Calculate area of the Yellow and Green shaded semicircles | Diameter of the largest semicircle is 4

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Calculate area of the Yellow and Green shaded semicircles | Diameter of the largest semicircle is 4

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[bigm]

Thanks, Professor, you seem to have developed a penchant for circles!🥂👍❤

[kennethstevenson]

I got the area of the yellow semicircle quickly but had trouble finding the correct radii for the green circle which was the same process but my labeling of the smaller triangles was faulty. Great practice and reason for step by step recording of information. I am still learning!

[thefunpolice]

Very nice problem. Used the Pythagorean theorem and that gets you yellow radius as √2 very easily and using the intersecting chords theorem generates a quadratic for the second radius: r^2 + r√2 - 1 = 0. One of the roots is negative and can be discarded. The other root is r = (√3 - 1)/√2 which is the same as (√6 - √2)/2.

With exact values for the two radii, it's trivial to get the yellow and green areas.

[solimana-soli]

Very interesting, It seems we have a fractal structure!!! We can add other semicircles smaller as many as we want. Would be interesting calcolate the function f(r) of radius

[theoyanto]

Great, got the yellow pretty quick, and I got close to the second bit, nearly did your method, but just got a bit lost, anyway great fun trying to solve.... I love those where a quadratic has to be derived, this is where I need practice keep 'em coming 🤓👍🏻

[neelaramramesh]

Mathacular professor, namaste from india

[SuperMtheory]

Would it be possible to calculate the area of an infinite number of semicircles, each smaller than the previous? Are the radii of the semicircles in a geometric progression? Great video!

[misterenter-izrz]

Thanks for interesting puzzle. Let R, r be the radii of large and small semicircles, so clearly R=2/root2, then the area of yellow semicircle is 2pi/2=pi=3.14 approximately, using pythagorean theorem, 4=r^2+(2/root2+r)^2, 4=r^2+2+4/root2 r+r^2, r^2+root2r-1=0, r=(-root2+root(2+4))/2=(root6-root2)/2=0.517638, hence the area of the small semicircle is 0.421 approximately. 🙂

[marioalb]

Yellow semi circle:
Taking the appropriate right triangle :
R²= r²+r²= 2 . r²
r² = R² / 2 = 2² /2 = 2
r = √2 = 1, 414 cm

Area = π r² / 2
Area = 3, 14 cm² ( Solved √ )

Green semi circle:
Same procedure
R²= r'²+(r+r')²=
2² = r'² + (√2)²+2√2r'+r'²
2²= 2 r'² + 2 + 2√2r'
r'² + √2 r' -1 = 0
r'= 0, 517 cm

Area = π.r'² / 2
Area = 0, 421 cm² ( Solved √ )

[princejag]

Could you show me how to work on "Negative Angles"?

[vidyadharjoshi]

Yellow is simple. Let OQ = R. Rsq + Rsq = 4. R = sqrt2, Yellow Area = pi.
Green is little more involved. OEsq = EPsq + POsq. Let PE = r. simplifying we get rsq + sqrt2 - 1 = 0. r = (sqrt 6 +/- sqrt 2)/2
Reject + (r=1.931). r = 0.517. Green Area = 0.5*pi*(0, 517sq) = 0.420.

[ybodoN]

The area of the white, yellow and green semicircles is in a ratio of 4∶2∶(2−√3).

[spiderjump]

Let a be the radius of yellow semi circle .
by Pythagorean theorem,
a^2 + a^2 = 2•2
a =sqroot2
Hence area of yellow semicircle = pi2/2
= pi

Let b be the radius of the green semicircle
b^2 + (b+sqrt2)^2=2•2
2b^2 +2b •sqrt2 =4
b^2+b•sqrt2=2
Use quadratic formula,
b =sqrt6-sqrt2/2
Area of green semi circle can then calculated using calculator

[mohamadtaufik]

Yellow Semicircle = pi = 3. 141 and Green Semicircle = 0.134*pi = 0.421

[giuseppemalaguti]

Dati R(yellow), r(green)... risulta R=1, 2^2=(R+r)^2+r^2...r=(sqrt7-1)/2...A(yel)=pi/2...A(gre)=(1-sqrt7/4)pi... Ho sbagliato, R=sqrt2....

[albertofernandez]

So isi.
Using the Pytagorean theorem un the triángulo OQC.
OC²=OQ²+CQ²
OQ=CQ
So, OC²=2OQ²
If OC=2
2²=2OQ²
OQ²=2
OQ=√2
The area of yellow semicircle is π√2²/2=π
Now using the Pytagorean theorem in the triangle OEP:
OE²=OP²+PE²
If OP=OQ+QP, and QP=PE
OP=√2+PE
OE=2:
Then, 2²=(√2+PE)²+PE²
4=2+2√2•PE+PE²+PE²
2PE²+2√2•PE-2=0
PE²+√2•PE-1=0
PE=(-√2+-√(2+4))/2
PE=(-√2+-√6)/2
PE=(-√2+√6)/2
The negative solution has no sense.
The area of Green semicircle is

[ramprasadp]

Sir, I not understood, why OQ = QD. Kindly clarify.

[qbertq]

As much as you use the 30-60-90 triangle properties, you never use the 45-45-90 triangle properties.