question_answer

\[In\,\,\Delta ABC,\,\,AB=BC=K,\,\,AC=\sqrt{2}\,\,K\]then is a:

A)  Isosceles triangle

B)  Right angled triangle

C)  Equilateral triangle

D)  Right isosceles triangle

Correct Answer: D

Solution :

[d] \[In\,\Delta ABC\] \[AC=\sqrt{2}K\] \[A{{C}^{2}}=2{{K}^{2}}\] \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\] So \[\Delta \]ABC is right angled triangle So in \[\Delta \]ABC \[\frac{AB}{AC}=\,\frac{K}{\sqrt{2}K}\,=\frac{1}{\sqrt{2}}\] So \[=\frac{1}{2}\times 24\times 32=384\,\,c{{m}^{2}}\] \[\theta =45{}^\circ \] \[So,\,\,ABC,\,\,\angle B=90{}^\circ \,\,;\,\,\angle C=45{}^\circ \,\,;\,\,\angle A=45{}^\circ \] So, ABC is right isoscles triangle.