Calculate area of Yellow shaded region | Radius of the big circle is 12 | Important skills explained

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Calculate area of Yellow shaded region | Radius of the big circle is 12 | Important skills explained

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Thank you 🙏🙏 professor wishes from Bharat

neelaramramesh
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After the Pythagorean Theorem, the problem became easy and paused the video and solved it on my own.

alster
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Generally, if R is the radius of the big circle, the radius of the smallest circle is R/3, thus the area is (R^2 pi=7/18 R^2 pi.

misterenter-izrz
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Radius of blue circles = 12/2=6
Let r be radius of smallest circle .
(6+r)^2= 6•6 + (12–r)^2
36+12r+r^2=36+144–24r+r^2
36r=144
r=4
Yellow area =pi144–pi16-pi36•2
= 56pi

spiderjump
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Yay! I solved it, 56*(pi). The hardest part for me was calculating the r value for the green circle at 4 units. I did not have a pencil or paper or calculator, but I was able to do it. I was lucky that r^2 and 36 were on both sides of the equation to cancel out with the math.

Copernicusfreud
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R=12 → Radio azul =a=R/2 =6 → Si radio verde =b; en el triángulo BAD: (a+b)²=a²+(12-b)² → b=12/3=4 → Area amarilla = Pi(12² - 2x6² - 4²) =56Pi =175.93
Gracias y un saludo.

santiagoarosam
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Suffice to determine the radius r of green circle, by pythagorean theorem, 6^2+(12-r)^2=(6+r)^2, 36+144-24r+r^2=36+12r+r^2, then 36r=144, r=4, therefore the area is 12^2 pi-2x 6^2 pi-4^2 pi=(144-72-16)pi=56pi=173.93 approximately. 😃

misterenter-izrz
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👍Prof, a great problem and a wonderful solution. Looking forward to viewing such interesting problems.

vara
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Не помнил точно про центры, пока не увидел. Теперь всё сходится.

zawatsky
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AD = 12 - r. AB = 6 + r. ABsq = BDsq + ADsq. r = 4. Area = 144pi - 2*36pi - 16pi = 175.93

vidyadharjoshi
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The area of yellow region is 56π, thank you for teaching me PreMath. (My real name is Andreu)

TonTin
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I love these, but I always struggle to begin, but once I see you first moves, I understand, so I need more like this one please, it's quite annoying, I can see my error, I guess I don't fully understand the methodology yet, hey ho, more practice, needed
I think the idea is to represent the unknown in a way that doesn't introduce another variable, I see it, but maybe I get it one day thanks again 🤓👍🏻

theoyanto
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This problem I feel is like the Hunt for Neptune's Gold...and in this case we found it. cool!🙂

wackojacko
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Question: why keep the pi in the results so late. If i know something is 16pi I'm multiplying 16 * 3.1415 as early as I can. Why do you keep them in right up to the end?

Asking for myself, I have no friends (obviously, I watch maths videos on YouTube)

DeltaMikeTorrevieja
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I took a different approach and came to a different conclusion. Where do I go wrong? Thanks in advance. I took the trangle ABC. AB and AC are 6+r, BC = 12. Thus, (6+r)^2 + (6+r)^2 = 144 -> r^2 + 12r + 36 + r^2 + 12r + 36 = 144 -> 2r^2 + 24r + 72 = 144 -> r^2 + 12r - 36 = 0; r = (-12 +/- sqrt(144 - 4*1*-36))/2*1 -> r = (-12 +/- sqrt(288))/2 -> r = 2.485, which leads to a yellow area of ca 206.8.

martinbrederoo
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How to solve this generally r=R/3, where r, R are radii of small and large circles without using pythagorean theorem?

misterenter-izrz
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If ck is common tangent to A and B then ck^2=12^2-6^2=108. So 108+r^2=(6+r)^2 resulting in r=6. Sir can you explain where I made a mistake?TIA

srinivasansundararajan
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Amazingly, this construction produces 3:4:5 right triangles...

ybodoN
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We can sauy that the area of the yellow region is equal of 176 square units because pi is also equal at 22/7

francismoles
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