A) 2 sq. units
B) 4 sq. units
C) \[\frac{3}{2}sq.\] units
D) 5 sq, units
Correct Answer: A
Solution :
| The coordinates of the vertices R, Band Pare \[(3,5),\,\,(3,7)\]and \[(1,7)\] respectively. Now, |
| \[RB=\sqrt{{{(3-3)}^{2}}+{{(7-5)}^{2}}}=\sqrt{0+4}=2\]units |
| \[PB=\sqrt{{{(3-1)}^{2}}+{{(7-7)}^{2}}}=\sqrt{4+0}=2\] units |
| and \[PR=\sqrt{{{(3-1)}^{2}}+{{(5-7)}^{2}}}=\sqrt{4+4}=2\sqrt{2}\] units |
| \[P{{R}^{2}}+R{{B}^{2}}+P{{B}^{2}}\] and \[RB=PB\] |
| \[\therefore \,\,\,\,\Delta RBP\] is a right angle isosceles triangle. |
| Area of triangle |
| \[\Delta RBP=\frac{1}{2}\times RB\times PB=\frac{1}{2}\times 2\times 2=2sq.\]units |
| So, option [a] is correct. |
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