Semi in Latin means half. Semicircle by itself means half circle.
In Geometry, semicircle can be defined as a 2-dimensional figure that forms half of a circle is called Semicircle. It is a closed shape that is half a circle but diameter of that circle. Which means the diameter of a circle cuts the circle into two equal semicircles.
Semi Circle

A circle makes a full turn of 360°. Semicircle being half a circle measures half of circle’s 360°, thus measuring 180°.

Thales' theorem

An angle inscribed in a semicircle is always 90 °.

Thales TheoremThale Theorem

Proof: We have to prove that ∠ACB = 90 °
Mark Centre O. Join OC.
Now, OA = OB =OC = radius.
Δ OCA and Δ OCB are isosceles triangles.
Since the base angles of an isosceles triangle are equal,
We have ∠ OCB = ∠ OBC and ∠ OCA = ∠ OAC.
Also, Sum of angles in a triangle = 180 °
Thus, ∠ OAC + ∠ ACB +∠ OBC = 180°.
∠ OAC + ∠ OCA + ∠ OCB +∠ OBC = 180° (Since ∠ ACB = ∠ OCA + ∠ OCB)
2∠ OCA + 2∠ OCB = 180°.
2(∠ OCA + ∠ OCB) = 180°.
∠ OCA + ∠ OCB = 90°.
∠ ACB =90° (Since ∠ ACB = ∠ OCA + ∠ OCB)

Hence proved.

Any triangle inscribed in semicircle is always a right triangle.

The area of the semicircle is found from the area of a circle. Since the semicircle is half circle, the area of the semicircle is half the area of the circle from which it is made.
The area of the circle = $\pi r^2$ where r is the radius of the circle and $\pi$ is pi which is approximately 3.142.The area of the semicircle = $\frac{1}{2}$$\pi r^2$ or $\frac{\pi r^2}{2}$ where r is the radius of the circle from which it is made and p is pi which is approximately 3.142.

Solved Example

Question: Find the area of the portion of the semi circle of radius 35cms that is outside the area of an inscribed square if the base of the square lies on the diameter of the semi-circle.

Area of the Semicircle
Solution:
 
Let midpoint of base of inscribed square be M; let side of the inscribed square be 2d.

Draw a line joining M to one of the top corners of the square. The length of this line forms a radius of the circle with centre M.

Length of this line from M = radius of the semicircle = $\sqrt{(d^2 + 4d^2)}$ = $\sqrt{5d}$

So, radius = 35 = $\sqrt{5d}$

Hence diameter, d = $\frac{35}{\sqrt{5}}$ = 7$\sqrt{5}$ cms.

So square is of side-length 14$\sqrt{5}$ cms and circle has radius 35cms.

The area of the semicircle is given by the formula $\frac{1}{2}$$\pi r^2$ where r is the radius of the semicircle.

Thus, the Semicircle area = $\frac{1}{2}$$\pi (35)^2$
                                                                 
                                                                     = $\frac{1}{2}$ $\times$ $\frac{22}{7}$ $\times$ 35 $\times$ 35
                              
                                               = 55 $\times$ 35

                                               = 1925 sq.cms.

Area of a square is given by $(side)^2$.

Thus,  Square area = (14 $\sqrt{5}$)^² = 196 $\times$ 5 = 980 sq.cms.
The area of the portion of the semi circle that is outside the area of an inscribed square is given by the difference of these two areas.
The area of the portion of the semi circle that is outside the area of an inscribed square = 1925 – 980 = 945 sq.cms.