question_answer

One acute angle of a right angled triangle is double the other. If the length of its hypotenuse is 10 cm, then its area is

A) \[\frac{25}{2}\sqrt{3}c{{m}^{2}}\]                    

B) \[25\,\,c{{m}^{2}}\]

C) \[25\sqrt{3}c{{m}^{2}}\]         

D)        \[\frac{75}{2}c{{m}^{2}}\]

Correct Answer: A

Solution :

From\[\Delta ABC\]             \[\sin {{30}^{o}}=\frac{AB}{AC}\Rightarrow \frac{1}{2}=\frac{AB}{AC}\] \[\Rightarrow \]   \[AB=\frac{1}{2}AC=\frac{1}{2}\times 10=5\,\,cm\] \[\therefore \]      \[BC=\sqrt{A{{C}^{2}}-A{{B}^{2}}}=\sqrt{{{10}^{2}}-{{5}^{2}}}\]             \[=\sqrt{100-25}=\sqrt{75}=5\sqrt{3}cm\] \[\therefore \]Area of\[\Delta ABC\]             \[=\frac{1}{2}\times AB\times BC\]             \[=\frac{1}{2}\times 5\times 5\sqrt{3}=\frac{25\sqrt{3}}{2}c{{m}^{2}}\]