A circle of radius $13$ cm touches the adjacent sides AB and BC of a square ABCD at M and N, respectively. If AB = $18$ cm and the circle intersects the other two sides CD and DA at P and Q, respectively, then the area, in sq. cm, of triangle PMD is

We need to find the area of triangle PMD in a square ABCD where a circle of radius $13$ cm touches sides AB and BC at points M and N respectively, and intersects sides CD and DA at points P and Q.

Let's place the square in a coordinate system with A at origin $(0,0)$, B at $(18,0)$, C at $(18,18)$, and D at $(0,18)$.

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The center of the circle must be at a distance of $13$ cm from both sides AB and BC.

Since the circle touches side AB (x-axis) and side BC (line $x = 18$), the center is at:

$(18-13, 13) = (5, 13)$

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Point M is where the circle touches side AB.

Since AB is along the x-axis, M is directly below the center: $M = (5, 0)$

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To find point P on side CD, we need to find where the circle intersects the line $y = 18$.

The equation of the circle is:

$(x-5)^2 + (y-13)^2 = 13^2$

Substituting $y = 18$:

$(x-5)^2 + (18-13)^2 = 13^2$

$(x-5)^2 + 25 = 169$

$(x-5)^2 = 144$

$x-5 = \pm 12$

So $x = 17$ or $x = -7$

Since P must be on CD, $P = (17, 18)$

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Now we have the coordinates:

$M = (5, 0)$

$P = (17, 18)$

$D = (0, 18)$

Using the formula for area of a triangle with coordinates:

Area = $\dfrac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$

Area = $\dfrac{1}{2}|5(18-18) + 17(18-0) + 0(0-18)|$

= $\dfrac{1}{2}|0 + 17(18) + 0|$

= $\dfrac{1}{2}(306)$

= $153$ sq. cm

Therefore, the area of triangle PMD is $153$ sq. cm.