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.
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 Theorem Thale 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.

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.

A centroid of any object in n-dimensional space is the intersection of all hyper planes that divide an object into two parts of equal moment about the hyper plane. Precisely, it is the "average" of all points of an object. The centroid of a body is also its centre of mass, for an object of uniform composition (mass, density, etc.).

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.

Unlike the area, the perimeter of a semicircle is not half the perimeter of a circle from which it is made. In the semicircle figure above, we can see that the perimeter is the curved line, which is half the circumference of a circle, along with the diameter line across the bottom.

The Perimeter which is also called as the Circumference of the circle is given by 2 $\pi$ r.

The curved part of this perimeter will be half of it, which is p r, and the diameter line is twice that of radius or 2 r.

Thus, the formula for Perimeter of Semicircle is given by :
Perimeter = $\pi$ r + 2r
= r ($\pi$ + 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: A piece of wire is 108 cm in length is bent into a semi circle and diameter. Find the diameter length taking $\pi$ = $\frac{22}{7}$.
Solution:

The length of the wire which is bent into a semicircle and diameter, gives the perimeter of the semicircle.
                   The Perimeter of a semicircle is given by the formula, P = r ($\pi$ + 2) where r is the radius.
               Thus,   r ($\pi$ + 2) = 108

                   r = $\frac{108}{\frac{22}{7} + 2}$

                    r = $\frac{108}{\frac{22 +14}{7}}$

                    r = $\frac{108}{\frac{36}{7}}$

                    r = 108 $\times$ $\frac{7}{36}$

                    r = 3 $\times$ 7

                    r = 21 cm

                   d = 2r = 2 $\times$ 21 = 42 cm

Thus the diameter of the semicircle is 42 cm.

Sector of a Circle

Angle Inscribed in a Semicircle	Part of a Circle	A Circle	How to do Integration by Parts
Corresponding Parts of Congruent Triangles	Parts of a Line Graph	A Chord of a Circle	A Sector of a Circle
An Arc of a Circle	Area Circle	Center of the Circle	Area of a Sector of a Circle

Integrate by Parts Calculator	Circle Solver	Circumference a Circle

Thales' theorem

Area of a Semicircle

Solved Example

Centroid of a Semicircle

Perimeter of Semicircle

Solved Example

Volume of Semicircle

Semicircle Equation

Sector of a Circle

Related Concepts

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