A **semicircle** is developed when a lining passing with the centre touches the 2 ends top top the circle.

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

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

Thus, the one is entirely identified by its centre (O) and also radius (r).

## Area of SemiCircle

The area that a semicircle is half of the area the the circle. Together the area of a one is πr2. So, the area that a semicircle is **1/2(πr2**** )**, wherein r is the radius. The value of π is 3.14 or 22/7.

Area that Semicircle = 1/2 (π r2) |

## Perimeter the Semicircle

The perimeter of a semicircle is the amount of the fifty percent of the circumference of the circle and diameter. Together 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 that Semicircle = (1/2) π d + d or (πr + 2r) |

### Semi one Shape

When a one is reduced into 2 halves or when the one of a circle is separated by 2, we obtain semicircular shape.

Since semicircle is fifty percent that that a circle, therefore the area will certainly be half that of a circle.

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

Let united state generate the above figure. This polygon can be damaged into n isosceles triangle (equal sides gift radius).

Thus, one together isosceles triangle have the right to be represented as presented below.

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The area that this triangle is provided as **½(h*s)**

Now because that n variety of polygons, the area that a polygon is offered as

**½(n*h*s)**

The ax **n×s** is same to the perimeter that the polygon. As the polygon it s okay to look more and much more like a circle, the value approaches the circle circumference, i m sorry is **2×π×r**. So, substituting **2×π×r** because that **n×s.**

**Polygon area = h/2(2×π×r)**

Also, together the number of sides increases, the triangle gets narrower and also so when s viewpoints zero, h and r have the exact same length. So substituting r because that h: