Tangents PA and PB are drawn from P(0,10) to the circle x^2+y^2=25, touching it at A and B. The area of triangle PAB is

With centre $O$ at the origin, $OP=10$ and the tangent length is $PA=\sqrt{100-25}=5\sqrt3$. The chord of contact $AB$ is perpendicular to $OP$ and is $y=\dfrac{25}{10}=2.5$, so its distance from $P$ is $10-2.5=7.5$. Half of $AB$ is $\sqrt{25-2.5^2}=\dfrac{5\sqrt3}{2}$, so $AB=5\sqrt3$. The area of triangle $PAB$ is $\dfrac12\cdot5\sqrt3\cdot7.5=\dfrac{75\sqrt3}{4}$. The kite $OAPB$ has area $25\sqrt3$, which is the tempting wrong choice. As a check, the kite $OAPB$ (area $25\sqrt3$) is split by $AB$ into triangle $PAB$ and triangle $OAB$ of area $\dfrac12\cdot5\sqrt3\cdot2.5=\dfrac{25\sqrt3}4$, and indeed $\dfrac{75\sqrt3}4+\dfrac{25\sqrt3}4=25\sqrt3$.