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Q.

A triangle ABC  is drawn to circumscribe a circle of radius 3 cm   so that the segments BD  and DC  into which BC   is divided by the point of contact D  , are of lengths 6 cm   and 9 cm   respectively. If the area of ΔABC=54  cm 2  then the lengths of sides AB  and AC  will be,


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a

9, 12

b

8, 14

c

6, 10

d

9, 13Question Image 

answer is A.

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Detailed Solution

Given are the following,
Here PA=PB  and OBBA  .
BD=6cm DC=9cm OD=3cm  
The area of ABC=54  cm 2  
Equating between BF   and BD  , CD   and CE  , AF   and AE  .
Question Image BF=6 cm BD=6 cm CD=9 cm CE=9 cm AF=X cm AE=X cm  
To calculate x   we now apply Heron’s formula,
Area of ΔABC=54  cm 2  
S=  semi-perimeter.
S= AB+BC+AC 2  
S= X+6+X+9+9+6 2   S=X+15   Now applying Heron's formula, we get,
Area= S(SAB)(SBC)(SAC)   54= (X+15)(X+15X6)(X+1515)(X+15X9)   54= (X+15)9×6×X   54= X X+15  
54= X 2 +15X  
On simplifying the above equation, we get,
  X 2 +15X54=0 X 2 +18X3X54=0 X(X+18)3(X+18)=0 (X3)(X+18)=0  
X=3cm   [Length cannot be negative]
From i) we have,
AB=X+6 AB=9cm AC=X+9 AB=12cm  
Thus the length of AB   is 9 cm   and the length of AC   is 12 cm  .
Therefore the correct option is 1.
 
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