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A triangle ABC is drawn to circumscribe ...

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. Find the sides AB and AC.

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Given, CD = 6 cm, BD = 8 cm and radius = 4 cm
Join OC, OA and OB.
We know that the tangents drawn from external point are equal in length
`:." "CD=CF=6 cm`
and `" "BD=BE=8cm`
Let `" "AF=AE=xcm`
In `triangleOCB,`
Area of triangle, `A_(1)=(1)/(2)xxBasexx Height=(1)/(2)xxCBXXOD`
`=(1)/(2)xx14xx4=28cm^(2)`
In `triangleOCA,`
Area of triangle,`A_(2)=(1)/(2)xxACxxOE=(1)/(2)(6+x)xx4=12+2x` ltbtgt In `triangleOBA,`
Area fo triangle,`A_(3)=(1)/(2)xxABxxOE=(1)/(2)(8+x)xx4=16+2x`
Now, semiperimeters of triangle `ABC=(1)/(2)(AB+BC+CA)` ltvrgt `s=(1)/(2)(x+6+14+8+x)=14+x`
Now, `" "`area of `triangleABC=A_(1)+A_(2)+A_(3)`
`=28+(12+2x)+(16+2x)`
`=56+4x" "...(1)`
Using Heron's formula,
Area of `triangleABC=sqrt(s(s-a)(s-b)(s-c))`
`=sqrt((14+x) (14+x-14)(14+x-x-6)(14+x-x-8))`
`=sqrt((14+x)(x)(8)(6))=sqrt((14+x)48x)" "...(2)`
`:.`From Eqs. (1) and (2), we get
`sqrt((14+x)48x)=56+4x`
On squaring both sides,
`(14+x)48x=4^(2)(14+x)^(2)" "implies" "3x=14+x`
`implies" "2x=14" "implies" "x=7`
`:.` Length ` AC=6+X=6+7=13cm`
Length of `AB=8+x=8+7=15cm`
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