A) 2
B) 3
C) 4
D) 5
Correct Answer: B
Solution :
[b] Let r be the inradius. Area of \[\Delta AIB,\] \[{{\Delta }_{1}}=\frac{1}{2}\times r\times AB=\frac{1}{2}rc\] Area of \[\Delta \,CID\], \[{{\Delta }_{2}}=\frac{1}{2}\times r\times CD=\frac{1}{2}r\left[ \frac{ba}{b+c} \right]\] \[\therefore \,\,\,\,\frac{{{\Delta }_{1}}}{{{\Delta }_{2}}}=\frac{(b+c)c}{ab}=\frac{(4+6)6}{4\times 5}=3\]You need to login to perform this action.
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