A semicircle is created when a lining passing v the centre touches the two ends on the circle.
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In the below figure, the line AC is referred to as the diameter the the circle. The diameter divides the circle into two halves such that they room equal in area. These two halves are described as the semicircles. The area that a semicircle is fifty percent of the area that a circle.
A one is a locus of point out equidistant from a given suggest which is the centre of the circle. The usual distance native the center of a circle come its allude is referred to as a radius.
Thus, the one is entirely identified by its center (O) and radius (r).
Area that Semi Circle
The area of a semicircle is fifty percent of the area the the circle. Together the area the a one is πr2. So, the area that a semicircle is 1/2(πr2 ), where r is the radius. The worth of π is 3.14 or 22/7.
Area the Semicircle = 1/2 (π r2) |
Perimeter that Semicircle
The perimeter of a semicircle is the amount of the half of the one of the circle and diameter. As the perimeter that a one 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 Circumference = (πr + 2r) |
Semi circle Shape
When a circle is reduced into 2 halves or as soon as the circumference of a circle is divided by 2, we get semicircular shape.
Since semicircle is fifty percent that the a circle, thus the area will certainly be fifty percent that that a circle.
The area that a one is the variety of square devices inside the circle.
Let united state generate the above figure. This polygon deserve to be damaged into n isosceles triangle (equal sides being radius).
Thus, one such isosceles triangle deserve to be represented as presented below.
The area the this triangle is offered as ½(h*s)
Now because that n variety of polygons, the area the a polygon is given as
½(n*h*s)
The hatchet n × s is same to the perimeter the the polygon. Together the polygon gets to look much more and much more like a circle, the value approaches the circle circumference, i beg your pardon is 2 × π × r. So, substituting 2×π×r because that n × s.
Polygon area = h/2(2 × π × r)
Also, as the variety of sides increases, the triangle gets narrower and so as soon as s approaches zero, h and also r have the very same length. For this reason substituting r for h:
Polygon area = h/2(2 × π × r)
= (2 × r × r × π)/2
Rearranging this we get
Area = πr2
Now the area that a semicircle is same to half of the of a full circle.
Therefore,
Area of a semicircle =(πr2)/2
Semi circle Formula
The below table reflects the formulas linked with the semicircle that radius r.
Area | (πr2)/2 |
Perimeter (Circumference) | (½)πd + d; when diameter (d) is known |
πr + 2r | |
Angle in a semicircle | 90 degrees, i.e. Appropriate angle |
Central angle | 180 degrees |
Semi circle Examples
Example 1:
Find the area of a semicircle that radius 28 cm.
Solution:
Given,
Radius that semi circle = r = 28 cm
Area of semi circle = (πr2)/2
= (½) × (22/7) × 28 × 28
= 1232
Therefore, the area of the semi circle is 1232 sq.cm.
Example 2:
What is the perimeter the a semicircle through diameter 7 cm?
Solution:
Given,
Diameter the semicircle = d = 7 cm
Formula for the circumference (perimeter) of a semicircle utilizing its diameter = (½)πd + d
Substitute the worth of d, we get;
= (½) × (22/7) × 7 + 7
= 11 + 7
= 18
Therefore, the perimeter that the semicircle is 18 cm.
See more: How Long Is A Furlong In Miles, Furlong To Miles Conversion (Furlong To Mi)
Frequently Asked concerns on Semicircle
Is a semicircle half the circle?
Yes, a semicircle is fifty percent the circle. The means, a circle deserve to be separated into two semicircles.