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Multiply Pi (3.14...) by the radius of the circle squared, then divide that number by 2.

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Q: How do you work out the area of a semicircle?
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Related questions

What is the equation to solve the area of a semicircle?

Area of a semicircle = (pi*radius^2)/2


What is the area of the semicircle if the length of the arc of a semicircle is 10 and pi feet?

It is 27.49 units.


What is area of semicircle?

The area of a semicircle is half of the area of a circle. The area of a circle is pi x radius x radius. So the area of a semicircle is 1/2 x pi x radius x radius or (pi x radius x radius) / 2.


What is the area of a semicircle when the radius is 21?

The equation for area of a circle is A=pi x R x R. A semicircle is half a circle, so we calculate the area of a circle and divide by 2. The area of the circle would be approximately 1384.74. Thus the area of the semicircle would be 692.37.


What is the formula to find the area of a semicircle?

Pi*radius squared is how to find the area of a semicircle


What is the area outside the triangle if the triangle is inscribed in a semicircle while the side length of triangle is 2underroot2?

Diameter of semicircle = 1 Area of semicircle = Pi/8 Area of triangle = 0.25 Area outside triangle = (Pi/8) - 0.25 = 0.1427 (rounded to 4th decimal).


How do you find the area of a semi-circle if you know the diameter?

A semicircle is 1/2 of a circle. Find the area with the diameter you are given as if you had a whole circle, then divide that answer by 2 to get the area of the semicircle.


Calculate the area of semicircle with radius 4 meters?

An area of a circle is pi*r2 - so an area of a semicircle is 0.5*pi*r2: 0.5*3.14*42 = 25.13 square meters


What is the area of a semicircle with a diameter of 77 centimeters?

2328.31286 cm2


Area of a semicircle with a diameter of 8 cm?

25


What is the area of a semicircle whose radius is 5 cm?

50.25cm


What is the diameter of a semicircle?

In a semicircle, the diameter is the straight line. It is also equal to double the radius, and that property allows it to be used when calculating area and circumference.