Calculate area of the Green shaded Region | Semicircle | Important Geometry skills explained

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Calculate area of the Green shaded Region | Semicircle | Important Geometry skills explained

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[jasonk]

This is quickly becoming one of my favorite channels. Great stuff. The thing I like most is that it brings the theorems to life with real world applications.

[harikatragadda]

Nice! This can also be solved another way.
If R is the radius of the big circle and r the radius of the semicircle, then R = ½(48+8+r) = ½(56+r)
Applying Pythagoras theorem to Triangle CPO, with PO = R-8
R² = r² + (R-8)²
r² - 8r - 384=0
r = 24

[lukeheatley]

First i found AC = √(8^2+r^2)
Then I made 2 right angle triangles ABC and PBC to get 2 expressions for BC in terms of r. Then solved.

Your way is much better.

[himo]

CP=PD=PQ=r r² = 8*(r+48) r²-8r-384=0
(r+16)(r-24)=0 r>0, r=24
(8+24+48)/2=80/2=40 40*40*π - 24*24*π*1/2 = 1600π - 288π = 1312π

[rishudubey]

thankyou so much for your hard work sir 😊

[murdock]

Nice! Many thanks, Sir!
AO = BO = r → AB = 2r = 56 + h
CP = DP = PQ = h
AP = 8; BQ = 48
h(h) = 8(h + 48) → h = 24 → r = 40 → π(r^2 - h^2/2) = 1312π

[theoyanto]

I actually managed to derive the quadratic, well pleased, but it is one of the more simple ones, but still did it!
Thanks again 🤓👍🏻

[srinivasansundararajan]

AB^2=AC^2+CB^2 (all lines in terms of r )as angle ACB=90°, gives r=24.

[ybodoN]

The radius of the semicircle is _½ (a + √(a² + 4ab))_ where _a_ is AP and _b_ is QB.

[danielwoody]

I thought there was an error neglecting the distance QO but it was included in the distance QB (48). I was not listening to the description of the problem so I stand corrected.

[giuseppemalaguti]

Ho il sistema 2R=48+8+r...R^2=r^2+(R-8)^2 che dà r=24 con

[seansreynou]

អរគុណលោកគ្រូសម្រាប់វីឌីអូ
Thanks you teacher for video 📘🙏🙏

[stephenbrand]

Enjoyed this problem. Thank you. Can you please recommend a clear proof of the intersecting chords theorem on You Tube.

[mohamadtaufik]

R=40 and r=24, from equation r^2-8r-384=0 exactly similar with other correct answers. Therefore, green shaded region is 1312pi = 4121.77

[ybodoN]

More fun with the same diagram:
When Q = O, the semicircle is ¼ the area of the circle (simply because R = r√2).
So two semicircles tangent at O will be exactly half the area of the green circle.
But when AP = QB, the area of the semicircle is exactly ⅖ the area of the circle.

[marioalb]

Green area :
Área= π R² - π r² /2

Adding segments:
2R=48+8+r
R=(48+8+r)/2=(56+r)2
R=28+r/2

Taking the appropriate right triangle:
and applying Pitágoras:
r²+(R-8)²=R²
r²+(28+r/2-8)²=(28+r/2)²
r²+(r/2+20)² = (r/2+28)²
r²+(r²/4+20r+20²) = r²/4+28r+28²
r²-8r-384=0
r = 24 cm

R=28+r/2
R = 40 cm

Area= π R² - π r² /2
Area = π 40² - ½ π 24²
Area = 4121, 76 cm² (Solved √ )

[vidyadharjoshi]

Although not given but APQOB must be colinear. so Let r be the radius of white semicircle. r*r = 8*(r+48) Solve for r, r = 24
White Area = 0.5*pi*24*24 = 904.78. Green Radius = 0.5*(8+24+48) = 40. Green Area = (pi*40*40) - 904.78 = 5026.55 - 904.78 = 4121.77

[albertofernandez]

Using Pitagoras theorem, if CP=r of semicircle
48²=40²+r²
(40+8)²=40²+r²
40²+640+64=40²+r²
704=r²
r=√704
r=8√11


Abigcircle=πR²=π(48)²=2304π

[DjappoMKS]

Dear sir...

For the next video can you make the picture more clear ?

In the picture the length of the line is 48 but it is not clear which line ? Is it OB = 48 or the QB = 48.

Maybe you can me it more clear in your next video ? Because I always do the calculations BEFORE i watch your videos. I got confused with the length 48.

Thank you very much sir.

[JSSTyger]

I figured the radius of the semi circle to be 8sqrt(11) and thus the area is (64)(11)π/2 = (32)(11)π = 352π. The area of the green circle is π(48²) = 2304π. That means the green area is π(2304-352) = 1952π