Answer
$$A=2.25\pi$$
Work Step by Step
$x^2+y^2\leq9$ Covers the area of a circle, with radius $3$ and center on the origin of coordinate plane. (See the Blue region on the image above)
$y\geq|x|$ Is the region above $y=|x|$ graph, including the line itself. (See the Red region on the image above)
We are asked to find the sector of the circle, that is intersection of both graphs. The graph of $y=|x|$ is bisector for the first and second quadrants. Which means that the sector is $\frac{1}{4}$ of the whole circle.
$A_{circle} = \pi \times r^2 = 9\pi$
$A_{sector}=\frac{1}{4}9\pi = 2.25\pi$
