I. Area\[=\frac{1}{2}{{x}^{2}}\] |
II. Perimeter\[=\left( 2+\sqrt{2} \right)x\] |
III. Hypotenuse\[=2x\] |
A) I only
B) II only
C) I and II
D) I and III
Correct Answer: C
Solution :
Consider \[\Delta \text{ }ABC\] in which \[\angle B=90{}^\circ \]and AB = BC. Then, we must haveI. Area\[=\frac{1}{2}\times BC\times AB=\frac{1}{2}{{x}^{2}}.\] |
II. \[AC=\sqrt{{{x}^{2}}+{{x}^{2}}}=\sqrt{2{{x}^{2}}}=\sqrt{2x}\] |
\[\therefore \]Perimeter |
\[=x+x+\sqrt{2x}=2x+\sqrt{2}x=x\left( 2+\sqrt{2} \right)\] |
III. \[AC=\sqrt{2x}\ne 2x\] |
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