Two Methods! Can You Find the Area of the Green Circle? | 2 Easy Methods

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Area =271 (approximately)
Let the length of the radius = 7+ n, then the length of the radius from any point = 7+ n, 7+n form a right triangle with sides, 9, 7+n, and n (with the hypotenuse =7+n), So 9^2+ n^2= (7+n)^2 (from a^2 + b^2= c^2)
81 + n^2 = 49 + n^2 +14n
81-49 =14n
32 = 14n
16=7n (divide both sides by 2)
n=16/7 hence radius= 7 + 16/7 (since the radius = 7 +n) or 49/7 + 16/7 = 65/7 so the radius of the circle =65/7
And it is home from there as pi r^2= Area of a circle hence 22/7 x 65/7 x 65/7 = approximately 270.992 9:08

devondevon
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Thanks! I did it another way: pythagorean theorem to the triangle with the hypotenuse as the diameter of the circle and got the same result

zampone
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love your problems and solutions. keep it up

richardparks
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good video sir, hope you can make videos about GEOMETRY PROBLEMS everyday. love from MALAYSIA.

haofengxd
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Interesting problem, thank you very much, Sir. Since AFB is a right triangle, I found EF using the Geometric mean theorem.

cinzia_f
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I got it by drawing a line from the center of the ○ to the bottom of the vertical cord. The diameter of the circle bisects the vertical cord, therefore each half is 18/2=9. From the right triangle with sides of 9, R-7 where R is the radius of the green circle, and the hypotenus is R, I have the equation R^2=81+(R-7)^2 ==> R=130/14 ==> Area of the green ○ = (130/14)^2 x Pie 🤭 ~ 270.8822.

hienvo
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Easy to understand your solution. Thank you very much, Sir.

luigipirandello
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Great method. I actually went for calling EC 'x' . That made right triangle ECD as 9^2 + x^2 = (7+x)^2 (7+x being r) and took it from there

MrPaulc
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great video and interesting solutions !
I used that according to Thales theorem the triangle ABF has a 90 degree angle in B.
With Euklid's theorem ( I'm not sure what's the English name - is it the "Geometric mean" ??? ) the solution is easy:
AE * EF = EB²
=> 7 * EF = 81 <=> EF = 81/7
r = 1/2 ( EF + 7 ) = 1/2 ( 81/7 + 49/7 ) = 130/14 = 65/7

thomast.
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Also solve by considering iso triangle ACD, base AB, and chord formula L=2rsin(ang/2), ang = subtended angle. 1. angle BAC=arctan(9/7), then 1/2 subtend ang = 180-2BAC. Finally chord formula to get r.

timmy---
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Woww👍🏻 superior & fantastic content 💯 i love it ☺💗
Stay blessed 💥🌷
Stay healthy 🌝🍹☀ waiting next vedio 🌹Stay connected

learnenglishwithsanafarhan
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👍👍Question was damn easy . Btw nice explanation

mohdzaheersiddiqui
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very well explained, thanks for sharing bro

math
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In Civil Engineering, practically We are using a formula Radius= L^2/8h +h/2, where L is length of Chord and h is the central vertical distance. We are using the formula to make arches. Radius = 18^2/8*7 + 7/2 = 9.28..

anandharamang
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Very good Sir. I understood the Pythogrean method well

Am still trying to 7nderstand the Intersecting Chords Theorem. I have several doubts in it

procash
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Two good methods. I used the first. My only doubt is about the rounding. Using pi to many more significant figures, I got an answer of about 270.8822, in which the first two digits after the decimal point differ from those in your rounded answer. The more accurate answer is actually closer to 271 than it is to 270.75. So, my question is: is 3.14 a sufficiently good approximation to pi for the purpose, or might it be better to add a couple more significant digits in order to minimize the overall rounding error?

AnonimityAssured
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Yeah, I massively overcomplicated this.

I used 2arctan(9/7) to find DAB, then the centre-double-circumference circle theorem to find BCD and then sine to work out BC, using angle identities to convert from tan to sin.

I'm pretty bad at noticing simple solutions to visual puzzles.

XLatMaths
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Very good and great question. I have solved the same approach. However, the answer is about 270.8822237 square units.

mustafizrahman
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Extend AC to make it Diameter. Call that end as F. So AE*EF = BE*ED. EF = 9*9/7 = 11.57. So Diameter = 18.57. Area = pi*sq(18.57/2) = 270.88

vidyadharjoshi
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Hah! Did it exactly the same way as you did it, with the intersecting chords theorem!

bentels