Question

The radius of the incircle of a triangle in 4cm and the segments into which one side is divided by the points of contact are 6cm and 8cm. Determine the other two sides of the triangle.

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Answer 1

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BF=BD=6cm

[length of tangent from external point are equal]

CE=CD=8cm

[length of tangent from external point are equal]

Let AF=AE=x

[length of tangent from external point are equal ]

Now,

AreaofΔABC=AreaofΔAOB+AreaofΔBOC+AreaofΔCOA

= 21 ×4(6+x)+ 21×4(14)+ 21 ×4(8+x)= 21 ×4(28+2x)=4(14+x)

Also, area of ΔABC by Heron's formula

S= 214+6+x+8+x=14+x

Area of ΔABC= (14+x)(8)(6)x

So, 4(14+x)= 48x(14+x)

16(14+x) 2=48x(14+x)

14+x=3 x 2x=14⇒x=7

So, AB=6+x=6+7=13cm

AC=8+x=8+7=15cm

Sum = 13+15= 28cm

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Answer 2

Answer:

BF=BD=6cm

[length of tangent from external point are equal]

CE=CD=8cm

[length of tangent from external point are equal]

Let AF=AE=x

[length of tangent from external point are equal ]

Now,

AreaofΔABC=AreaofΔAOB+AreaofΔBOC+AreaofΔCOA

= 21 ×4(6+x)+ 21×4(14)+ 21 ×4(8+x)= 21 ×4(28+2x)=4(14+x)

Also, area of ΔABC by Heron's formula

S= 214+6+x+8+x=14+x

Area of ΔABC= (14+x)(8)(6)x

So, 4(14+x)= 48x(14+x)

16(14+x) 2=48x(14+x)

14+x=3 x 2x=14⇒x=7

So, AB=6+x=6+7=13cm

AC=8+x=8+7=15cm

Sum = 13+15= 28cm

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