A **semicircle** is formed when a lining passing through the centre touches the two ends on the circle.

In the below figure, the line AC is called the diameter of the circle. The diameter divides the circle into two halves such that they are equal in area. These two halves are referred to as the semicircles. The area of a semicircle is half of the area of a circle.

A circle is a locus of points equidistant from a given point which is the centre of the circle. The common distance from the centre of a circle to its point is called a radius.

Thus, the circle is entirely defined by its centre (O) and radius (r).

## Area of SemiCircle

The area of a semicircle is half of the area of the circle. As the area of a circle is πr^{2}. So, the area of a semicircle is **1/2(πr ^{2}**

**)**, where r is the radius. The value of π is 3.14 or 22/7.

Area of Semicircle = 1/2 (π r^{2}) |

## Perimeter of Semicircle

The perimeter of a semicircle is the sum of the half of the circumference of the circle and diameter. As the perimeter of a circle is 2πr or πd. So, the perimeter of a semicircle is** 1/2 (πd) + d or πr + 2r, **where r is the radius.

Therefore,

The perimeter of Semicircle = (1/2) π d + d or (πr + 2r) |

### Semi Circle Shape

When a circle is cut into two halves or when the circumference of a circle is divided by 2, we get semicircular shape.

Since semicircle is half that of a circle, hence the area will be half that of a circle.

The area of a circle is the number of square units inside that circle.

Let us generate the above figure. This polygon can be broken into n isosceles triangle (equal sides being radius).

Thus, one such isosceles triangle can be represented as shown below.

The area of this triangle is given as **½(h*s)**

Now for n number of polygons, the area of a polygon is given as

**½(n*h*s)**

The term **n×s** is equal to the perimeter of the polygon. As the polygon gets to look more and more like a circle, the value approaches the circle circumference, which is **2×π×r**. So, substituting **2×π×r** for **n×s.**

**Polygon area = h/2(2×π×r)**

Also, as the number of sides increases, the triangle gets narrower and so when s approaches zero, h and r have the same length. So substituting r for h:

Polygon area = h/2(2×π×r)

= (2×r×r×π)/2

Rearranging this we get

Area = πr^{2}

Now the area of a semicircle is equal to half of that of a full circle.

Therefore,

**Area of a semicircle** =(πr^{2})/2

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