Question

Q47. The radius of the inner circle of a triangle
is 4 cm and the segments into which one side is
divided by the point of contact are 6 cm and
cm, Then other two sides of the triangle are.​

Answer 1

Answer:

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=3x

2x=14⇒x=7

So, AB=6+x=6+

Step-by-step explanation:

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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

=

2

1

×4(6+x)+

2

1

×4(14)+

2

1

×4(8+x)

=

2

1

×4(28+2x)

=4(14+x)

Also, area of ΔABC by Heron's formula

S=

2

14+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=3x

2x=14⇒x=7

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

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

Sum = 13+15=28cm

Step-by-step explanation:

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