Answer:
Part a) The area of the figure is 
Part b) The perimeter of the figure is 
Step-by-step explanation:
Step 1
Find the area of the figure
In this problem we have that
The figure ABC is the half of a square and the other figure is a semicircle
<u>Find the area of the figure ABC</u>
we have

The area of the half square ABC is equal to find the area of triangle ABC
so

<u>Find the area of the semicircle</u>
The area of the semicircle is equal to

we have that

substitute


The area of the figure is equal to

Step 2
Find the perimeter of the figure
The perimeter of the figure is equal to

we have

Applying Pythagoras theorem

Remember that
the circumference of a semicircle is equal to




The perimeter of the figure is equal to

Simplify
