| In given figure, ABC is an isosceles triangle, right-angled at C. Therefore |
 |
A) \[A{{B}^{2}}=2A{{C}^{2}}\]
B) \[B{{C}^{2}}=2A{{B}^{2}}\]
C) \[A{{C}^{2}}=2A{{B}^{2}}\]
D) \[A{{B}^{2}}=4A{{C}^{2}}\]
Correct Answer: B
Solution :
| In right \[\Delta ACB\], use Pythagoras theorem |
|
| \[A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}}\] |
| \[=A{{C}^{2}}+{{\left( AC \right)}^{2}}\] \[\left[ \because \,\,\,BC=AC=given \right]\] |
| \[=2A{{C}^{2}}\] |
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