Calculate Area of the Blue shaded region between semicircles | Important Geometry skills explained

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Calculate Area of the Blue shaded region between semicircles | Important Geometry skills explained

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[lukeheatley]

why is it assumed that the chord is parallel to the diameters?

[mattsta]

This is an interesting problem because the radius of each arc segment can vary and as long as the chord segment is 16 units long, the blue area will remain the same.

[garyspencer]

In the beginning, you state that the chord is tangent to small semi-circle. However, you don't define the chord as a parallel line to the "line" that is the
diameter of large semi-circle. This is a remarkably good geometric problem. But, the chord geometry needs to be fully given. The tangency to small
semi-circle can take on infinitely many orientations. In the "middle" of the talk, you ASSUME the chord is parallel to diameter line. Love your channel and
solution methods. Thanks !

[williamwingo]

The chord is 16 units. Let the white semicircle shrink down to nothing and the 16-unit chord becomes the diameter of the blue semicircle.
The radius is 8 and the area is ((8^2)*pi)/2 = 64 pi /2 =32 pi. This is true for any value of the small semicircle radius r.
There is no maximum value for r; but r = sqrt(R^2 -- 64). Setting r = 0 gives the same answer as above.
Cheers. 🤠

[snowmanking]

the first two lines you draw are not necessarily the same length, for them to be the same length we would need to know that the chord is parallel to the base of the larger semi-circle but we are not told this and it is not possible to prove from the given information

[AnonimityAssured]

Another in-the-head quickie.

Spoiler alert.

No variables needed. Simply imagine the smaller semicircle shrinking to nothing, so that 16 becomes the diameter of the larger semicircle. Then its radius is 8 and its area is 8²π /2 = 64π /2 = 32π square units.

[philipkudrna]

One believes at first it‘s impossible to solve, but no matter what the two radii are, the area is always 32pi!

[marioalb]

Blue shaded Area :
A = ⅛ π C²
A = ⅛ π 16²
A = 32π cm² ( Solved √ )

[HappyFamilyOnline]

Very well explained👍
Thanks for sharing😊😊

[joseeduardomachado]

Exercício parece ser difícil, mas na realidade é fácil

[egillandersson]

This type of puzzle with very few data is often simple. In this case, let's assume there is no missing data. So, the absolute radius of the two circle don't matter. Suppose the little one be zero, then 16 is the diameter of the great and the blue shaded region is just a semi-circle of radius 8.

[rezagoli]

It's wrong
We cannot be sure that the side of the triangle is 8 because we do not know that the chord is parallel to the diameter of the great circle
It's could be bigger or less than 8

[СергейКовалев-тдм]

Just let r = 0. Then 16 is the diameter of the bigger circle. Hence, the area is
Pi times 8^2/2

[tontonbeber]

No need for Pythagore, just trigonometry ...
R cos a = 8 and R sin a = r
PI/2 (R2 - r2) = PI/2 R2 (1 - sin2 a) = PI/2 R2 cos2 a = PI/2 64 = 32 PI
(took me 5 seconds)

[TomasPböckerlyftningschack]

If there is one solution it must be independant of the size of the smaller semicirkle. So, by putting the smaller cirkle =0 the blue region equals rhe area of a semicirkle with radius=8 i.e. 32xPI.

[andriibibik]

Considering that small semicorcle can have any radius, imagine that small circle extremely small. Then shaded area will equal to area of large semicircle with radius = 8. =} 32π

[rickdesper]

Very neat problem. If you try to solve for r or R, you'll get nowhere. Neither r nor R can be pinned to a specific value, even though (R^2 - r^2) is itself determined.

[kennethstevenson]

You should mention what number you are using for Pi, you mention 3.14 but you used the Pi key on your calculator to create the answer 100.53 sq. un.

[maxxflyer]

too complicated. push down the 16 segment, until the tiny circle becomes 0 radius. and just calculate the are a of the semicircle of diameter 16.

[mr.hemanta]

Hi PRE MATH, I have a question for you . Solve it √(129-90√2)=?