question_answer

A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length\[x\]. The maximum area enclosed by the park is     AIEEE  Solved  Paper-2006

A) \[\sqrt{\frac{{{x}^{3}}}{8}}\]                      

B) \[\frac{1}{2}{{x}^{2}}\] 

C)        \[\pi {{x}^{2}}\]                

D)        \[\frac{3}{2}{{x}^{2}}\]

Correct Answer: B

Solution :

The area of isosceles triangle is maximum when it is right angled triangle and then use area of\[\Delta =\frac{1}{2}\times b\times h\] Given that \[AB=AC=x\] We know that area of isosceles triangle is maximum, if it is right angled triangle. \[\therefore \]Maximum area of triangle \[=\frac{1}{2}{{x}^{2}}\] Alternate Solution \[\therefore \]Area\[=\frac{1}{2}(2x\,\cos \theta )(x\sin \theta )\] \[=\frac{1}{2}{{x}^{2}}\sin 2\theta \] \[\therefore \]\[{{(Area)}_{\max }}=\frac{1}{2}{{x}^{2}}\]