A boy was playing with a rectangular cardboard of dimensions 13 cm x 6 cm. While playing, he sliced off identical triangles from the corners of the cardboard in such a manner that a figure having all its sides equal was generated (as shown in the adjoining figure). The area of this six-sided figure is:

Setting up the equations:

Let each right triangle have horizontal leg = $a$ and vertical leg = $b$.

After cutting all four corners, the hexagon will have:

• Top and bottom edges: $13 - 2a$ (we remove $a$ from each end)
• Left and right edges: $6 - 2b$ (we remove $b$ from top and bottom)
• Slanted edges (hypotenuse): $\sqrt{a^2 + b^2}$

Since all sides of the hexagon are equal, let this common length be $s$.

$13 - 2a = s \quad ...(1)$

$6 - 2b = s \quad ...(2)$

$\sqrt{a^2 + b^2} = s \quad ...(3)$

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Solving for a and b:

From equations (1) and (2):

$13 - 2a = 6 - 2b$

$13 - 6 = 2a - 2b$

$7 = 2(a - b)$

$a - b = 3.5 \quad ...(4)$

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From equation (3): $a^2 + b^2 = s^2$

From equation (1): $a = \frac{13 - s}{2}$

From equation (2): $b = \frac{6 - s}{2}$

Substituting into $a^2 + b^2 = s^2$:

$\left(\frac{13-s}{2}\right)^2 + \left(\frac{6-s}{2}\right)^2 = s^2$

$\frac{(13-s)^2 + (6-s)^2}{4} = s^2$

$(13-s)^2 + (6-s)^2 = 4s^2$

$169 - 26s + s^2 + 36 - 12s + s^2 = 4s^2$

$205 - 38s + 2s^2 = 4s^2$

$2s^2 + 38s - 205 = 0$

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Using the quadratic formula:

$s = \frac{-38 \pm \sqrt{1444 + 1640}}{4} = \frac{-38 \pm \sqrt{3084}}{4} = \frac{-38 \pm 55.53}{4}$

Taking the positive value: $s = 5$ cm

Therefore: $a = 4$ cm and $b = 3$ cm

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Calculating the area:

$\text{Area of hexagon} = \text{Area of rectangle} - \text{Area of 4 triangles}$

$= (13 \times 6) - 4 \times \left(\frac{1}{2} \times 4 \times 3\right)$

$= 78 - 24 = 54 \text{ cm}^2$

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The area is 54 cm².