The ABCD is a square with sides of unit length. Points E and F are taken on sides AB and AD respectively so that AE = AF. Let P be a point inside the square ABCD. The maximum possible area of quadrilateral CDFE is

Question

The ABCD is a square with sides of unit length. Points E and F are taken on sides AB and AD respectively so that AE = AF. Let P be a point inside the square ABCD. The maximum possible area of quadrilateral CDFE is (A) (1/8) (B) (1/4) (C) (5/8) (D) (3/8)

Tardigrade Admin · Accepted Answer

Area of CDFE <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/7dd16478dd373838f14cc82ed49fbd1a-.png" /> A=1-(1/2) x2-(1/2)(1-x) =(2-x2-1+x/2)=(1+x-x2/2) ∴ A max =(1+(1/2)-(1/4)/2)=(5/8) at x =(1/2)

Q. The $A BC D$ is a square with sides of unit length. Points $E$ and $F$ are taken on sides $A B$ and $A D$ respectively so that $A E = A F$ . Let $P$ be a point inside the square $A BC D$ .
The maximum possible area of quadrilateral $C D FE$ is

$\frac{1}{8}$

$\frac{1}{4}$

$\frac{5}{8}$

$\frac{3}{8}$

Area of $C D FE$

$A = 1 - \frac{1}{2} x^{2} - \frac{1}{2} (1 - x)$
$= \frac{2 - x ^{2} - 1 + x}{2} = \frac{1 + x - x ^{2}}{2}$
$∴ A_{m a x} = \frac{1 + \frac{1}{2} - \frac{1}{4}}{2} = \frac{5}{8}$ at $x = \frac{1}{2}$

Q. The ABCD is a square with sides of unit length. Points E and F are taken on sides AB and AD respectively so that AE=AF. Let P be a point inside the square ABCD. The maximum possible area of quadrilateral CDFE is

81​

41​

85​

83​