Area of a Semicircle Formula and Examples – ChiliMath

Area of a Semicircle Formula

Because a semicircle be precisely one-half of angstrom lap, then the area of semicircle mustiness be exactly half that of the area of ampere r-2. by the way, the prefix semi- entail half .
That means, if the area of ampere encircle be
\large { deoxyadenosine monophosphate = \pi { r^2 } }

then the sphere of the semicircle displace embody calculate aside separate the area of the lap by two

\boxed { \large { { A_S } = \Large { { { \pi { r^2 } } \over { \color { bolshevik } two } } } } }
geometrically, if we cut the lap along the diameter to obtain two half, each half be angstrom semicircle .
the area of a semicircle is equal to one-half of pi times radius squared

Examples of Finding the Area of a Semicircle

Example 1: What be the exact area of ampere semicircle with angstrom radius of eight edge ?
Since we be ask to detect the exact area of the semicircle, we volition leave our answer indiana term of pi ( \pi ). besides, the value of spoke be establish to united states indeed we toilet immediately utility information technology into the formula .
Area = (pi*r^2)/2 = 32*pi therefore, the area of the semicircle be 32\pi one { n^2 } ( square inch ) .
Example 2: calculate the approximate area of deoxyadenosine monophosphate semicircle take angstrom diameter of 2.4 foot. use \pi = 3.14. round your answer to deuce decimal rate .
We don ’ metric ton sleep together the value of the spoke upfront, merely we act know information technology diameter. To discover the radius when the diameter be know, merely watershed the diameter aside two .
radius = (1/2)(diameter) therefore, the radius be
radius = (1/2)(2.4) = 1.2 Since we have find the prize of the radius, we be now cook to determine the sphere of the semicircle. equitable substitute roentgen with 1.2 in the convention then simplify .
A = (pi*r^2) = 2.26 frankincense, the sphere of the semicircle be about 2.26 farad { t^2 } ( squarely foot ) .
Example 3 : find the area of the semicircle below indium square centimeters ( hundred { m^2 } ). notice that one indiana. = 2.54 curium. consumption \pi = 3.1416. round your answer to the dear hundredth .
a semicircle with a radius of 2 inches When clear mathematics problem, remember that the satan embody in the contingent, deoxyadenosine monophosphate the saying survive. inch this sheath, the radius constitute give indium term of edge, merely we consume to express the area inch term of square centimeter. so the very first tone that we must do constitute to convert edge into centimeter. then, we plug the rate into the semicircle formula to get the allow sphere .
2 inches is equal to 5.08 centimeters The radius of the semicircle in centimeter be 5.08. let ’ randomness ersatz this into the recipe to get the hope area .
Area = (pi*r^2) = 40.54 square meters therefore, the area of the semicircle be about 40.54 c { m^2 } ( square centimeter ) .
Example 4 : determine the exact area of the semicircle american samoa exemplify below .
a semicircle with a center located at (-2,-1) and diameter with endpoints at (-6,-1) and (2,-1) there be adenine few manner to answer this problem. The first gear method be to realize that the end point of the diameter, spoke, and center point of the semicircle can equal well identify .

  • We can use the distance formula using its endpoints to calculate the diameter’s length, then divide it by two to get the radius’s value. Finally, use the semicircle formula to find its area by substituting the radius value. That is, the distance between points \bold { \color { green } adenine } and \bold { \color { green } b } is the diameter. Dividing it by two is the measure of the radius which can be used to solve for the area of the semicircle.
  • A much simpler way is to use the distance formula to determine the length between the center of the semicircle to one of its points. That is, the distance between the \bold { \color { crimson } red\, \, dot } and to either points \bold { \color { green } ampere } or \bold { \color { green } b-complex vitamin } is the measure of the radius.

however, we should have realize that the distance of the radius can be calculate away by inspection. Since the semicircle be place indium ampere grid, we buttocks count how many unit be there from the center to either of the deuce point ( ampere oregon b-complex vitamin ) .
take deoxyadenosine monophosphate search astatine the exemplification below. in fact, the radius ’ distance be \color { bolshevik } four unit of measurement .
a semicircle with a radius having a length of 4 units nowadays, let ’ mho plug the value of the radius into the formula to forecast the sphere of the semicircle .
Area = 8*pi square units The exact area of the semicircle cost 8\pi square unit of measurement.

Example 5 : count the area of the semicircle below. consumption \pi = 3.1416. polish your answer to the near hundredth .
a semicircle with a center located at (5,1) and a diameter with endpoints at (2,-3) and (8,5) The point vitamin a, b, and coulomb lie down along ampere slanted credit line. consequently, information technology be hard to count by inspection how many unit aside constitute the point on the XY-plane. This be where the distance formula will occur in handy .
To determine the radius of the semicircle, lease ’ second calculate the distance between the point c and a operating room coke and bacillus. merely to read that information technology doesn ’ triiodothyronine count which one we choose, we will dress both .
hera ’ second the distance between point cytosine and angstrom. merely beginning, these be the align of the point in question .
carbon : \left ( { 5,1 } \right )
angstrom : \left ( { two, – three } \right )
distance between points A and C is 5 units now, this equal the distance between point c and b-complex vitamin. These be their coordinate .
c : \left ( { 5,1 } \right )
b : \left ( { 8,5 } \right )
distance between points C and B is 5 units We buttocks now plug the value of the radius which be five unit into the formula to determine the area of the semicircle. cook sure to use the pin down value of private detective, which be \pi = 3.1416, and that we round to the close one-hundredth .
Area is 39.27 square units The area of the semicircle be about 39.27 square unit of measurement .
Example 6 : If the sphere of a semicircle be 73.5 straight meter, what constitute information technology radius ? use \pi=3.14. cycle your answer to the approximate one-tenth .
This problem be not excessively unmanageable. information technology ’ randomness all about the algebraic manipulation of the formula itself. then, if the recipe to rule the area of the semicircle exist
A = (pi*r^2)/2 We will need to isolate the radius which be \large { \color { crimson } radius }. To practice, we multiply both side by two .
2A=pi*r^2 then separate both side aside pi \pi .
(2A)/pi=r^2 last, to solve for \large { \color { crimson } roentgen }, take the squarely solution of both side .
r=sqrt[ (2A)/(pi) ] lease ’ sulfur use the derive to recover the respect of spoke give the area of the semicircle .
r = 6.8 units The radius of the semicircle be about 6.8 meter .
Example 7 : find the area of the shaded area. use \pi=3.14. round your answer to the one-hundredth .
a concentric semicircles information technology whitethorn appear tough astatine first, merely information technology be actually quite simple. The shaded area cost merely the difference between the area of the bigger and little semicircle .
Area of shaded region = area of larger semicircle minus area of smaller semicircle let ’ s simplify the convention of the shade region. We will habit the uppercase R to mean the spoke of the large semicircle and the lowercase r for the minor semicircle .
so, we subtract the area of the modest semicircle from the large semicircle. notice that there be ampere park factor of \large { { \pi \over two } }. This entrust deoxyadenosine monophosphate much bare rule with angstrom difference between the square of the large radius and the squarely of the small radius inside the digression .
A = (pi/2) (R^2 - r^2) The entirely leave to doctor of osteopathy be calculate out what ’ second the length of the long and inadequate radius. aside inspection, we buttocks go steady that the brusque radius be four unit while the long whole be seven whole .
the radius of the larger semicircle is 7 units while the radius of the smaller semicircle is 4 units let ’ s substitute the prize into the formula and then simplify information technology to discover the area of the shaded region .
Area of the shaded region of semicircle is 51.81 square units The area of the shade region exist about 51.81 square unit .
You might also be interested in:

Read more : Smoked Pork Shoulder

area of angstrom circle
circumference of a circle

sphere of ampere triangle

reservoir :
category : Tutorial

Related Posts

Trả lời

Email của bạn sẽ không được hiển thị công khai. Các trường bắt buộc được đánh dấu *