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Question

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.

A
13 and 15
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B
15 and 13
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C
16 and 12
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D
None of These
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Solution

The correct option is B 15 and 13
Let there is a circle having center O touches the sides AB and AC of the triangle at point E and F respectively.

Let the length of the line segment AE is x.

Now in ABC,
CF=CD=6 (tangents on the circle from point C)
BE=BD=6 (tangents on the circle from point B)
AE=AF=x (tangents on the circle from point A)

Now AB=AE+EB
AB=x+8=c

BC=BD+DC
BC=8+6=14=a

CA=CF+FA
CA=6+x=b

Now
Semi-perimeter, s=(AB+BC+CA)2
s=(x+8+14+6+x)2
s=(2x+28)2
s=x+14

Area of the ABC=s(sa)(sb)(sc)

=(14+x)((14+x)14)((14+x)(6+x))((14+x)(8+x))
=(14+x)(x)(8)(6)
=(14+x)(x)(2×4)(2×3)

Area of the ABC=43x(14+x) .............................(1)

Area of OBC=12×OD×BC
=12×4×14=28

Area of OBC=12×OF×AC
=12×4×(6+x)
=12+2x

Area of ×OAB=12×OE×AB
=12×4×(8+x)
=16+2x


Now, Area of the ABC=Area of OBC+Area of OBC+Area of OAB

43x(14+x)=28+12+2x+16+2x

43x(14+x)=56+4x=4(14+x)

3x(14+x)=14+x

On squaring both sides, we get

3x(14+x)=(14+x)2

3x=14+x ------------- (14+x=0x=14 is not possible)

3xx=14
2x=14
x=142
x=7

Hence
AB=x+8
AB=7+8
AB=15

AC=6+x
AC=6+7
AC=13

So, the value of AB is 15 cm and that of AC is 13 cm. Option B.

953702_866886_ans_a2cdf058f6064dbe99883b65a164aa2b.JPG

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