Let’s call the sides of the rectangle a and b, and the radius of the circle r.

We know from Pythagoras that a^2+b^2=21^2=441

Since we know the value of a^2+b^2, if we know either the sum or the product of a and b, we can find the other. This is because if we square the sum we get the sum of the squares (which we know) plus twice the product.

Since the area of the rectangle is simply the product of a and b, if we can’t work it out directly, we can do it by first learning the sum of a and b.

Let’s call that sum ‘s’.

Since the incircle is in a right angled triangle we can deduce that the radius is r=(s-21)/2

If we were to draw a line from the centre of the circle to the opposite corner and call that length x, it would simply be the diagonal of a rectangle that was smaller by r in each direction, so again using Pythagoras: x^2=(a-r)^2+(b-r)^2

The reason that line is useful is that it forms a right angled triangle with the radius and the 15 unit line, so using Pythagoras again we get: x^2=r^2+15^2=r^2+225.

To eliminate x we can use this expression for x^2 in the previous equation, so now we have:

a^2+b^2=441

s=a+b

r=(s-21)/2

r^2+225=(a-r)^2+(b-r)^2

Taking the final equation and expanding it:

r^2+225=a^2-2ar+r^2+b^2-2br+r^2

We can replace (a^2+b^2) by 441, (a+b) by s, and r by (s-21)/2:

225=441+(-s)(s-21)+((s-21)^2)/4

s(s-21)-((s-21)^2)/4=216

4s^2-84s-s^2+42s-441=864

3s^2-42s-1305=0

s^2-14s-435=0

The positive value of which is s=29

So now we have the sum of a and b we can find the product:

s^2=(a+b)^2=29^2

a^2+b^2+2ab=841

But we know a^2+b^2 is 441

2ab=400

ab=200

So the area of the rectangle is 200 square units.

If you wanted to you could use ab=200 and a+b=29 to find exact values for a and b, and if you were to do that you would get that they are (29 +/- sqrt(41))/2, which are about 17.7 and 11.3.