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.You need to login to perform this action.
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