ABC\] in which \[\angle B=90{}^\circ \]and AB = BC. Then, we must have I. 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. \[AC=\sqrt{2x}\ne 2x\] \[\therefore \]I and II are true

"> ABC\] in which \[\angle B=90{}^\circ \]and AB = BC. Then, we must have I. 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. \[AC=\sqrt{2x}\ne 2x\] \[\therefore \]I and II are true

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For an isosceles right - angled triangle having each of equal sides 'x', We have I. Area\[=\frac{1}{2}{{x}^{2}}\] II. Perimeter\[=\left( 2+\sqrt{2} \right)x\] III. Hypotenuse\[=2x\] Which of the following is true?

10th Class Mathematics in 10th Class 11 months ago

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Consider \[\Delta \text{ }ABC\] in which \[\angle B=90{}^\circ \]and AB = BC. Then, we must have I. 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. \[AC=\sqrt{2x}\ne 2x\] \[\therefore \]I and II are true

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