Let's Think outside the Box! | Find the chord length AB | (Simple explanation) |

Learn how to calculate the Chord AB length. Area of the Blue rectangle is 7. Important Geometry skills are also explained: area of the rectangle formula; Pythagorean Theorem; Intersecting chords theorem. Step-by-step tutorial by PreMath.com
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Let's Think outside the Box! | Find the chord length AB | (Simple explanation) | #math #maths
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#ChordAB #FindChordAB #Square #ThalesTheorem #Radius #GeometryMath #PythagoreanTheorem #QuarterCircle #Rectangle #IntersectingChordsTheorem
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[@MrPaulc222]

Excellent, I tried all manner of solutions before watching the video but didn't get it. And, yes, I did expand the circle and tried working with intersecting chords, but still couldn't see it. Thank you. You are an education.

[@PreMath]

Glad it helped! 🌹 Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@ajbonmg]

We are not told the dimensions of the square, or the radius. So presumably, the length AB is independent of r, and only related to the blue area. So we can assume the blue area fills the whole square. Hence each side (and the radius) can be √7. And AB becomes the diagonal of the square, which, by Pythagoras, is √14.

[@nishagupta4628]

What a marvelous argument

[@harikatragadda]

Since the circle has an indeterminate radius, it can be resized to fit the blue area, which becomes a Blue Square. Hence, diagonal of the Square = AB AB² = 2*(Side)² = 2*7 = 14 AB = √14

[@engralsaffar]

Let's lable the sides of the triangle in the blue rectangle as a and b Therefore, using pythagoras AB^2=a^2+b^2 - - - (1) Also we can note from the area of the rectangle that r=7/a - - - (2) Next we connect point B to the edge of the square, and apply pythagoras again to get r^2=b^2+(r-a)^2 =b^2+r^2-2ar+a^2 We cancele r^2 from both sides and rearrange to get a^2+b^2=2ar From (1) and (2) AB^2=2a*7/a=14 AB=sqrt(14)

[@pwmiles56]

ar = 7 As B is on the circumference (a-r)^2 + b^2 = r^2 a^2 - 2ar + r^2 + b^2 = r^2 Cancel r^2 and rearrange a^2 + b^2 = 2ar AB^2 = 14 AB = sqrt(14)

[@Abby-hi4sf]

Do you mean ((r-a)^2 + b^2 = r^2) when you put (a-r)^2 + b^2 = r^2

[@pwmiles56]

@@Abby-hi4sf Yes, sorry, I did a dumb edit and I forgot r was greater than a.

[@chanpangchin9744]

The way of thinking by Waldlaeufer70 is the neatest. After all, a square is but a "unique" rectangle. Cheers from Malaysia!

[@mathbynisharsir5586]

Fantastic video sir

[@PreMath]

So nice of you dear ❤️

[@JLvatron]

Wow, nice!

[@Waldlaeufer70]

Since it seems that the side length of the square doesn't matter, I let the blue area fill all the square. In this scenario, AB is equal to the diagonal of the square, which is: AB = √7*√2 = √14 cm

[@bigm383]

Thanks, Professor, lovely work!❤🥂

[@vcvartak7111]

This problem is based on our of Box thinking. Since merely looking at the quarter circle and rectangle we don't get result.we have to think circle as whole, join the chords and use chords intersection theorem. Nice sum

[@kevinn1158]

Great! But I would have just called AB line C so as to not get confused.

[@tellerhwang364]

1.作OH丄AB→AH=BH=AB/2 2.△AHO~△ACB(AA) AO:AB=AH:AC→r:AB=(AB/2):a AB^2=2ar=14→AB=sqrt14😊

[@mibsaamahmed]

Elegant solution, sir!

[@PreMath]

Many thanks! ❤️🌹

[@MihaiKusko]

From B to center of quarter circle we draw a line of r length so from Pythagorean theorem we have r^2 = (r-a)^2+b^2 and by expanding this equation we have r^2 = r^2 - 2ra +a^2+b^2. We drop the r^2 terms on the both sides, so we obtain a^2+b^2=2ra, i,d AB^2=14, AB =square root of 14.

[@WernHerr]

My solution: (1) r*a=7 ACB: CB² = AB² - a² MCB: CB² = r² - (r-a)²; together: AB² = 2*2*x = (1) =2*7 = 14, AB=sqr(14)

[@WernHerr]

sorry: Last line: together: AB² = 2*a*r = (1) =2*7 = 14, AB=sqr(14)

[@MrMichelX3]

Awesome, dear master !!

[@PreMath]

Many thanks!! ❤️🌹

[@ybodoN]

Generalizing: AB = √(2K) where K is the area of the blue rectangle. Indeed, the length of the sides of the blue rectangle does not matter.

[@ramanivenkata3161]

Very well explained. 👍

[@PreMath]

Glad you liked it ❤️🌹

[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!!!

[@unknownidentity2846]

Let's face this nice challenge: . .. ... .... ..... May s be the side length of the square and the radius of the quarter circle as well. The horizontal line within the square divides the vertical sides of the square into two parts with lengths l (lower part) and s−l (upper part). Additionally we assume that the lower side of the square is located on the x-axis and that the left side of the square is located on the y-axis. Then we have: AB² = (y(B) − y(A))² + (x(B) − x(A))² = (l − s)² + x²(B) = (l − s)² + s² − y(B)² = (l − s)² + s² − l² = l² − 2ls + s² + s² − l² = 2s² − 2ls = 2s(s − l) = 2A(blue) = 2*7cm² = 14cm² ⇒ AB = (√14)cm

[@nehronghamil4352]

Alternate Solution:: (I see someone in the comments has adopted my style of presentation.. Hopefully more will do so. Its makes the solution easier to follow) Circle at center O, height of the rectangle "b", distance from point B to right side of square "a", side of square "s" = radius of circle, Area of rectangle "A" and AB is "x" A = bS =7 (1) (s - a)^2 + b^2 = x^2 s^2 - 2as + a^2 + b^2 = x^2 (2) (s - a)^2 + (s - b)^2 = s^2 (s^2 - 2as + a^2) + ( s^2 - 2bs + b^2) = s^2 (3) (3)-(2): s^2 - 2bs = s^2 - x^2 2bs = x^2 (4) substitute (1) in (4): 14 = x^2 x = 14^.5

[@alexkirchoff5286]

I got same result by using trigonometry. Any circumference can be considered a trigonometric circumference ,then in my solution b becomes b = r * cos θ and a = r - r * sin θ . So 7 = r * a which at my case equals to 7 = r^2 * (1 - sin θ). Being AB^2 = a^2 + b^2 , you finally get AB^2 = 2 * r^2 * (1 - sin θ) . Except for the factor 2 it's exactly the formula for the blue area. Therefore you get AB^2 = 2 * 7 and hence the solution. Anyway I don't know if this is enough of the box😊

[@netopierro10]

i could find 2 equations, but not the last one... nice

[@comdo777]

asnwer=13cm skin over isit

[@wackojacko3962]

Fantissamo! 🙂

[@PreMath]

❤️ Thank you! Cheers! 😀

[@giuseppemalaguti435]

I didnt Remember this teorem...😂

[@DB-lg5sq]

AB^2=R^2-[R-(7/R)]^2=14

[@rakeshkumarrkk5498]

First

[@PreMath]

Thanks dear ❤️