Trapezoid

In the above figure, we note that $\triangle AEF$ is similar to $\triangle AOC$ and $AE=EF= 5\space cm$ .

Therefore, $OE = 9-5= 4 \space cm$ .

Then, the area of the $\color{#3D99F6}{blue}$ -shaded region is $=\dfrac {5+9}{2}\times 4 = \color{#3D99F6}{\boxed{28}} \space cm^2$ .

It's the difference of the areas of two isosceles right triangles, $\frac{9\times9}{2}-\frac{5\times5}{2}=\boxed{28}$

Is AO PERPENDICULAR to BC?

if AO . CO are radii that means AO=OC the area=9 9 0.5-5 5 0.5 // // =28cm^2

let the line equivalent to 5cm be BT Therefore; Triangle AOC ||| Triangle ABT(similar) AO/AB = OC/BT 9/AB =9/5 9AB=45 AB =5cm BO=9-5 BO=4 AREA OF TRAPEZOIDAL=1/2 *(9+5) * 4 =28cm2

In the problem it has been mentioned that the shaded area is trapezoidal.

Since it is a trapezoid two sides has to be parallel.

Therefore the smaller triangle and triangle AOC are similar.

If the triangles are similar, the corresponding sides are proportional.

Hence we get one side of the trapezium as 4 cm.

Since one more side is known the trapezium can be splitted as a rectangle of area 5*4 sq cm and a right angled triangle with height 4cm and base 4 cm

Adding the area of the rectangle and that of the triangle gives the total area of the trapezium I.e,

Area of trapezium in sq cm= (5 * 4)+(1/2 * 4 * 4)= 28 sq cm

Also this applies if AO is perpendicular to OC