Quadrilateral PQRS is inscribed inside a rectangle of dimensions $10 \mathrm{~cm} \times 8 \mathrm{~cm}$. The value of ' $x$ ', if the area of the quadrilateral is minimum is

Let the rectangle be $ABCD$ with $AB = CD = 10$ cm and $AD = BC = 8$ cm. Let $AQ = x$, $BR = x$, $CS = x$, and $DP = x$.

The points are located as follows: $Q$ is on $AB$ such that $AQ = x$, so $QB = 10 - x$. $R$ is on $BC$ such that $BR = x$, so $RC = 8 - x$. $S$ is on $CD$ such that $CS = x$, so $DS = 10 - x$. $P$ is on $AD$ such that $DP = x$, so $AP = 8 - x$.

The area of the quadrilateral $PQRS$ is the area of the rectangle $ABCD$ minus the sum of the areas of the four right-angled triangles at the corners: $\triangle APQ$, $\triangle BQR$, $\triangle CRS$, and $\triangle DSP$.

Area of $\triangle APQ = \dfrac{1}{2} \times AP \times AQ = \dfrac{1}{2}(8 - x)x = 4x - \dfrac{x^2}{2}$.
Area of $\triangle BQR = \dfrac{1}{2} \times BQ \times BR = \dfrac{1}{2}(10 - x)x = 5x - \dfrac{x^2}{2}$.
Area of $\triangle CRS = \dfrac{1}{2} \times RC \times CS = \dfrac{1}{2}(8 - x)x = 4x - \dfrac{x^2}{2}$.
Area of $\triangle DSP = \dfrac{1}{2} \times DS \times DP = \dfrac{1}{2}(10 - x)x = 5x - \dfrac{x^2}{2}$.

Sum of areas of the four triangles $= 2(4x - \dfrac{x^2}{2}) + 2(5x - \dfrac{x^2}{2}) = 8x - x^2 + 10x - x^2 = 18x - 2x^2$.

Area of $PQRS = 10 \times 8 - (18x - 2x^2) = 80 - 18x + 2x^2$.

To find the minimum area, differentiate with respect to $x$ and set to zero:
$\dfrac{d}{dx}(80 - 18x + 2x^2) = -18 + 4x = 0$.

$4x = 18 \Rightarrow x = 4.5$ cm.

Answer: 4.5 cm