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:

Question

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: (A) √(x3/8) (B) (1/2)x2 (C) π x2 (D) (3/2)x2

Tardigrade Admin · Accepted Answer

Given that AB = AC = x We know that area of isosceles triangle is maximum, if it is right angled triangle. ∴ Maximum area of triangle =(1/2) x2

Q. 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:

$\frac{x ^{3}}{8}$

$\frac{1}{2} x^{2}$

$π x^{2}$

$\frac{3}{2} x^{2}$

Given that $A B = A C = x$
We know that area of isosceles triangle is maximum, if it is right angled triangle.
$∴$ Maximum area of triangle $= \frac{1}{2} x^{2}$

Q. 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:

8x3​​

21​x2

πx2

23​x2

Given that AB=AC=x We know that area of isosceles triangle is maximum, if it is right angled triangle. ∴ Maximum area of triangle =21​x2

$\frac{x ^{3}}{8}$

$\frac{1}{2} x^{2}$

$π x^{2}$

$\frac{3}{2} x^{2}$

Given that $A B = A C = x$
We know that area of isosceles triangle is maximum, if it is right angled triangle.
$∴$ Maximum area of triangle $= \frac{1}{2} x^{2}$