What is the area of this triangle?

First, length of OA is \(\sqrt50\)as it's a radius of the circle. No need to simplify this,soon see why..

Now since line AP lies on the tangent line L ,then AP/OA   = tan 90 deg

So AP/OA = 1

AP/\(\sqrt50 = 1\)

So AP is also \(\sqrt50 \)

Area of triangle is 1/2 base  X height

          = 1/2   X \(\sqrt50*\sqrt50\)

          =  1/2  X 50

          =  25

Let's look at this a little closer......I don't know if "Point A"  is some specific point   or any random point on the circle.....but....if it's a random point.......look at this graph

Let  " Point  A "  be on the on the circle   =  ( a , sqrt (r^2 - a^2))

The slope of  the tangent line to this circle will be   =    -a / sqrt (r^2 - a^2)

So....the equation of a  line tangent to  the circle at  this point  is

y =  [ -a / sqrt (r^2 - a^2)] ( x - a) +  sqrt (r^2  - a^2)

To find "Point P"  let  y  = 0   we have that

0 =   [ -a / sqrt (r^2 - a^2)] ( x - a)  +  sqrt (r^2  - a^2)

 [ a / sqrt (r^2 - a^2)] ( x - a)   =   sqrt (r^2  - a^2)

a (x - a)   =   r^2 - a^2

ax - a^2  =  r^2  - a^2

ax =  r^2

x  = r^2 / a   ...   so  "Point P"   (  r^2/ a , 0 )

So.......  the   the base of triangle OAP will  be  r^2 / a      and the height will be sqrt ( r^2 - a^2)

So....the area of this triangle will be    (1/2) (r^2 / a) sqrt (r^2 - a^2)  =    (r^2) sqrt (r^2 - a^2) / [ 2a ]

So....suppose  r  = sqrt (50)  and  a  = 4

Then the area of triangle OAP  will  be    50 sqrt (34) / 8  ≈  36.44 sq units

But suppose that   r =sqrt (50) and  a  = 2

Then the area of triangle OAP  will be  50 sqrt ( 46) / 4  ≈ 84.78 sq units

As an addendum to my answer.....I believe  that frasinscotland's  answer will be true if "Point A"  is (5,5)

Point P will be located at  ( 50/5, 0)  = (10,0)

So AP  will have a length of   sqrt   [ (10 - 5)^2  + ( 5 - 0)^2 ]  =  sqrt ( 25 + 25)  = sqrt (50)

And this will be  equal  to  OA