First find the area of the sector, which includes the shaded region's area and the area of the triangle JKL. Let [tex]A[/tex] be the area of the sector. This area occurs in a fixed ratio with the area of the entire circle based on the measure of the central angle subtended by the arc LK:
[tex]\dfrac A{54^\circ}=\dfrac{27^2\cdot3.14}{360^\circ}\implies A\approx343.359[/tex]
To get the area of the shaded region, subtract from [tex]A[/tex] the area of the triangle.
The area of a triangle is 1/2 the base times the height. If we bisect the central angle with a line segment that meets the side KL at its midpoint, then we get a right triangle with hypotenuse 27 and one angle of measure 27º (half of the central angle). This triangle has base [tex]b[/tex] and height [tex]h[/tex] such that
[tex]\sin27^\circ=\dfrac b{27}\implies b\approx12.258[/tex]
[tex]\cos27^\circ=\dfrac h{27}\implies h\approx24.057[/tex]
So the right triangle has area approximately 1/2*12.258*24.057, or about 147.443. Triangle JKL is made up of two of these right triangles, so it has area of 294.887.
Subtracting this from [tex]A[/tex] gives an area of the shaded region of about 48.472, which we round up to 48.5.
