Let $P$ be an interior point of a right-angled isosceles triangle $ABC$ with hypotenuse $AB$ . If the perpendicular distance of P from each of $A B, B C$, and $C A$ is $4(\sqrt{2}-1) \mathrm{cm}$, then the area, in sq cm, of the triangle $A B C$ is

We have a right-angled isosceles triangle ABC with hypotenuse AB, and point P inside the triangle is at the same distance from all three sides. This distance is $4(\sqrt{2}-1)$ cm.

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When a point inside a triangle is equidistant from all three sides, that point is the incenter of the triangle. The incenter is where all three angle bisectors meet, and the common distance from the incenter to each side is the inradius (r).

Since P is equidistant from AB, BC, and CA, P is the incenter of triangle ABC.

Therefore: inradius r = $4(\sqrt{2}-1)$ cm

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In a right-angled isosceles triangle:

Two sides are equal (let's call them 'a' each)

The hypotenuse is $a\sqrt{2}$ (using Pythagorean theorem: $\sqrt{a^2 + a^2} = a\sqrt{2}$)

The right angle is between the two equal sides

So our triangle has sides: a, a, $a\sqrt{2}$

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For any triangle, there's a relationship:

$\text{Area} = \text{inradius} \times \text{semi-perimeter}$

$A = r \times s$

Where semi-perimeter $s = \frac{\text{perimeter}}{2}$

This works because when we connect the incenter to each vertex, we divide the triangle into three smaller triangles. Each has the same height (the inradius r) and bases equal to the sides of the original triangle.

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Finding the semi-perimeter:

$s = \frac{a + a + a\sqrt{2}}{2} = \frac{a(2 + \sqrt{2})}{2}$

Finding the area using two methods:

Method 1: $A = \frac{1}{2} \times a \times a = \frac{a^2}{2}$

Method 2: $A = r \times s = 4(\sqrt{2}-1) \times \frac{a(2 + \sqrt{2})}{2}$

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Since both expressions equal the area:

$\frac{a^2}{2} = 4(\sqrt{2}-1) \times \frac{a(2 + \sqrt{2})}{2}$

$\frac{a^2}{2} = 2(\sqrt{2}-1) \times a(2 + \sqrt{2})$

$a = 4(\sqrt{2}-1)(2 + \sqrt{2})$

Expanding:

$a = 4[(\sqrt{2}-1)(2 + \sqrt{2})]$

$a = 4[\sqrt{2} \times 2 + \sqrt{2} \times \sqrt{2} - 1 \times 2 - 1 \times \sqrt{2}]$

$a = 4[2\sqrt{2} + 2 - 2 - \sqrt{2}]$

$a = 4[\sqrt{2}] = 4\sqrt{2}$

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Now that we know $a = 4\sqrt{2}$:

$\text{Area} = \frac{a^2}{2} = \frac{(4\sqrt{2})^2}{2} = \frac{16 \times 2}{2} = \frac{32}{2} = 16$

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The area of triangle ABC is 16 sq cm.