I. \[AP+BQ+CR=PB+QC+RA\] |
II.\[AP+BQ+CR=\frac{1}{2}\](perimeter of \[\Delta ABC\]) |
III. \[AP+BQ+CR=3(AB+BC+CA)\] |
A) I, II and III
B) only I
C) II and III only
D) I and II
Correct Answer: D
Solution :
As the tangents drawn from an external point to a circle are equal. \[\therefore \] \[AP=AR,\,\,BQ=BP\]and\[CR=QC\] \[\therefore \]\[AP+BQ+CR=BP+QC+RA\] Here I is correct. Perimeter of \[\Delta ABC=AB+BC+AC\] \[=AP+PB+BQ+QC+RC+RA\] \[=2(AP+BQ+CR)\] \[\therefore \] \[AP+BQ=CR=\frac{1}{2}\] (Perimeter of \[\Delta ABC\]) II is correct.You need to login to perform this action.
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