(14+x14)(14+xx6)(14+xx8)
=(14+x)(x)(8)(6)
=(14+x)48x...(i)
Also, area of Δ ABC = 2 × (area of ΔAOF+ΔCOD+ΔDOB)
=2×[(12×OF×AF)+(12×CD×OD)+(12×DB×OD)]
=2×12(4x+24+32)=56+4x...(ii)
Equating equations (i) and (ii) we get,
(14+x)48x=56+4x
Squaring both 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
So, AB = x + 8 = 7 + 8 = 15 cm
CA = 6 + x = 6 + 7 = 13 cm.

Hence, AB=15 cm , CA = 13 cm


flag
Suggest Corrections
thumbs-up
929
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon