Question

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

Answer 1

Answer:

AB=15  & AC=13

Step-by-step explanation:

Let the length of the line segment AE is x.

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=2

(AB+BC+CA)  

s=  2 (x+8+14+6+x)

s= 2 (2x+28)

⟹s=x+14

Area of the △ABC=  

s(s−a)(s−b)(s−c) =  (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=4  

3x(14+x)  .............................(1)

Area of △OBC= 2 1   ×OD×BC= 2 1  ×4×14=28

Area of △OBC=  2 1   ×OF×AC =  2 1​ ×4×(6+x) =12+2x Area of ×OAB=  2 1   ×OE×AB =2 1    ×4×(8+x) =16+2x

Now, Area of the △ABC=Area of △OBC+Area of △OBC+Area of △OAB 4  

3x(14+x)    =28+12+2x+16+2x 4  

3x(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=0⟹x=−14 is not possible)

3x−x=14

2x=14

x=  2

14 x=7

Hence  

AB=x+8

AB=7+8

AB=15

AC=6+x

AC=6+7

AC=13

Answer 2

Answer:

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Step-by-step explanation:

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