A) \[\angle A=90{}^\circ \]
B) \[\angle B=90{}^\circ \]
C) \[\angle C=90{}^\circ \]
D) ABC is not a right angled triangle
Correct Answer: B
Solution :
| It is given that in triangle ABC, \[BD\bot AC\] such that |
| \[B{{D}^{2}}=DC.AD.\] |
| Therefore, applying Pythagoras theorem in triangles ABD and BCD, we get |
| \[A{{B}^{2}}=B{{D}^{2}}+A{{D}^{2}}\] ...(1) |
| \[B{{C}^{2}}=B{{D}^{2}}+D{{C}^{2}}\] ...(2) |
| Adding equations (1) and (2), we get, |
| \[A{{B}^{2}}+B{{C}^{2}}=2B{{D}^{2}}+A{{D}^{2}}+D{{C}^{2}}\] |
| \[=2(DC.AD)+A{{D}^{2}}+D{{C}^{2}}\] \[[B{{D}^{2}}=DC.AD]\] |
| Therefore, \[A{{B}^{2}}+B{{C}^{2}}={{(AD+DC)}^{2}}=A{{C}^{2}}\] |
| This means that by converse of Pythagoras theorem, ABC is a right triangle with AC os the hypotenuse and angle opposite to AC is \[90{}^\circ \]. Therefore, \[\angle B=90{}^\circ \]. |
| So, option [b] is correct. |
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