A) \[\frac{5\left( 2+\sqrt{2} \right)}{\sqrt{2}}m\]
B) \[\frac{2+\sqrt{2}}{\sqrt{2}}m\]
C) \[17\sqrt{2}\,m\]
D) \[16\sqrt{2}\,m\]
E) None of these
Correct Answer: A
Solution :
Explanation Option [a] is correct. Let the base and height of the right angled triangle is x. By Pythagorean Theorem \[{{(Base)}^{2}}={{(Hypotenuse)}^{2}}-{{(Height)}^{2}}\] \[\Rightarrow {{x}^{2}}={{5}^{2}}-{{x}^{2}}\] \[\Rightarrow {{x}^{2}}+{{x}^{2}}={{5}^{2}}\Rightarrow 2{{x}^{2}}={{5}^{2}}\Rightarrow x=\sqrt{\frac{{{5}^{2}}}{2}}=\frac{5}{\sqrt{2}}\] Hence, the perimetre of the triangle \[=\frac{5}{\sqrt{2}}+\frac{5}{\sqrt{2}}+5=\frac{5+5+5\sqrt{2}}{\sqrt{2}}=\] \[\frac{10+5\sqrt{2}}{\sqrt{2}}=\frac{5\left( 2+\sqrt{2} \right)}{\sqrt{2}}m.\]
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