| ABC is an isosceles triangle, right-angled at B. Similar triangles ACD and ABE are constructed on sides AC and AB respectively. Then the ratio between the areas of\[\Delta ABE\] and \[\Delta ACD\] is: |
 |
A) \[2:1\]
B) \[1:2\]
C) \[1:3\]
D) \[3:1\]
Correct Answer: B
Solution :
| [b] In \[\Delta ABC,\] by Pythagoras theorem |
| \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}=2A{{B}^{2}}\] ...(1) |
| [\[\Delta ABC\] is isosceles, \[\therefore \,\,AB=BC\]] |
| We have, \[\Delta ACD\tilde{\ }\Delta ABE\] |
| Since, ratio of areas of two similar triangles is equal to the square of the ratio of any two corresponding sides. |
| \[\therefore \,\,\,\,\,\,\,\frac{ar(\Delta ABE)}{ar(\Delta ACD)}=\frac{A{{B}^{2}}}{B{{C}^{2}}}=\frac{A{{B}^{2}}}{2A{{B}^{2}}}\] [From (1)] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\frac{ar(\Delta ABE)}{ar(\Delta ACD)}=\frac{1}{2}\] |
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