Semicircle
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. 
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 °.
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 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}$$\pi r^2$ or $\frac{\pi r^2}{2}$ where r is the radius of the circle from which it is made and p is pi which is approximately 3.142.
Solved Example
Solution:
Let midpoint of base of inscribed square be M; let side of the inscribed square be 2d.
Draw a line joining M to one of the top corners of the square. The length of this line forms a radius of the circle with centre M.
Length of this line from M = radius of the semicircle = $\sqrt{(d^2 + 4d^2)}$ = $\sqrt{5d}$
So, radius = 35 = $\sqrt{5d}$
Hence diameter, d = $\frac{35}{\sqrt{5}}$ = 7$\sqrt{5}$ cms.
So square is of side-length 14$\sqrt{5}$ cms and circle has radius 35cms.
The area of the semicircle is given by the formula $\frac{1}{2}$$\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.
The Centroid of any semicircle lies in the middle of the semicircle.
In this figure the centroid of the semicircle is the point G.
The centroid of a semicircle is given by the formula,
G = $\frac{4r}{3\pi}$
If Point P is assumed to be at the lower left corner of the wall where the semicircle is resting, the centroid of this semicircle lies at a distance of half of its diameter (d) along the horizontal direction, as in the case of a regular circle. However, the centroid lies at a distance of ($\frac{4d}{6\pi}$) up along the vertical direction from Point P.