question_answer

What is the area of an isosceles triangle having base 'Y' cm and one side 'Y' cm?

A)  \[\frac{x}{2}\sqrt{{{y}^{2}}+\frac{{{x}^{2}}}{4}}c{{m}^{2}}\]

B)  \[\frac{x}{2}\sqrt{\frac{4{{y}^{2}}-{{x}^{2}}}{4}}c{{m}^{2}}\]

C)  \[\frac{{{x}^{2}}-{{y}^{2}}}{4}c{{m}^{2}}\]

D)  \[\frac{{{x}^{2}}+{{y}^{2}}}{4}c{{m}^{2}}\]

Correct Answer: B

Solution :

Consider the isosceles triangle as shown in the figure. Drop a perpendicular AD to base BC. It bisects the base. \[\therefore \] \[BD=\frac{x}{2}\] Now, by Pythagoras' theorem, we have \[A{{D}^{2}}=A{{B}^{2}}-B{{D}^{2}}\] \[\Rightarrow \]               \[{{y}^{2}}-\frac{{{x}^{2}}}{4}=\frac{4{{y}^{2}}-{{x}^{2}}}{4}\] \[\Rightarrow \]               \[AD=\sqrt{\frac{4{{y}^{2}}-{{x}^{2}}}{4}}\] Area of triangle \[=\frac{1}{2}\times \]x base \[\times \]height \[\therefore \] Area of \[\Delta \,\,ABC\]                 \[=\frac{1}{2}\times x\times \sqrt{\frac{4{{y}^{2}}-{{x}^{2}}}{4}}\,c{{m}^{2}}\]